2170 + Interview Questions in PROBABILITY IN GENERAL AWARENESS Page 10 InterviewSolution

451.	A pair of die is rolled six times. Find the probability that a third sum of seven is obtained at the sixth throw.
Answer» `5C_2(1/6)^2(5/6)^31/3` `(5!)/((2!3!))1/36(5/6)^31/6` `54/21/36(5/6)^31/6` `5^4/(6^53)`.

Discussion

452.	If `A and B` are independent events of a random experiment such that `P(A nn B) = 1/6 and P( overlineA nn overlineB)=1/3` then `P(A)=`A. `(1)/(4)`B. `(1)/(3)`C. `(1)/(6)`D. `(2)/(3)`
Answer» Correct Answer - B We have, `P(A cap B)=(1)/(6) " and " (overline(A) cap overline (B))=(1)/(3)` `implies P(A) P(B)=(1)/(6) " and " P(overline(A))P(overline(B))=(1)/(3)` `implies xy=(1)/(6) " and" (1-x)(1-y)=(1)/(3)`, where P(A)=x, P(B)=y `implies xy=(1)/(6) " and " 1-x-y+(1)/(6)=(1)/(3)` `implies xy=(1)/(6) " and " x+y=(5)/(6)` `implies x=(1)/(2) " and " y=(1)/(3) " or" x=(1)/(3) " and " y=(1)/(2)`

Discussion

453.	Let A and B be two events such that `P (bar(AuuB))=1/6, P(AnnB)=1/4and P(barA)=1/4,where barA` stands for the complement of the event A. Then the events A and B areA. mutually exclusive and independentB. equally likely but not independentC. Independent but not equally likelyD. independent and equally likely
Answer» Correct Answer - C `P (bar(AuuB))=1/6` `P(AuuB)=5/6,P(A)=3/4` `P(AuuB)-P(A)+P(B)-P(AnnB)=5/6` `impliesP(B)=5/6-3/4+1/4=1/3` `impliesP(AnnB)=P(A).P(B)=1/4`

Discussion

454.	Let A and B be two events such that `p( bar AuuB)=1/6, p(AnnB)=1/4`and `p( bar A)=1/4`, where ` bar A`stands forthecomplement of the event A. Then the events A and Bare(1) mutually exclusive and independent (2) equally likely but not independent(3) independentbut not equally likely (4)independent and equally likelyA. equally likely but not independent equally likely and mutually exclusiveB. equually like and mutually exclusiveC. Mutually exclusive and independentD. independent but not equally likely
Answer» Correct Answer - D We have, `P(bar(AuuB))=1/4,P(AnnB)=1/4` `P(barA)=1/4impliesP(A)=1-1/4=3/4` `thereforeP(bar(AuuB))=1-P(auuB)=1-[P(A)+P(B)-P(AnnB)]` `impliesimplies1/6=1-3/4-P(B)+1/4` `impliesP(B)=1-1/2-1/6=(6-3-1)/(6)=2/6=1/3` Since `P(AnnB)=P(A)P(B)and P(A)neP(B),` therefore A and B are independent but not equally likely.

Discussion

455.	Probability if `n`heads in `2n`tosses of a fair coin can be given by`prod_(r=1)^n((2r-1)/(2r))`b. `prod_(r=1)^n((n+r)/(2r))`c. `sum_(r=0)^n((^n C_r)/(2^n))`d. `(sumr=0n(^n C_r)^2)/(sumr=0 2n(^(2n)C_r)^)`A. `underset(r = 1)overset(n)prod((2r-1)/(2r))`B. `underset(r = 1)overset(n)prod((n+r)/(2r))`C. `underset(r = 0)overset(n)Sigma((.^(n)C_(r))/(2^(n)))^(2)`D. `(underset(r = 0)overset(n)Sigma(.^(n)C_(r))^(2))/((underset(r=0)overset(2n)Sigma.^(2n)C_(r)))`
Answer» Correct Answer - A::C::D `P(E) = (.^(2n)C_(n))/(2^(2n)) = ((2n)!)/(n! n! 2^(n) 2^(n))` `=(1xx2xx3xx...xx(2n))/(n!n!2^(n)2^(n))` `=(1xx3xx5...xx(2n - 1))/(n! 2^(n))` Now, `underset(r = 1)overset(n) prod((2r-1)/(2r))=(1xx3xx5xx...xx(2n-1))/(2xx4xx6xx...xx(2n))` `=(1xx3xx5xx...xx(2n-1))/((1xx2xx3xx...xxn)2^(n))` `underset(r = 0)overset(n)sum((.^(n)C_(r))/(2^(n)))^(2) = (1)/(2^(n)2^(n)) underset(r = 0)overset(n)sum (.^(n)C_(r))^(2) = (1)/(2^(n)2^(n)) .^(2n)C_(n)` Also, `(underset(r = 0)overset(n)sum(.^(n)C_(r))^(2))/((underset(r = 0)overset(2n)sum .^(2n)C_(r))) = (.^(2n)C_(n))/(2^(2n))`

Discussion

456.	Let A and B be two events such that `p( bar AuuB)=1/6, p(AnnB)=1/4`and `p( bar A)=1/4`, where ` bar A`stands forthecomplement of the event A. Then the events A and Bare(1) mutually exclusive and independent (2) equally likely but not independent(3) independentbut not equally likely (4)independent and equally likely
Answer» `P( bar (A uu B) = 1/6` `1- P(A uu B) =1/6` `P(A uu B) = 5/6` `P(bar A) = 1/4` `1- P(A) = 1/4` `P(A) = 3/4` `P(A uu B) = P(A) + P(B) - P(A nn B)` `5/6 = 3/4 + P(B) - 1/4` `P(B) = 5/6 - 1/2 = 2/6 = 1/3` `P(A nn B) = P(A)P(B)` `= 3/4 xx 1/3 = 1/4 ` A & B are independent events `P(A) = 3/4` `P(B)= 1/3` `P(A) != P(B) ` option 3 is correct

Discussion

457.	If `P(A cap B)=1//3, P(A cup B)=5//6 " and " P(A)=1//2`, then which one of the following is correct ?A. A and B are independent eventsB. A and B are mutually exclusive eventsC. P(A)=P(B)D. `P(A) lt P(B)`
Answer» Correct Answer - A We have, `P(A cup B)=P(A)+P(B)-P(A cap B)` `implies (5)/(6)=(1)/(2)+P(B)-(1)/(3) implies P(B)=(2)/(3)` Clearly, `P(A cap B)=P(A)P(B)` So, A and B are independent events.

