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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 301. |
In a particular section of class IX, 40 students were asked about the month of their birth and the following graph was prepared for the data so obtained. Find the probability that a student of the class was born in August. |
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Answer» Total number of students in class IX, n(S) = 40 Number if student born in the month of August, n(E) = 6 `therefore" Probability, that the students of the class IX was born in August" = (n(E))/(n(S))=(6)/(40)=(3)/(20)`Ans. |
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| 302. |
Following table shows the birth month of 40 students of class IX.Jan.Feb.MarchAprilMayJuneJulyAug.Sep.Oct.Nov.Dec.342251263444Find the probability that a student was born in August. |
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Answer» Given, Probability (students was born in August) \(=\frac{favourable\,outcome}{total\,outcome}\) \(=\frac{6}{40}=\frac{3}{20}\) |
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| 303. |
A box `B_1`, contains 1 white ball, 3 red balls and 2 black balls. Another box `B_2`, contains 2 white balls, 3 red balls and 4 black balls. A third box `B_3`, contains 3 white balls, 4 red balls and 5 black balls.A. `(116)/(181)`B. `(126)/(181)`C. `(65)/(181)`D. `(55)/(181)` |
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Answer» Correct Answer - D Let `E_(i)(i=1,2,3)` denote the event of selecting `i^(th)` bag and A denote the event of drawing one white and one red ball. Then, `P(E_(1))=P(E_(2))=P(E_(3))=(1)/(3)` `P(A//E_(1))=(1)/(6)xx(3)/(6)+(3)/(6)xx(1)/(5)=(1)/(5)` `P(A/E_(2))=(2)/(9)xx(3)/(8)+(3)/(9)xx(2)/(8)=(1)/(6)` `P(A//E_(3))=(3)/(12)xx(4)/(11)+(4)/(12)xx(3)/(11)=(2)/(11)` Required probability `=P(E_(2)//A)` `=(P(E_(2))P(A//E_(2)))/(P(E_(1))P(A//E_(1))+P(E_(2))P(A//E_(2))+P(E_(3))P(A//E_(3)))` `((1)/(3)xx(1)/(6))/((1)/(3)xx(1)/(5)+(1)/(3)xx(1)/(6)+(1)/(3)xx(2)/(11))=(55)/(181)` |
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| 304. |
The probability that randomly selected calculator from a store is of brand r is proportional to r, r=1,2,..,6. Further, the probability of a calucltor of brand r being defective is `(7-r )/(21), r=1,2,..,6`. Then the probability that a calculator randomly selected from the store being defective isA. `(8)/(63)`B. `(13)/(63)`C. `(55)/(63)`D. `(50)/(63)` |
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Answer» Correct Answer - A Let `C_(r )` denote the event that a calculator of brand r is selected and `D_(r )` denote the event that a calculate of brand r is defective. Then, `P(C_(r ))=lamda r " and " P(D_(r )//C_(r ))=(7-r)/(21), r=1, 2,..,6`. Now, `underset(r=1)overset(6)(sum)P(C_(r )=1` `implies underset(r=1)overset(6)(sum) lamda r=1 implies lamda(underset(r=1)overset(6)(sum) r)=1 implies 21 lamda =1 implies lamda =(1)/(21)` Required probability `= underset(r=1)overset(6)(sum) P(C_(r ) cap D_(r ))` `= underset(r=1)overset(6)(sum) P(C_(r ))P(D_(r )//C_(r ))` `= underset(r=1)overset(6)(sum) lamda r xx(7-r)/(21)` `=(lamda)/(21) underset(r=1)overset(6)(sum) (7r - r^(2))` `=(1)/(21^(2)){ underset( r=1)overset(6)(sum) 7r- underset(r=1)overset(6)(sum) r^(2)}` `=(1)/(21^(2)){7xx21-(6xx7xx13)/(6)}=(8)/(63)` |
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| 305. |
Six dice are thrown simultaneously. The probability that exactly three of them show the same face and ramining three show different faces, isA. `((5!)^(2))/(6^(5))`B. `(5!)/(2!6^(6))`C. `((5!)^(2))/(2(6^(6))`D. `(5!)/(2(6^(6)))` |