Discussion

458.	If `pa n dq`are chosen randomly from the set `{1,2,3,4,5,6,7,8,9, 10}`with replacement, determine the probability that the roots of theequation `x^2+p x+q=0`are real.A. 0.58B. 0.55C. 0.38D. 0.03
Answer» Correct Answer - (d)

Discussion

459.	If `pa n dq`are chosen randomly from the set `{1,2,3,4,5,6,7,8,9, 10}`with replacement, determine the probability that the roots of theequation `x^2+p x+q=0`are real.A. 0.38B. 0.03C. 0.59D. 0.89
Answer» Correct Answer - (c)

Discussion

460.	If A and B are mutually exclusive events, thenA. `P(A) le P(barB)`B. `P(A) gt P(B)`C. `P(B) le P(barA)`D. `P(A) gt P(B)`
Answer» Correct Answer - A::C Given that A and B are mutually exclusive events. `therefore A nn B = phi` implies `A sube barB and B sube barA` `implies P(A) le P (barB) and P(B) le P(barA)`

Discussion

461.	Let A and B be two events such that `p( bar AuuB)=1/6, p(AnnB)=1/4`and `p( bar A)=1/4`, where ` bar A`stands forthecomplement of the event A. Then the events A and Bare(1) mutually exclusive and independent (2) equally likely but not independent(3) independentbut not equally likely (4)independent and equally likelyA. mutually exclusive and independentB. independent but not equally likelyC. equally likely but not independentD. equally likely and mutually exclusive
Answer» Correct Answer - B We have, `P(A cup B)=(1)/(4) ne 0` So, A and B are not mutually exclusive events. Now, `P(overline(A cup B))=(1)/(6)` `implies 1-P(A cup B)=(1)/(6)` `implies P(A cup B)=(5)/(6)` `implies P(A)+P(B)-P(A cap B)=(5)/(6)` `implies (3)/(4)+P(B)-(1)/(4)=(5)/(6) implies P(B)=(1)/(3)` Clearly, `P(A cup B)=(1)/(4)=(3)/(4)xx(1)/(3)=P(A)P(B)` So, A and B are independent. Also, `P(A) ne P(B)`. So, A and B are not equally likely.

Discussion

462.	If `P(A)=(2)/(3), P(B)=(1)/(2) " and " P(A cup B)=(5)/(6)`, then events A and B areA. mutually exclusiveB. independent as well as mutually exclusiveC. independentD. dependent only on A
Answer» Correct Answer - C We have, `P(A)=(2)/(3), P(B)=(1)/(2) " and " P(A cup B)=(5)/(6)` `therefore P(A cup B)=P(A)+P(B)-P(A cup B)` `implies P(A cup B)=(2)/(3)+(1)/(2)-(5)/(6)=(1)/(3)` Clearly, P(AB)=P(A)P(B) Hence, A and B are independent events. Since independent events are never mutually exclusive. So, option (c ) is correct.

Discussion

463.	If M and N are any two events, then the probability that exactly one of them occurs isA. `P(A)+P(B)+2P(A cap B)`B. `P(A)+P(B)-P(A cap B)`C. `P(overline(A))+P(overline(B))+P(overline(A) cap overline(B))`D. `P(A cap overline(B)) + P(overline(A) cap overline(B))`
Answer» Correct Answer - D

Discussion

464.	If A and B are two events, the probability that exactly one of them occurs is given byA. `P(A) + P(B) - 2P(A nn B)`B. `P(A nn barB) + P(barA nn B)`C. `P(A uu B) - P(A nn B)`D. `P(barA) + P(barB) - 2P(barA nn barB)`
Answer» Correct Answer - A::B::C::D We have, P(exactly one of A, B occurs) `= P [(A nn barB) uu (barA nn B)]` `=P(A nn barB) + P(barA nn B)` `=P(A) - P(A nn B) + P(B) - P(A nn B)` `=P(A) + P(B) - 2P(A nn B)` `=P(A uu B) - P(A nn B)` Also, P(exactly one of A, B occurs) `=[1 - P(barA nn barB)] - [1 - P(barA uu barB)]` `=P(barAuubarB) - P(barA nn barB) = P(barA) + P(barB) - 2P(barA nn barB)`

Discussion

465.	Twoaeroplanes I and II bomb a target in succession. The probabilities of I andII scoring a hit correctly are 0.3 and 0.2, respectively. The second planewill bomb only if the first misses the target. The probability that thetarget is hit by the second plane is(1)0.06(2) 0.14(3) 0.2(3) 0.7A. 0.2B. 0.7C. 0.06D. 0.14
Answer» Correct Answer - D Let `A_(i)(i=1,2)` denote the event that `i^(th)` plane hits the target. Clearly, `A_(1) " and " A_(2)` are independent events. Required probability `=P(overline(A_(1)) cap A_(2))` `=P(overline(A_(1)))P(A_(2))=(1-0.3)x0.2=0.14`

Discussion

466.	P(A cup B)=P(A cap B)` if and only if the relation between P(A) and P(B) is …………
Answer» Correct Answer - A::B `P(AuuB)=P(A)+P(B)-P(AnnB)` If `P(AuuB)=P(AnnB)`, then P(A) and P(B) are equals. Since, `P(AuuB)=P(AnnB)rArrA`and B are equals sets Thus, P(A) and P(B) is equal to `P(AnnB)`.

Discussion

467.	If M and N are any two events, then the probability that exactly one of them occurs isA. `P(M)+P(N)-2P(M cap N)`B.C.D.
Answer» Correct Answer - A::C P(exactly one of M,N occurs) `P{(MnnbarN)uu(barMnnN)}=P(MnnbarN)+P(barMnnN)` `=P(M)-P(MnnN)+P(N)-P(MnnN)` `=P(M)+P(N)-2P(MnnN)` Also, P(exactly one of them occurs) `={1-P(barMnnbarN)}{1-P(barMuubarN)}` `=P(barMuubarN)-P(barMnnbarN)=P(barM)+P(barN)-2P(barMnnbarN)` Hence, (a) and (c ) are correct answers.

Discussion

468.	Two aeroplanes I and II bomb a target in succession. The probabilities of I andII scoring a hit correctly are 0.3 and 0.2, respectively. The second planewill bomb only if the first misses the target. The probability that thetarget is hit by the second plane is(A) `0.06` (B) `0.14`(C) `7/22` (D) `0.7`
Answer» `P=0.7xx0.2+(0.7xx0.8)(0.7xx0.2)+(0.7xx0.8)^2(0.7xx0.2)+..........` `=0.14[1+0.56+0.56^2+...........]` `=0.14[(1)/(1-056)]` `=0.14/(0.14)` `=7/22`

Discussion

469.	Three numbers are chosen at random without replacement from {1,2,3,....10}. The probability that the minimum of the chosen number is 3 or their maximum is 7 , is:
Answer» Correct Answer - A Let `E_(1)` be the event getting minimum number 3 and `E_(2)` be the event getting maximum number 7. Then, `P(E_(1))=P` (getting one number 3 and other two from numbers 4 to 10) `(overset(1)""C_(1)xxoverset(7)""C_(2))/(overset(10)""C_(3))=(7)/(40)` `P(E_(2))=P` (getting one number 3, second number 1 to 6) `(overset(1)""C_(1)xxoverset(1)""C_(1)xxoverset(3)""C_(1))/(overset(10)""C_(3))=(1)/(10)` `:. P(E_(1)uuE_(2))=P(E_(1))+P(E_(2))-P(E_(1)nnE_(2))` `=(7)/(40)+(1)/(8)-(1)/(40)=(11)/(40)`

Discussion

470.	Three numbers are chosen at random without replacement from {1, 2, 3,...... 8}. The probability that their minimum is 3, given that their maximumis 6, is(1) `3/8`(2) `1/5`(3) `1/4`(4) `2/5`
Answer» let event A= minimun is 3 let event B= maximum is 6 total ways = `.^8C_3` `P(A nn B) = 2/(.^8C_3)= 2/((8!)/(3!5!))` `= (2 xx 3!)/(8 xx 7 xx6) = 1/28` `P(A/B) = (P(A nn B))/(P(B))` `= (1/28)/((.^5C_2)/(.^8C_3))` `= (2/(.^8C_3))/((.^5C_2)/(.^8C_3)` `= 2/(.^5C_2)` `= ( 2 xx2)/(5 xx4) = 1/5 ` option 1 is correct

Discussion

471.	Three numbers are chosen at random without replacement from {1, 2, 3,...... 8}. The probability that their minimum is 3, given that their maximumis 6, is(1) `3/8`(2) `1/5`(3) `1/4`(4) `2/5`A. `(1)/(4)`B. `(2)/(5)`C. `(3)/(8)`D. `(1)/(5)`
Answer» Correct Answer - D

Discussion

472.	Four numbers are chosen at random (without replacement) from the set{1, 2, 3, ....., 20}.Statement-1:The probability that the chosen numbers whenarranged in some order will form an APIs `1/(85)`.Statement-2:If the four chosen numbers from an AP, then the setof all possible values of common difference is {1, 2, 3, 4, 5}.(1)Statement-1 istrue, Statement-2 is true; Statement-2 is not the correct explanation forStatement-1(2)Statement-1 istrue, Statement-2 is false(3)Statement-1 isfalse, Statement-2 is true(4)Statement-1 istrue, Statement-2 is true; Statement-2 is the correct explanation forStatement-1A. Statement -1 is true , Statement -2 is true, Statement-2 is a correct explanation for Statement -1.B. Statement -1 is true, Statement -2 is falseC. Statement -1 is false , Statement -2 is trueD. Statement-1 is true, Statement-2 is true, Statement-2 is not a correct explanation for Staement-1.
Answer» Correct Answer - B

Discussion

473.	For two events `A` and `B`, if `P(A)P((A)/(B))=(1)/(4)` and `P((B)/(A))=(1)/(2)`, then which of the following is not true ?A. A and B are mutually exclusive eventsB. A and B are independent events such that `P(overline(A)//B)=3//4`C. A and B are independent events such that `P(overline(A)//B)=1//2`D. A and B are in independent events such that `P(overline(A)//B)=3//4`
Answer» Correct Answer - B We have, `P(A)=P(A//B)=1//4` This shows that A and B are independent events. `therefore P(B)=P(B//A)=1//2` Now, `P(A//B)=(1)/(4) implies P(overline(A)//B)=1-(1)/(4)=(3)/(4)` Hence, alternatives (b) is correct.

Discussion

474.	If ` 0 lt P(A) lt 1, 0 lt P(B) lt 1` and `P(A cup B)=P(A)+P(B)-P(A)P(B)`, thenA. `P(A//B)=P(A)+P(B)`B. `P(A cup B)^(c )=P(A^(c ))P(B^(c ))`C. `P(A^(c )-B^(c ))=P(A^(c ))P(B^(c ))`D. `P(B//A)=P(B)-P(A)`
Answer» Correct Answer - B We have, `P(A cup B)=P(A)+P(B)-P(A)P(B)` `implies P(A)+P(B)-P(A cap B)=P(A)+P(B)-P(A)P(B)` `implies P(A cap B)=P(A)P(B)` `implies` A and B are independent events `implies A^(c ) " and " B^(c )` are independent events `implies (A^(c ) cap B^(c ))=P(A^(c ))P(B^(c ))` `implies P(A cup B)^(c ) =P(A^(c ))P(B^(c ))` Hence, option (b) is true.