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Answer» Correct Answer - C Six dice when thrown simultaneously can result in `6^(6)` ways. `therefore` Total number of elementary events `=6^(6)`. Select a number which occurs on three dice out of six numbers 1,2,3,4,5,6. This can be done in `.^(6)C_(1)` ways. Now, select three numbers out of the remaining 5 numbers. This can be done in `.^(5)C_(3)` ways. Now, we have 6 numbers like 1,2,3,4,4,4, 2,3,6,1,1,1 etc. These digits can be arranged in `(6!)/(3!)` ways. So, the number of ways in which three dice show the same face and the remaining three show distinct faces is `.^(6)C_(1)xx .^(5)C_(3)xx(6!)/(3!)` `therefore` Favourable number of elementary events `= (.^(6)C_(1)xx .^(5)C_(3)xx(6!)/(3!))/(6^(6))=((5!)^(2))/(2(6^(6)))` |
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| 306. |
Six dice are thrown simultaneously. The probability that all of them show the same face, isA. `(1)/(6^(6))`B. `(1)/(6^(5))`C. `(1)/(6)`D. none of these |
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Answer» Correct Answer - B The total number of elementary events associated to the random experiment of throwing six dice is `6xx6xx6xx6xx6xx6=6^(6)` All dice show the same face means we are getting same number on all six dice. So, this number can be any one of the six numbers 1,2,..6. The number of ways of selecting a number is ` .^(6)C_(1)`. `therefore` Favourable number of elementary events `= .^(6)C_(1)`. Hence, required probability `= (.^(6)C_(1))/(6^(6))=(1)/(6^(5))` |
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| 307. |
Six dice are thrown simultaneously. The probability that all of them show the different faces, isA. `(1)/(6^(5))`B. `(6!)/(6^(6))`C. `(1)/(6!)`D. `(5!)/(6!)` |
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Answer» Correct Answer - B The total number of ways in which all dice show different faces is same as the number of arrangements of 6 numbers 1,2,3,4,5,6 by taking all at a time. So, favourable number of elementary events -6 ! Hence, required probaility `=(6!)/(6^(6))` |
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| 308. |
Four persons independently solve a certain problem correctly with proabilities `1/2,3/4,1/4,1/8.` Then the probability that the problem is solved correctly by at least one of them isA. `235/256`B. `21/256`C. `3/256`D. `253/256` |
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Answer» Correct Answer - A P (at least one of them solves correctly) = 1 -P (none of them solves correctly) `=1-(1/2xx1/4xx3/4xx7/8)=235/256` |
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| 309. |
If e1, e2, e3, e4 are the four elementary outcomes in a sample space and P(e1) = 0.1, P(e2) = 0.5, P(e3) = 0.1, then the probability of e4 is _____________. |
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Answer» 0.3 If e1, e2, e3, e4 are the four elementary outcomes in a sample space and P(e1) = 0.1, P(e2) = 0.5, P(e3) = 0.1, then the probability of e4 is 0.3 . |
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| 310. |
Four person independently solve a certain problem correctly withprobabilities `1/2,3/4,1/4,1/8dot`Then the probability that he problem is solve correctly by at least oneof them is`(235)/(256)`b. `(21)/(256)`c. `3/(256)`d. `(253)/(256)`A. `(235)/(256)`B. `(21)/(256)`C. `(3)/(256)`D. `(253)/(256)` |
| Answer» Correct Answer - A | |
| 311. |
There are three events E1, E2 and E3, one of which is must and only one can happen. The odds are 7 to 4 against E1 and 5 to 3 against E2. Find the odds against E3.? |
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Answer» Since one and only one of the three events E1, E2 and E3 can happen, i.e, they are mutually exclusive. Therefore, P(E1) + P(E2) + P(E3) = 1 ...(i) Odds against E1 are 7 : 4 ⇒ Odds in favour of E1 are 4 : 7 ⇒ P(E1) = \(\frac{4}{4+7}\) = \(\frac{4}{11}\) ....(ii) Odds against E2 are 5 : 3 ⇒ Odds in favour of E2 are 3 : 5 ⇒ P(E2) = \(\frac{3}{3+5}\) = \(\frac{3}{8}\) ....(iii) ∴ From (i), (ii) and (iii) \(\frac{4}{11}\) + \(\frac{3}{8}\) + P(E3) = 1 ⇒ P(E3) = 1 - \(\bigg(\)\(\frac{4}{11}\)+\(\frac{3}{8}\)\(\bigg)\) = 1 - \(\bigg(\frac{32+33}{88}\bigg)\) = 1 - \(\frac{65}{88}\) = \(\frac{23}{88}\) ∴ Odds against E3 are \(\frac{1-P(E_3)}{P(E_3)}\) = \(\frac{1-\frac{23}{88}}{\frac{23}{88}}\) = \(\frac{\frac{65}{88}}{\frac{23}{88}}\) = 65 : 23. |