Discussion

475.	If `((1-3p))/2,((1+4p))/3,((1+p))/6`are theprobabilities of three mutually excusing and exhaustive events, then the setof all values of `p`isa. (0,1)b. (-1/4,1/3) c.(0,1/3) d.A. (0, 1)B. `[-1//4, 1//3]`C. `(0, 1//3)`D. `(0, prop)`
Answer» Correct Answer - B Since `(1-3p)/(2),(1+4p)/(3),(1+p)/(6)` are the probabilities of three mutually exculsive and exhaustive events. Therefore, `(1-3p)/(2) ge 0, (1+4p)/(3) ge 0, (1+p)/(6) ge 0` and, `(1-3p)/(2)+(1+4p)/(3)+(1+p)/(6)=1` `implies -(1)/(4) le p le (1)/(3) implies p in [-1//4, 1//3]`

Discussion

476.	Events A, B, C are mutually exclusive events such that `P(A)=(3x+1)/(3), P(B)=(1-x)/(4) " and " P(C )=(1-2x)/(2)`. The set of all possible values of x are in the intervalA. [0, 1]B. `[-1//3, 1//2]`C. `[1//3, 2//3]`D. `[1//3, 13//3]`
Answer» Correct Answer - B It is given that the events A, B, C, are mutually exclusive. Therefore, `P(A) ge 0, P(B) ge 0, P(C ) ge 0, " and " P(A cup B cup C) ge 0` `implies P(A) ge 0, P(B) ge 0, P(C ) ge 0, " and "P(A)+P(B)+P(C ) ge0` `implies (3x+1)/(3) ge 0, (1-x)/(4) gt 0, (1-2x)/(2) ge 0`, and, `(3x+1)/(3)+(1-x)/(4)+(1-2x)/(2) ge 0` `implies x gt -(1)/(3), x le 1, x le (1)/(2) " and" 1-3x gt 0` `implies -(1)/(3) le x le (1)/(2) " and " x le (1)/(3)` `implies -(1)/(3) le x le (1)/(2) implies x in [-1//3, 1//2]`

Discussion

477.	Two natural numbers x and y are chosen at random from the set `{1,2,3,4,...3n}`. find the probability that `x^2-y^2` is divisible by 3.A. `(5n-3)/(2(3n-1))`B. `(5n-3)/(3(3n-1))`C. `(3n-1)/((5n-3))`D. `(3n-1)/(2(5n-3))`
Answer» Correct Answer - B The number of ways of choosing two numbers from the given set is `.^(3n)C_(2)`. Let us divide given 3n numbers into three groups `G_(1),G_(2) " and " G_(3)` as follows : `G_(1) : 3,6,9,.., 3n` `G_(2) : 1,4,7,10, .., 3n-1` `G_(3) : 2,5,8,11,..,3n-2` We have, `a^(2)-b^(2)=(a-b)(a+b)` Therefore, `a^(2)-b^(2)` will be divisible by 3 if either a and b are chosen from the same group or one of them is chosen from group `G_(2)` and the other from group `G_(3)`. Therefore, the number of favourable elemnetary events is `(.^(n)C_(2)+ .^(n)C_(2)+ .^(n)C_(2))+ .^(n)C_(1) xx .^(n)C_(1)=3 .^(n)C_(2)+n^(2)` `therefore` Required probability `=(3xx .^(n)C_(2)+n^(2))/(.^(3n)C_(2))=(5n-3)/(3(3n-1))`

Discussion

478.	If two distinct numbers m and n are chosen at random form the set {1, 2, 3, …, 100}, then find the probability that `2^(m) + 2^(n) + 1` is divisible by 3.
Answer» Correct Answer - `(49)/(198)` `2^(m) + 2^(n) + 1 = (3 - 1)^(m) + (3 - 1)^(n) + 1` `=3k + (-1)^(m) + (-1)^(n) + 1` This is divisible by 3 if both m and n are even. `therefore` Required probabaility = `(.^(50)C_(2))/(.^(100)C_(2)) = (50 xx 49)/(100 xx 99) = (49)/(198)`

Discussion

479.	Three distinct numbers are chosen at random from the first 15 natural numbers. The probability that the sum will be divisible by 3 isA. `(30)/(91)`B. `(31)/(91)`C. `(60)/(91)`D. none of these
Answer» Correct Answer - B Out of first 15 natural numbers, 3 natural numbers can be chosen is `.^(15)C_(3)` ways. When we divide a natural number by 3, it leaves either 0 or 1 or 2 as the remainder, so, let us divide 15 natural numbers into the following groups : `G_(1) : 1 4 7 10 13` `G_(2) : 2 5 8 11 14` `G_(3) : 3 6 9 12 15` The sum of three natural numbers chosen from these 15 natural numbers will be divisible by 3 if either all the three natural numbers are chosen from the same group or one natural is chosen from each group. `therefore` Favourable number of elementary events `=(.^(5)C_(3)+ .^(5)C_(3)+ .^(5)C_(3))+(.^(5)C_(1)xx.^(5)C_(1)xx .^(5)C_(1))=155` Hence, required probability `=(155)/(.^(15)C_(3))=(155xx6)/(15xx14xx13)=(31)/(91)`

Discussion

480.	A binary operation is chosen at random from the set of all binary operations on a set A containing n elements. The probability that the binary operation is commutative, isA. `(n^(n))/(n^(n^(2)))`B. `(n^(n//2))/(n^(n^(2)))`C. `(n^(n//2))/(n^(n^(2//2)))`D. none of these
Answer» Correct Answer - C

Discussion

481.	Two number `aa n db`aer chosen at random from the set of first 30 naturalnumbers. Find the probability that `a^2-b^2`is divisible by 3.
Answer» Correct Answer - `(47)/(87)` The total number of ways of choosing two numbers out of 1, 2, 3, …, 30 is `.^(30)C_(2)` = 435. Now, 30 natural numbers are categorized as 3k type: 3, 6, 9, … 30 `rarr` 10 numbers 3k + 1 type: 1, 4, 7, …, 28 `rarr` 10 numbers 3k + 2 type: 2, 5, 8, ..., 29 `rarr` 10 numbers `a^(2) - b^(2)` is divisible by 3 iff either both of a and b are So, number of favourable cases = `.^(10)C_(2) + .^(20)C_(2) = 235` `therefore` Required probability = `(235)/(435) = (47)/(87)`

Discussion

482.	If the events A and B are mutually exclusive events such that P(A) = `(3x + 1)/(3)` and P(B) = `(1 - x)/(4)`, then the set of possible real values of x lies in the intervalA. [0, 1]B. `[-(1)/(3), (5)/(9)]`C. `[-(7)/(9),(4)/(9)]`D. `[(1)/(3),(2)/(3)]`
Answer» Correct Answer - B We must have, `0 le P(A) le 1` implies `0 le (3x + 1)/(3) le 1` implies `(-1)/(3) le x le (2)/(3)" (1)"` and `0le P(B) le 1` implies `0 le (1-x)/(4) le 1` implies `-3 le x le 1 " (2)"` and `0le P(A uu B) = P(A) + P(B) le 1` `implies 0 le (3x + 1)/(3) + (1 - x)/(4) le 1` implies `(-7)/(9) le x le (5)/(9) " (3)"` From (1), (2) and (3), we get x `in [(-1)/(3), (5)/(9)]`.

Discussion

483.	Two natural numbers x and y are chosen at random. What is the probability that `x^(2) + y^(2)` is divisible by 5?
Answer» Let r and s be the remainders when x and y are divided by 5, so `x = 5p + r, y = 5q + s,p,q, in N, 0 le r le4, 0lesle4` `therefore x^(2) + y^(2) = 25(p^(2) + q^(2))+ 10(pr+qs)+r^(2)+s^(2) = 5k + (r^(2) + s^(2)), k in N` Thus, `x^(2) + y^(2)` will be divisible by 5 if `r^(2) + s^(2)` is divisible by 5. Therefore, total number of ways is equal to the number of ways of selecting r and s. which is `5^(2) = 25` For favorable ways, we should have `r^(2) + s^(2)` is divisible by 5 and `0 le r^(2) + s^(2) le 32 so r^(2) + s^(2) = 0, 5, 10, 20, 25, or 30`. Thus `{:(r,0,1,2,3,4),(s,0,"2,3","1,4","1,4","2,3"):}` So number of favourable ways is 9. Hence, the probability is 9/25.

Discussion

484.	A dice is thrown six times, it being known that each time a differentdigit is shown. The probability that a sum of 12 will b e obtained in thefirst three throws is`5//24`b. `25//216`c. `3//20`d. `1//12`A. `(5)/(24)`B. `(25)/(216)`C. `(3)/(20)`D. `(1)/(12)`
Answer» Correct Answer - C Since it is known that each time a different digit appears, total number of cases is `6!` The sum in the first three throws is 12 if we get (1, 5, 6), (2, 4, 6) or (3, 4, 5) in any order. So, number of cases for first three throws = `3 xx 3!` ways. `therefore` Number of favourable cases = `3 xx 3! xx 3!` `therefore` Required probability = `(3 xx 3! xx 3!)/(6!) = (3)/(20)`

Discussion

485.	A natural number is chosen at random from the first 100 naturalnumbers. The probability that `x+(100)/x > 50`is`1//10`b. `11//50`c. `11//20`d. none of theseA. `1//10`B. `11//50`C. `11//20`D. none of these
Answer» Correct Answer - C

Discussion

486.	A natural number is chosen at random from the first 100 naturalnumbers. The probability that `x+(100)/x > 50`is`1//10`b. `11//50`c. `11//20`d. none of theseA. `1//10`B. `11/50`C. `11/20`D. none of these
Answer» Correct Answer - C We have, `x + (100)/(x) gt 50` or `x^(2) + 100 gt 50x` or `(x - 25)^(2) gt 525` `implies x - 25 lt - sqrt(525) or x - 25 sqrt(525)` `implies x lt 25 - sqrt(525) or x gt 25 + sqrt(525)` As x is a positive integer and `sqrt(525)` = 22.91, we must have `x le 2` or `x ge 48`. Thus, the favorable number of cases is 2 + 53 = 55. Hence, the required probability is `55//100` = `11//20`.

Discussion

487.	Five persons entered the lift cabin on the groundfloor of an 8-floor house. Suppose that each of them independently and withequal probability can leave the cabin at any floor beginning with the first.Find out the probability of all five persons leaving at different floors.
Answer» Correct Answer - `(.^(7)P_(5))/(7^(5))` In an 8-floor house, there are 7 floors above the ground floor. Each person can leave the cabin at any of the seven floors, i.e., each person can leave the cabin in 7 ways. Thus, total number of ways into which 5 persons can leave the cabin is `7^(5)`. Now number of the ways of leaving the cabin by 5 person each at different floor is `.^(7)P_(5)`. Hence, the required probability is `.^(7)P_(5)//7^(5)`.

Discussion

488.	Five persons entered the lift cabin on the groundfloor of an 8 floor house. Suppose that each of them independently and withequal probability can leave the cabin at any floor beginning with the first,then the probability of al 5 persons leaving at different floor isa. `( ^7P_5)/(7^5)`b. `(7^5)/( ^7P_5)`c. `6/( ^6P_5)`d. `( ^5P_5)/(5^5)`A. `(.^(7)C_(5))/(7^(5))`B. `(.^(7)C_(5)xx5!)/(5^(7))`C. `(.^(7)C_(5)xx5!)/(7^(5))`D. `(5!)/(7^(5))`
Answer» Correct Answer - C Besides ground floor, there are 7 floors. Since a person can leave the cabin at any of the seven floors, therefore the total number of ways in which each of the five persons can leave the cabin at any of the 7 floors `=7^(5)`. `therefore` Total number of ways `=7^(5)`. Five persons can leave the cabin at five different floors in `.^(7)C_(5)xx5!` ways. Therefore, `therefore` Favourable number of ways `=.^(7)C_(5)xx5!` Hence, required probability `=(.^(7)C_(5)xx5!)/(7^(5))`

Discussion

489.	Five persons entered the lift cabin on the ground floor of an 8 floored house. Suppose that each of them independently and with equal probability can leave the cabin at any floor beginning with the first, then the probability of all 5 persons leaving at different floor isA. \(\frac{^7P_5}{7^5}\) B. \(\frac{7^5}{^7P_5}\) C. \(\frac{6}{^6P_5}\) D. \(\frac{^5P_5}{5^5}\)
Answer» As building has 8 floors including ground. So, they have 7 options to leave the lift. ∴ total ways in which persons can leave the lift = 7⁵ As the number of ways in which all persons leave in a different floor can be given by: 7× 6× 5× 4× 3 = ⁷P₅ ∴ P(that all leave lift in different floor) = \(\frac{^7P_5}{7^5}\) Our answer matches with option (a) ∴ Option (a) is the only correct choice

489.