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| 312. |
Fill in the blanks:If e1, e2, e3, e4 are the four elementary outcomes in a sample space and P(e1) =.1, P(e2) = .5, P (e3) = .1, then the probability of e4 is ______. |
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Answer» Given: P(e1) = 0.1, P(e2) = 0.5, P(e3) = 0.1 We know that, Sum of all probabilities = 1 ∴ P(e1) + P(e2) + P(e3) + P(e4) = 1 ⇒ 0.1 + 0.5 + 0.1 + P(e4) = 1 ⇒ P(e4) = 1 – 0.7 ⇒ P(e4) = 0.3 Ans. If e1, e2, e3, e4 are the four elementary outcomes in a sample space and P(e1) =.1, P(e2) = .5, P (e3) = .1, then the probability of e4 is 0.3 |
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| 313. |
If the three independent events E1, E2 and E3, the probability that only E1 occurs is \(\alpha\), E2 occurs is \(\beta\), only E3 occurs is \(\gamma\). Let the probability p that none of the event E1, E2 or E3 occurs satisfy the equations (\(\alpha\) – 2\(\beta\)) p = \(\alpha\)\(\beta\) and (\(\beta\) – 3\(\gamma\)) p = 2\(\beta\)\(\gamma\). All the given probabilities are assumed to lie in the interval (0, 1).Then, \(\frac{\text{Probability of occurrence of}\,E_1}{\text{Probability of occurrence of}\,E_3}\) is equal to(a) 3 (b) 5 (c) 6 (d) 8 |
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Answer» (c) 6 Let \(x\), y, z be the probabilities of happening of events E1, E2 and E3 respectively. Then, \(\alpha\) = P (occurrence of E1 only) = x (1 – y) (1 – z) \(\beta\) = P (occurrence of E2 only) = (1 – \(x\)) y (1 – z) \(\gamma\) = P (occurrence of E3 only) = (1 – \(x\)) (1 – y) z p = P (not occurrence of E1, E2, E3) = (1 – \(x\)) (1 – y) (1 – z). ∴ (\(\alpha\) – 2\(\beta\)) p = \(\alpha\)\(\beta\) ⇒ [\(x\)(1 – y) (1 – z) – 2 (1 – \(x\))y (1 – z)] (1 – \(x\)) (1 – y) (1 – z) = \(x\) (1 – y) (1 – z) (1 – \(x\)) y (1 – z) ⇒ (1 – z) [\(x\) (1 – y) – 2y (1 – \(x\))] = \(x\)y (1 – z) ⇒ \(x\) – \(x\)y – 2y + 2\(x\)y = \(x\)y ⇒ \(x\) = 2y ...(i) Also, (\(\beta\) – 3\(\gamma\)) p = 2\(\beta\)\(\gamma\) ⇒ y = 3z ...(ii) ⇒ y = 3z ...(ii) ∴ \(x\) = 6z (From (i) and (ii)) ⇒ \(\frac{x}{z}\) = 6 ⇒ \(\frac{P(E_1)}{P(E_3)}\) = 6. |
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| 314. |
A Jar contains 3 mangoes and x guavas. Two fruits are pulled from the jar without replacement. An expression that represents the probability one fruit is mango and the next fruit is guava isA) \((\frac{3}{x+3})(\frac{x-1}{x+2})\)B) \((\frac{3}{x+3})(\frac{x-1}{x-2})\)C) \(\frac{3\times2}{(x+3)(x+2)}\)D) \((\frac{3}{x+3})(\frac{x}{x-2})\) |
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Answer» Correct option is: D) \((\frac{3}{x+3})(\frac{x}{x-2})\) |
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| 315. |
A Class IX English book contains 200 pages. A page is selected at random. What is the probability that the number on the page is divisible by 10? |
| Answer» Correct Answer - `(1)/(10)` | |
| 316. |
A social studies text book contains 250 pages. A page is selected at random. What is the probability that the number on the page selected is a perfect square?A) 6/53B) 8/51C) 3/50D) 7/50 |
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Answer» Correct option is: C) \(\frac{3}{50}\) Total No of pages = 250 Perfect squares from 1 to 250 are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225. Hence, total No of perfect squares from 1 to 250 = 15 \(\therefore\) Probability that number on the selected page is a perfect square = \(\frac {Total \,No \,of\, perfect \,square \,from \,1 \,to \,250}{Total \,No \,of \,pages}\) = \(\frac {15}{250} = \frac {3}{50}\) Correct option is: C) \(\frac{3}{50}\) |
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| 317. |
A book contains 1000 pages. A page is chosen at random. Find the probability that the sum of the digits of the marked number on the page is equal to 9.A. `(23)/(500)`B. `(11)/(200)`C. `(7)/(100)`D. None of these |
| Answer» Correct Answer - (b) | |
| 318. |
A page is opened at random from a book containing 100 pages. Find the probability that the page number is a perfect square. |
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Answer» Number of pages in given book = 100 The page numbers that will be perfectly a square number (If randomly that are selected) are 1,4,9, 16, 25, 36, 49, 64, 81 and 100. ∴ Number of favourable outcomes = 100 Number of all possible outcomes = its number of pages = 100 ∴ Probability of getting a perfect number = 10/100 = 0.1 |
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| 319. |
Let `n_1, and n_2`, be the number of red and black balls, respectively, in box I. Let `n_3 and n_4`,be the number one red and b of red and black balls, respectively, in box II. A ball is drawn at random from box 1 and transferred to box II. If the probability of drawing a red ball from box I, after this transfer, is `1/3` then the correct option(s) with the possible values of `n_1 and n_2` , is(are)A. `n_(1)=4, n_(2)=6`B. `n_(1)=2, n_(2)=3`C. `n_(1)=10, n_(2)=20`D. `n_(1)=3, n_(2)=6` |
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Answer» Correct Answer - C::D Consider the following events: `E_(1)`=Transferring a red ball from box-I to box-II `E_(2)`=Transferring a black ball from box-I to box-II A=Drawing a red ball from box-I, after transferring a ball from box-I to box-II By, addition theorem of probability `P(A)=(n_(1))/(n_(1)+n_(2))xx(n_(1)-1)/(n_(1)+n_(2)-1)+(n_(2))/(n_(1)+n_(2))xx(n_(2))/(n_(1)+n_(2))xx(n_(1))/(n_(1)+n_(2)-1)` `implies (1)/(3)=(n_(1))/(n_(1)+n_(2))xx(n_(1)-1)/(n_(1)+n_(2)-1)+(n_(2))/(n_(1)+n_(2))xx(n_(1))/(n_(1)+n_(2)-1)` We observe that the values in options (c ) and (d) satisfy the above relation. Hence, options (c ) and (d) are correct. |
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| 320. |
An experiment has 10 equally likely outcomes. Let A and B be two non-empty events of the experiment. If A consists of 4 outcomes, the number of outcomes that B must have so that A and B are independent, is (A) 2, 4 or 8 (B) 3, 6 or 9A. 2,4 or 8B. 3,6, or 9C. 4 or 8D. 5 or 10. |
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Answer» Correct Answer - D We have, `P(A)=(4)/(10)=(2)/(5)` Suppose B has n outcomes. Then, `P(B)=(n)/(10), 0 lt n lt 10`. Since A and B are independent events. `therefore P(A cap B)=P(A)P(B)` `implies P(A cap B)=(2)/(5)xx(n)/(10)=(2n//5)/(10)` `implies (2n)/(5)` is an integer between 0 ans 10 `implies (n)/(5)` is an integer between 0 and 5 `implies n=5, 10`. |
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| 321. |
In Example 99, the probability that `X ge 3` equalsA. `(125)/(216)`B. `(25)/(36)`C. `(5)/(36)`D. `(25)/(216)` |
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Answer» Correct Answer - B Required probability `=P(X ge 3)` `=1-P(X lt 3)` `=1-P(X le 2)` `=1-((1)/(6)+(5)/(6)xx(1)/(6))=(25)/(36)` ALITER Required probability `=P(X ge 3)` =Probability of not getting six in first two trials and getting any number in subsequent trials `=(5)/(6)xx(5)/(6)xx1=(25)/(36)` |
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| 322. |
Let `Ea n dF`be tow independent events. The probability that exactly one of themoccurs is 11/25 and the probability if none of them occurring is 2/25. If `P(T)`deontes the probability of occurrence of the event `T ,`then`P(E)=4/5,P(F)=3/5``P(E)=1/5,P(F)=2/5``P(E)=2/5,P(F)=1/5``P(E)=3/5,P(F)=4/5`A. `P(E )=(4)/(5),P(F)=(3)/(5)`B. `P(E )=(1)/(5),P(F)=(2)/(5)`C. `P(E )=(2)/(5),P(F)=(1)/(5)`D. `P(E )=(6)/(5),P(F)=(1)/(5)` |