Five persons entered the lift cabin on the ground floor of an 8 floored house. Suppose that each of them independently and with equal probability can leave the cabin at any floor beginning with the first, then the probability of all 5 persons leaving at different floor isA. \(\frac{^7P_5}{7^5}\) B. \(\frac{7^5}{^7P_5}\) C. \(\frac{6}{^6P_5}\) D. \(\frac{^5P_5}{5^5}\)

Answer»

As building has 8 floors including ground.

So, they have 7 options to leave the lift.

∴ total ways in which persons can leave the lift = 7⁵

As the number of ways in which all persons leave in a different floor can be given by:

7× 6× 5× 4× 3 = ⁷P₅

∴ P(that all leave lift in different floor) = \(\frac{^7P_5}{7^5}\)

Our answer matches with option (a)

∴ Option (a) is the only correct choice

Discussion

490.	If X and Y are two events such that `PX//Y)=(1)/(2), P(Y//X)=(1)/(3)` and `P(XcapY)(1)/(6)`. Then, which of the following is/are correct ?A. `P(XcupY)=2//3`B. X and Y are independentC. X and Y are not independentD. `P(X^(c )capY)=1//3`
Answer» Correct Answer - A::B Plan (i) Conditional probability, i.e., P(A/B) `= (P(A nn B))/(P(B))` (ii) P (A uu B) = P(A) + P(B) - P(A nn B)` `(iii) Independent event, then `P(A nn B) = P(A) * P(B) ` Here, `P(X//Y) = (1)/(2) , P ((Y)/(X)) = (1)/(3)` and `P (X nn Y) = 6` `therefore P ((X)/(Y)) = (P (X nn Y))/(P(Y))` `rArr (1)/(2) = (1//6)/(P (Y)) rArr P(Y)) = (1)/(3) " "... (i)` `P ((X)/(Y)) = (1)/(3) rArr (P(X nn Y))/(P(X)) = (1)/(3)` `rArr (1)/(6) = (1)/(3) P(X)` `therefore P(X) = (1)/(2) " "...(ii)` `P(X nn Y) = P(X) + P(Y) - P(X nn Y)` ` = (1)/(2) + (1)/(3) - (1)/(6) = (2)/(3) " " .... (iii)` `P(X nn Y) = (1)/(6) "and " P(X) * P(Y) = (1)/(2) * (1)/(3) = (1)/(6)` `rArr P(X nn Y) = P(X) * P(Y)` i.e. independent events `therefore P(X^(e) nn Y) = P(Y) - P(X nn Y)` ` = (1)/(3) - (1)/(6) = (1)/(6)`

Discussion

491.	A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out(i) an orange flavoured candy?(iii) a lemon flavoured candy?
Answer» (i) As the bag contains only lemon flavored candies. So, the event related to the experiment of taking out an orange flavored candy is an impossible event. So, its probability is 0. (ii) As the bag contains only lemon flavored candies. So, the event related to the experiment of taking out lemon flavored candies is certain event. So, its probability is 1.

Discussion

492.	The probability that an event A happens in one trial of an experiment, is 0.4 There independent trials of the experiments are performed. The probability that the event A happens atleast once, isA. 0.936B. 0.784C. 0.904D. None of these
Answer» Correct Answer - B Given that , `P(A)=0.4 , P(bar(A))=0.6 ` P(the event A happens at least once ) =1-p(none of the event happens ) 1-(0.6)(0.6)(0.6) =1-0.2165 =0.784

Discussion

493.	a bag contains lemon- flavoured candies only. Hema takes out one condy without looking into the bag. What is the probability that she takes out (i) an orange-flavoured candy? (ii) a lemon-flavoured candy?
Answer» Correct Answer - ` (i) 0 (ii) 1` Suppose there are x candies in the bag. Then, number of orange-flavoured candies in the bag = 0. And, the number of lemon-flavoured candies in the bag = x. ` :. ` (i) P(getting an orange-flavoured candy) ` = 0/x = 0`. (ii) P(getting a lemon-flavoured candy) = ` x/x = 1`.

Discussion

494.	(i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?(ii) Suppose the bulb drawn in(a) is not defective and nõt replaced. Now bulb is drawn at random from the rest. What is the probability that this bulb is not defective?
Answer» Total no. of possible outcomes = 20 {20 bulbs} (i) E ⟶ be event of getting defective bulb. No. of favourable outcomes = 4 {4 defective bulbs} Probability, P(E) = (No.of favorable outcomes)/(Total no.of possible outcomes) = 4/20 = 1/5 (ii) Bulb drawn in is not detective & is not replaced remaining bulbs = 15 good + 4 bad bulbs = 19 Total no. of possible outcomes = 19 E ⟶ be event of getting defective No. of favorable outcomes = 15 (15 good bulbs) P(E) = 15/19

494.

(i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?(ii) Suppose the bulb drawn in(a) is not defective and nõt replaced. Now bulb is drawn at random from the rest. What is the probability that this bulb is not defective?

Answer»

Total no. of possible outcomes = 20 {20 bulbs}

(i) E ⟶ be event of getting defective bulb.