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Answer» Correct Answer - A We have, `P(E )+P(F)-2P(E cap F)=(11)/(25) " and" P(overline(E ) cap overline(F))=(2)/(25)` `implies P(E )+P(F)-2P(E )P(F)=(11)/(25) " and "P(overline(E ))P(overline(F))=(2)/(25)` `implies x+y-2xy=(11)/(25) " and "1-x+xy=(2)/(25)`, where P(E ) =x and P(F)=y `implies x+y+2-2x-2y=(11)/(25)+2xx(2)/(25)` [ On eliminating xy] `implies x+y=(7)/(5) implies y=(7)/(5) -x` Substituting `y=(7)/(5)-x " in " 1-x-y+xy=(2)/(25)`, we get `1-(7)/(5)+x((7)/(5)-x)=(2)/(25)` `implies 25x^(2)-35x+12=0 implies x=(3)/(5), (4)/(5)` When `x=(3)/(5), y=(4)/(5) " and " y=(3)/(5) " for " x=(4)/(5)` Hence, `P(E )=(3)/(5), P(F)=(4)/(5) " or " P(E )=(4)/(5), P(F)=(3)/(5)` |
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| 323. |
If C and D are two events such that `CsubD""a n d""P(D)!=0`, then the correct statementamong the following is :(1) `P(C|D)""=""P(C)`(2) `P(C|D)geqP(C)`(3) `P(C|D)"" |
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Answer» Correct Answer - D From multiplication Theorem on probability, we have `P(C//D)=(P(C cap D))/(P(D))=(P(C ))/(P(D)) " " [because C sub D therefore C cap D = C]` |
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| 324. |
A fair die is tossed repeatedly until a six is obtained Let X denote the number of tosses required The probability that `X = 3` equalsA. `(25)/(216)`B. `(25)/(36)`C. `(5)/(36)`D. `(125)/(216)` |
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Answer» Correct Answer - A `P(X=3)=(5)/(6)*(5)/(6)*(1)/(6)=(25)/(216)` |
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| 325. |
Given that the drawn ball from U2 is white, the probability that head appeared on the coinA. `(17)/(23)`B. `(11)/(23)`C. `(15)/(23)`D. `(12)/(23)` |
| Answer» Correct Answer - D | |
| 326. |
Given that the drawn ball from U2 is white, the probability that head appeared on the coinA. `17/23`B. `11/23`C. `15/23`D. `12/23` |
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Answer» Correct Answer - D `P((H)/(W))=(P(W//H)xxP(H))/(P(W//T).P(T)+(W//H).P(H))` `=(1/2((3)/(2)xx1+2/5xx1/2))/(23//30)=12/23` |
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| 327. |
A box `B_(1)` contains 1 white ball, 3 red balls, and 2 black balls. An- other box `B_(2)` contains 2 white balls, 3 red balls. A third box `B_(3)` contains 3 white balls, 4 red balls, and 5 black balls. If 1 ball is drawn from each of the boxes `B_(1),B_(2) and B_(3),` the probability that all 3 drawn balls are of the same color isA. `82//648`B. `90//648`C. `558//648`D. `566//648` |
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Answer» Correct Answer - A P (required) = P (all are white) + P (all are red) + P (all are black) `=1/6xx2/9xx3/12+3/6xx3/9xx4/12+2/6xx4/9xx2/12=6/648+36/648+40/648=82/648` |
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| 328. |
A fair die is tossed repeatedly until a 6 is obtained. Let X denote the number of tosses rerquired. The probability that `ge3` equalsA. `125//216`B. `25//36`C. `5//36`D. `25//216` |
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Answer» Correct Answer - B Given that `P(Xle2)=1/6+5/6xx1/6=11/36` Hence, the required probability is `1-(11//36)=25//36.` |
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| 329. |
The operates if all of its three components function. The probability that thefirst component fails during the year is 0.14, the second component fails is0.10 and the third component fails is 0.05. What is the probability that themachine will fail during the year? |
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Answer» Correct Answer - 0.2647 Let `E_(1), E_(2), E_(3)` be the respective events that the 1st, 2nd and 3rd components function. Then, `P(bar(E)_(1))=0.14, P(bar(E)_(2))=0.10` and `P(bar(E)_(3))=0.05` `implies P(E_(1))=(1-0.14)=0.86, P(E_(2))=(1-0.10)=0.90` and `P(E_(3))=(1-0.05)=0.95` `implies` P(machine fails) `=1 -P` (machine functions) `=1-P [(E_(1) and E_(2) and E_(3))]` `=1-[P(E_(1))xxP(E_(2))xxP(E_(3))]`. |
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| 330. |
A fair die is tossed repeatedly until a 6 is obtained. Let X denote the number of tosses rerquired. The probability that `X=3` equalsA. `25//216`B. `25//36`C. `5//36`D. `125//216` |
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Answer» Correct Answer - A `P(X=3)=((5)/(6))((5)/(6))1/6=25/216` |
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| 331. |
A town has two fire-extinguishing engines, functioning independently. The probability of availability of each engine when needed is 0.95. What is the probability that (i) neither of them is available when needed ? (ii) an engine is available when needed? |
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Answer» Correct Answer - (i) `1/400` (ii) `19/200` Let `E_(1)=` event of availability of the first engine. And, `E_(2) =` event of availability of the second engine. Then, `P(E_(1))=P(E_(2))=0.95` and `P(bar(E)_(1))=P(bar(E)_(2))=(1-0.95)=0.05`. (i) P( neither of them is available when needed) `=P (bar(E)_(1) and bar(E)_(2))=P(bar(E)_(1))xxP(bar(E)_(2))`. (ii) P( an engine is available when needed) `=P [(E_(1)" and not "E_(2)) or (E_(2)" and not "E_(1))]` `=P[(E_(1) and bar(E)_(2)) or (E_(2) and bar(E)_(1))]` `=P (E_(1) nn bar(E)_(2))+P(E_(2) nn bar(E)_(1))` `={P (E_(1))xxP(bar(E)_(2))}+{P(E_(2))xxP(bar(E)_(1))}`. |
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| 332. |
A box `B_(1)` contains 1 white ball, 3 red balls, and 2 black balls. An- other box `B_(2)` contains 2 white balls, 3 red balls. A third box `B_(3)` contains 3 white balls, 4 red balls, and 5 black balls. If 2 balls are drawn (without replecement) from a randomly selected box and one of the balls is white and the other ball is red the probability that these 2 balls are drawn from box `B_(2)` isA. `116//182`B. `126//181`C. `65//181`D. `55//181` |
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Answer» Correct Answer - D Let A: one ball is white and other is red `E_(1):` both balls are from box `B_(1),` `E_(2): ` both balls are from box `B_(2),` `E_(3):` both balls are form box `B_(3)` Hence, P (required) `=P((E_(2))/(A))` `=(P((A)/(E_(2))).P(E_(2)))/(P((A)/(E_(1))).P(E_(1))+P((A)/(E_(2))).P(E_(2))+P((A)/(E_(3))).P(E_(3)))` `=((""^(2)C_(1)xx""^(3)C_(1))/(""^(9)C_(2))xx1/3)/((""^(1)C_(1)xx""^(3)C_(1))/(""^(6)C_(2))xx1/3+(""^(2)C_(1)xx""^(3)C_(1))/(""^(9)C_(2))xx1/3+(""^(3)C_(1)xx""^(4)C_(1))/(""^(12)C_(2))xx1/3)` `(1/6)/(1/5+1/6+2/11)=55/181` |
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| 333. |
If the independent events A and B are such that `0 lt P(A) lt 1 " and " P(B)lt 1`, then which of the following alternatives is not correct ?A. A and B are mutually exclusiveB. A and `overline(B)` are independentC. `overline(A) " and " overline (B)` are independentD. `P(A//B)+P(overline(A)//B)=1`. |
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Answer» Correct Answer - A Since A and B are independent events. `therefore P(A cap B)=P(A)P(B) ne 0` So, A and B cannot be mutually exclusive events. Now, `P(A cap overline(B))=P(A)-P(A cap B)` `implies P(A cap overline(B))=P(A)-P(A)P(B)` `implies P(A cap overline(B))=P(A){1-P(B)}=P(A)P(overline(B))` and, `P(overline(A) cap overline(B))=1-P(A cup B)` `implies P(overline(A) cap overline(B))=1-{1-P(overline(A))P(overline(B))} [because` A and B are independent] `implies P(overline(A) cap overline(B))=P(overline(A))P(overline(B))` So, A and `overline(B)` as well as `overline(A) " and " overline (B)` are independent events. Finally, `P(A//B)+P(overline(A)//B)=(P(A cap B))/(P(B))+(P(overline(A) cap B))/(P(B))` `implies P(A//B)+P(overline(A)//B)=(P(A cap B)+P(overline(A) cap B))/(P(B))=(P(B))/(P(B))=1` Hence, alternative (a) is incorrect. |