No. of favourable outcomes = 4 {4 defective bulbs}

Probability, P(E) = (No.of favorable outcomes)/(Total no.of possible outcomes) = 4/20 = 1/5

(ii) Bulb drawn in is not detective & is not replaced remaining bulbs = 15 good + 4 bad bulbs = 19

Total no. of possible outcomes = 19

E ⟶ be event of getting defective

No. of favorable outcomes = 15 (15 good bulbs)

P(E) = 15/19

Discussion

495.	(i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective? (ii) Suppose the bulb drawn in (i) in not defective and not replaced. Now, bulb is drawn at random from the rest. What is the probability that this bulb is not defective ?
Answer» Correct Answer - ` (i) 1/5 (ii) 15/19` (i) Total number of bulbs = 20. Number of defective bulb = 4. Number of non-defective bulbs = 20- 4= 16. `:. ` P(getting a defective bulb) = ` 4/20 = 1/5`. (ii) After removing 1 non-defective bulb, we have remaining number of bulbs = 20-1 = 19. Out of these, the number of non-defective bulb = 16-1=15. ` :. ` P(getting a non=defective bulb) = ` 15/19`.

Discussion

496.	(i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective ?
Answer» (i) Here, total number of bulbs `= 20` Number of defective bulbs `=4` So, probability of drawing a defective bulb `= 4/20 = 1/5` (ii)If drawn bulb is not defective and not replaced, then Remaining number of total bulbs `= 19` Remaining number of defective bulbs `=4` So,remaining number of non-defective bulbs `=15` Probability of drawing a non-defective bulbs `= 15/19`

Discussion

497.	An amobeba either splits into two or remains the same or eventually dies out immediately after completion of evary second with probabilities, respectively, 1/2, 1/4 and 1/4. Let the initial amoeba be called as mother amoeba and after every second, the amoeba, if it is distinct from the previous one, be called as 2nd, 3rd,...generations. The probability that after 2 s exactly 4 amoeba are alive isA. `1//16`B. `1//8`C. `3//4`D. `1//2`
Answer» Correct Answer - B After 2 s, exctly 4 amoeba are alive, i.e., initially amoeba must split into two and in 2nd second again amoeba must split into two. Hence, the required propability is `(1//2)xx(1//2)xx(1//2)=1//8.`

Discussion

498.	A lot contains 50 defective and 50 non defective bulbs. Two bulbs are drawn at random, one at a time, with replacement. The events A, B, C are defined asA = (the first bulb is defective)B = (the second bulb is non – defective)C = (the two bulbs are both defective or both non defective) Determine whether (i) A, B, C are pair wise independent (ii) A, B, C are independent.
Answer» Let S = defective and Y = non defective. Then all possible outcomes are {XX, XY, YX, YY} Also P (XX) = 50/100 x 50/100 = 1/4, P(XY) = 50/100 x 50/100 = 1/4, P(YX) = 50/100 x 50/100 = 1/4, P(YY) = 50/100 x 50/100 = 1/4 Here, A =XX ∪ XY; B = XY ∪ YY; C = XX ∪ YY ∴ P (A) = P(XX) + P(XY) = 1/4 + 1/4 = 1/2 ∴ P (B) = P (XY) + P (YX) = 1/4 + 1/4 = 1/2 P (C) = P (XX) + P (YY) = 1/4 + 1/4 =1/2 Now, P (AB) = P (XY) = 1/4 = P (A). P (B) ∴ A and B are independent events. P (BC) = P (YX) = 1/4 = P (B). P (C) ∴ B and C are independent events. P (CA) = P (XX) = 1/4 = P (C). P (A) ∴ C and A are independent events. P (ABC) = 0 (impossible event) ≠ P (A) P (B) P (C) ∴ A, B, C are dependent events, Thus we can conclude that A, B, C are pair wise independent bet A, B, C are dependent events.

498.

A lot contains 50 defective and 50 non defective bulbs. Two bulbs are drawn at random, one at a time, with replacement. The events A, B, C are defined asA = (the first bulb is defective)B = (the second bulb is non – defective)C = (the two bulbs are both defective or both non defective) Determine whether (i) A, B, C are pair wise independent (ii) A, B, C are independent.

Answer»

Let S = defective and Y = non defective. Then all possible outcomes are {XX, XY, YX, YY}

Also P (XX) = 50/100 x 50/100 = 1/4,

P(XY) = 50/100 x 50/100 = 1/4,

P(YX) = 50/100 x 50/100 = 1/4,

P(YY) = 50/100 x 50/100 = 1/4

Here, A =XX ∪ XY; B = XY ∪ YY; C = XX ∪ YY

∴ P (A) = P(XX) + P(XY) = 1/4 + 1/4 = 1/2

∴ P (B) = P (XY) + P (YX) = 1/4 + 1/4 = 1/2

P (C) = P (XX) + P (YY) = 1/4 + 1/4 =1/2

Now, P (AB) = P (XY) = 1/4 = P (A). P (B)

∴ A and B are independent events.

P (BC) = P (YX) = 1/4 = P (B). P (C)

∴ B and C are independent events.

P (CA) = P (XX) = 1/4 = P (C). P (A)

∴ C and A are independent events.

P (ABC) = 0 (impossible event)

≠ P (A) P (B) P (C)

∴ A, B, C are dependent events,

Thus we can conclude that A, B, C are pair wise independent bet A, B, C are dependent events.

Discussion

499.	An unbiased coin is tossed. If the result is a head, a pair of unbiased dice is rolled and the number obtained by adding the numbers on the two faces is noted. If the result is a tail, a card from a well shuffled pack of eleven cards numbered 2, 3, 4 . . . . 12 are picked and the number on the card is noted. What is the probability that the noted number is either 7 or 8?
Answer» E₁ ≡ number noted is 7, E₂ ≡ number notes is 8, H ≡ getting head on coin, T ≡ getting tail on coin. Then by total probability theorem, P (E₁) = P (H) P (E₁\| H) + P (T) P (E₁\| T) And P (E₂) = P (H) P (E₂\| H) + P (T) P (E₂\| T) Where P (H) = 1/2 ; P (T) = 1/2 P(E₁\| H) = prob. of getting a sum of 7 on two dice. Here favorable cases are {(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)} ∴ P (E₁\| H) = 6/36 = 1/6 Also P (E₁\| T) = prob. if getting '7' numbered card out of 11 cards = 1/11. P(E₂\|H) = Prob. of getting a sum of 8 on two dice. Here favorable cases are {(2, 6), (6, 2), (4, 4), (5, 3), (3, 5)} ∴ P (E₂\| H) = 5/36 P (E₂\| T) = prob. of getting '8' numbered card out of 11 cards = 1/11 P (E₁) = 1/2 x 1/6 + 1/2 x 1/11 = 1/12 + 1/22 = 11 + 6/132 = 17/132 P (E₂) = 1/2 x 5/36 + 1/2 x 1/11 = 1/2 [55 + 36/396] = 91/792 Now E₁ and E₂ are mutually exclusive events therefore P (E₁ or E₂) = P (E₁) + P (E₂) = 17/132 + 91/792 = 102 + 91/792 = 193/792 = 0.2436

499.