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| 334. |
A ship is fitted with three engines `E_(1),E_(2),and E_(3)` the engines function independently of each othe with respectively probability `1//2,1//4, and 1//4.` For the ship to be operational at least two of its engines must function. Let X denote the event that the ship is operational and let `X_(1),X_(2), and X_(3)` denote, respectively, the events that the engines `E_(1),E_(2),and E_(3)` are functioning. Which of the following is (are) true?A. `P(X_(1)^(C)//X)=3/16`B. P (exactly two engines of the ship are functioning X `=7/8`C. `P(X|X_(1))=5/16`D. `P(X|X_(1))=7/16` |
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Answer» Correct Answer - B::D `P(X_(1))=1/2,P(X_(2))=1/4,P(X_(3))=1/4` `P(X)=P(X_(1)nnX_(2)nnX_(3))+P(X_(1)nnX_(2)^(C)nnX_(3))` `+P(X_(1)^(C)nnX_(2)nnX_(3))+P(X_(1)nnX_(2)nnX_(3)^(C))=1/4` (1) `P(X_(1)^(C)//X)=(P(XnnX_(1)^(C)))/(P(X))=(1//32)/(1//4)=1/8` (2) P [exactly two engines of the shp are functioning `|X]=(7//32)/(1//4)=7/8` (3) `P((X)/(X_(2)))=(P(XnnX_(2)))/(P(X_(2)))=(5//32)/(1//4)=5/8` (4) `P((X)/(X_(1)))=(P(XnnX_(1)))/(P(X_(1)))=(7//32)/(1//2)=7/16` |
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| 335. |
Let `n_(1)and n_(2)` be the number of red and black balls, respectively, in box I. Let `n_(3) and n_(4)` be the numbers of red and black balls, respectively, in the box II. One of the two boxes, box I and II, was selected at random and a ball was drawn randomly our of this box. The ball was found to be red. If the probability that this red with the possible values of `n_(1),n_(2),n_(3)and n_(4)` is (are)A. `n_(1)=3,n_(2)=3,n_(3)=5, n_(4)=15`B. `n_(1)=3,n_(2)=6,n_(3)=10, n_(4)=50`C. `n_(1)=8,n_(2)=6,n_(3)=5, n_(4)=20`D. `n_(1)=6,n_(2)=12,n_(3)=5, n_(4)=20` |
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Answer» Correct Answer - A::B `Red ton_(1)` ` "Box" IltBlack ton_(2)` `Red to n_(3)` `"Box II"lt"Black" to n_(4)` `P(R)=1/2.(n_(1))/(n_(1)+n_(2))+1/2.(n_(3))/(n_(3)+n_(4))` `P (II//R) =(1/2(n_(3))/(n_(3)+n_(4)))/(1/2(n_(1))/(n_(1)+n_(2))+1/2.(n_(3))/(n_(3)+n_(4)))` `((n_(3))/(n_(3)+n_(4)))/((n_(1))/(n_(1)+n_(2))+(n_(3))/(n_(3)+n_(4)))` by option `n_(1)=3,n_(2)=3,n_(3)=5,n_(4)=15` `P(II//R)=(5/20)/(3/6+5/20)=(1/4)/(1/2+1/4)=1/4xx(4)/(2+1)=1/3` Also, when `n_(1)=3, n_(2),=6,n_(3)=10, n_(4)=50,P(II//R)=1//3` |
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| 336. |
If A and B are two independent events such that `P(overline(A) cap B)=2//15 " and " P(A cap overline(B))=1//6`, then P(B), isA. `(1)/(5) " or ", (4)/(5)`B. `(1)/(6) " or", (5)/(6)`C. `(4)/(5) " or", (1)/(6)`D. `(5)/(6) " or", (1)/(5)` |
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Answer» Correct Answer - C Let P(A)=x and P(B)=y. Since A and B are independent events. Therefore, `P(overline(A) cap B)=2//15` `implies P(overline(A))P(B)=2//15` `implies {1-P(A)}P(B)=2//15` `implies (1-x)y=2//15` `implies y=xy=2//15` and, `P(A cap overline(B))=(1)/(6)` `implies P(A)P(overline(B))=(1)/(6)` `implies x(1-y)=(1)/(6) implies x-xy=(1)/(6)` Subtracting (i) from (ii), we get `x-y=(1)/(30) implies x=(1)/(30)+y` Putting this value of x in (i), we get `y-y ((1)/(30)+y)=(2)/(15)` `implies 30y-y-30y^(2)=4` `implies 30y^(2)-29y+4=0` `implies (6y-1)(5y-4)=0 implies y=1//6 " or ", y=4//5` `therefore P(B)=1//6 " or ", P(B)=4//5` |
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| 337. |
For the three events `"A","B","andC","P"("e x a c t l yo n eo ft h ee v e n t sAorBo c c u r s")="P"("e x a c t l yo n eo ft h ee v e n t sBorCo c c u r s")="P"("e x a c t l yo n eo ft h ee v e n t sCandAo c c u r s")="p"`and`"P"("a l lt h et h r e ee v e n t so c c u rs i m u l t a n e o u s l y")="p"^2,"w h e r e"0 |