An unbiased coin is tossed. If the result is a head, a pair of unbiased dice is rolled and the number obtained by adding the numbers on the two faces is noted. If the result is a tail, a card from a well shuffled pack of eleven cards numbered 2, 3, 4 . . . . 12 are picked and the number on the card is noted. What is the probability that the noted number is either 7 or 8?

Answer»

E₁ ≡ number noted is 7, E₂ ≡ number notes is 8,

H ≡ getting head on coin, T ≡ getting tail on coin.

Then by total probability theorem,

P (E₁) = P (H) P (E₁| H) + P (T) P (E₁| T)

And P (E₂) = P (H) P (E₂| H) + P (T) P (E₂| T)

Where P (H) = 1/2 ; P (T) = 1/2

P(E₁| H) = prob. of getting a sum of 7 on two dice. Here favorable cases are

{(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)}

∴ P (E₁| H) = 6/36 = 1/6

Also P (E₁| T) = prob. if getting '7' numbered card out of 11 cards = 1/11.

P(E₂|H) = Prob. of getting a sum of 8 on two dice. Here favorable cases are

{(2, 6), (6, 2), (4, 4), (5, 3), (3, 5)}

∴ P (E₂| H) = 5/36

P (E₂| T) = prob. of getting '8' numbered card out of 11 cards = 1/11

P (E₁) = 1/2 x 1/6 + 1/2 x 1/11 = 1/12 + 1/22 = 11 + 6/132 = 17/132

P (E₂) = 1/2 x 5/36 + 1/2 x 1/11 = 1/2 [55 + 36/396] = 91/792

Now E₁ and E₂ are mutually exclusive events therefore

P (E₁ or E₂) = P (E₁) + P (E₂) = 17/132 + 91/792

= 102 + 91/792 = 193/792 = 0.2436

Discussion

500.	Numbers are selected at randm, one at time, from the two-digit numbers `00,01,02…,99` with replacement. An event E occurs if the only product of the two digits of a selected number is 18. If four members are selected, find the probability that the event E occurs at least 3 times.
Answer» Correct Answer - `(97)/((25)^(4))` The given numbers are `00,01,02…,99.` There are total 100 numbers, out of which the numbers, the product of whose digits is 18, are `29,36,63,and 92.` `thereforep=P(E)=(4)/(100)=1/25` ` impliesq=1-p =24/25` From binomial distribution, P(E occuring at least 3 times) =P(E occuring 3 times)+ P (E occuring 4 times) `=""^(4)C_(3)p^(3)q+""^(4)C_(4)p^(4)` `=4xx((1)/(25))^(3)((24)/(25))+((1)/(25))^(4)=(97)/((25)^(4))`

Discussion

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Two number `aa n db`aer chosen at random from the set of first 30 naturalnumbers. Find the probability that `a^2-b^2`is divisible by 3.

If the events A and B are mutually exclusive events such that P(A) = `(3x + 1)/(3)` and P(B) = `(1 - x)/(4)`, then the set of possible real values of x lies in the intervalA. [0, 1]B. `[-(1)/(3), (5)/(9)]`C. `[-(7)/(9),(4)/(9)]`D. `[(1)/(3),(2)/(3)]`

Two natural numbers x and y are chosen at random. What is the probability that `x^(2) + y^(2)` is divisible by 5?

A dice is thrown six times, it being known that each time a differentdigit is shown. The probability that a sum of 12 will b e obtained in thefirst three throws is`5//24`b. `25//216`c. `3//20`d. `1//12`A. `(5)/(24)`B. `(25)/(216)`C. `(3)/(20)`D. `(1)/(12)`

A natural number is chosen at random from the first 100 naturalnumbers. The probability that `x+(100)/x > 50`is`1//10`b. `11//50`c. `11//20`d. none of theseA. `1//10`B. `11//50`C. `11//20`D. none of these

A natural number is chosen at random from the first 100 naturalnumbers. The probability that `x+(100)/x > 50`is`1//10`b. `11//50`c. `11//20`d. none of theseA. `1//10`B. `11/50`C. `11/20`D. none of these

Five persons entered the lift cabin on the groundfloor of an 8-floor house. Suppose that each of them independently and withequal probability can leave the cabin at any floor beginning with the first.Find out the probability of all five persons leaving at different floors.

If X and Y are two events such that `PX//Y)=(1)/(2), P(Y//X)=(1)/(3)` and `P(XcapY)(1)/(6)`. Then, which of the following is/are correct ?A. `P(XcupY)=2//3`B. X and Y are independentC. X and Y are not independentD. `P(X^(c )capY)=1//3`

A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out(i) an orange flavoured candy?(iii) a lemon flavoured candy?

The probability that an event A happens in one trial of an experiment, is 0.4 There independent trials of the experiments are performed. The probability that the event A happens atleast once, isA. 0.936B. 0.784C. 0.904D. None of these

a bag contains lemon- flavoured candies only. Hema takes out one condy without looking into the bag. What is the probability that she takes out (i) an orange-flavoured candy? (ii) a lemon-flavoured candy?

Numbers are selected at randm, one at time, from the two-digit numbers `00,01,02…,99` with replacement. An event E occurs if the only product of the two digits of a selected number is 18. If four members are selected, find the probability that the event E occurs at least 3 times.