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Answer» Correct Answer - A We know that, P(exactly one of A or B occurs) `P(A)+P(B)-2P(A nn B)` `:. P(A)+P(B)-2P(A nn B)=p " "`….(i) Similarly, `P(B)+P(B)-2P(B nn C)=p" "`….(ii) and `P(C )+P(A)-2(C nn A)=p" "`....(iii) On adding Eqs. (i),(ii) and (iii), we get `2[P(A)+P(B)+P(C )-P(A nn B)-P(B nnC)-P(C nn A)]=3p` `rArr P(A)+P(B)+P(C )-P(A nn B)-P(B nn C)-P(C nn A)=(3p)/(2)" "`....(v) It also given that, `P(A nn B nn C)=p^(2) " "`....(vi) `:. ` P(at least one of the events A,B and C occurs) `=P(A)+P(B)+P(C )-P(A nn B)-P(B nn C)-P(C nn A)+(Ann B nn C)` `=(3p)/(2)+p^(2) " "` [from Eqs. (iv) and (v)] `=(3p+2p^(2))/(2)` |
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| 338. |
Let `n_(1)and n_(2)` be the number of red and black balls, respectively, in box I. Let `n_(3) and n_(4)` be the numbers of red and black balls, respectively, in the box II. A ball is drawn at random from box I and transferred to box II. If the probability of drawing a red ball from box I, after this transfer, is 1/3, then the correct options (s) with the possible values of `n_(1) and n_(2)` is (are)A. `n_(1)=4andn_(2)=6`B. `n_(1)=2andn_(2)=3`C. `n_(1)=10andn_(2)=20`D. `n_(1)=3andn_(2)=6` |
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Answer» Correct Answer - C::D Given `(n_(1))/(n_(1)+n_(2)).(n_(1)-1)/(n_(1)+n_(2)-1)+(n_(2))/(n_(1)+n_(2)).(n_(1))/(n_(1)+n_(2)-1)=1/3` `implies3(n_(1)^(2)-n_(1)+n_(1)n_(2))=(n_(1)+n_(2))(n_(1)+n_(2)-1)` `implies3n_(1)(n_(1)+n_(2)-1)=(n_(1)+n_(2))(n_(1)+n_(2)-1)` `implies2n_(1)=n_(2)` |
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| 339. |
If `E` and `F` are events with `P(E)0` then which one is not correct?A. occurrence of `E rArr ` occurrence of FB. occurrence of `F rArr` occurrence of EC. non-occurrence of `E rArr` non-occurrence of FD. None of the above |
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Answer» Correct Answer - C It is given that, `P(E )leP(F)rArr E subeF " "`…..(i) and `P(e nn F)gt0 rArr E sub F " "`….(ii) (a) occurrence of `E rArr ` occurrence of F [from Eq. (i)] (b) occurrence of `F rArr` occurrence of E [from Eq. (ii)] (c ) non-occurrence of `E rArr ` occurrence of F Hence, option (c )is not correct. [from Eq. (i)] |
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| 340. |
Football teams `T_(1)and T_(2)` have to play two games are independent. The probabilities of `T_(1)` winning, drawing and lossing a game against `T_(2)` are `1/6,1/6and 1/3,` respectively. Each team gets 3 points for a win, 1 point for a draw and 0 point for a loss in a game. Let X and Y denote the total points scored by teams `T_(1) and T_(2)` respectively, after two games. `P(XgtY)` isA. `1/4`B. `5/12`C. `1/2`D. `7/12` |
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Answer» Correct Answer - B `P(XgtY)=T_(1)T_(1)+DT_(1)+T_(1)D("where"T_(1)" represents wins and D represents draw")` `=((1)/(2)xx(1)/(2))+((1)/(6)xx(1)/(2))+((1)/(2)xx(1)/(6))=5/12` |
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| 341. |
Let A and B be two events such that ,P(A cap B)=(1)/(4)` and `Poverset(-)(A)=(1)/(4)`, where `overset(-)A` stands for the complement of the event A. Then, the events A and B areA. independent but not equally likelyB. independent nad equally likelyC. mutually exclusive and independentD. equally likely but not independent |
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Answer» Correct Answer - A Given , `P(bar(A cup B)) =(1)/(6) , P(A cap B) =(1)/(4) ,P(bar(A)) =(1)/(4)` `:. P(A cup B )=1-P(bar(A cup B)) =1-(1)/(6)=(5)/(6)` ` and P(A) =1-P(bar(A)) =1-(1)/(4)=(3)/(4)` `:. P(A cup B ) =P(A) + P(B) -P( cap B ) ` `(5)/(6) =(3)/(4) +P(B)-(1)/(4)` `implies P(B)=(1)/(3) implies ` A and are not equally likely `P( A cap B ) =P(A) * P(B) =(1)/(4)` So , events are independent . |
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| 342. |
If `P(A)=(3)/(8), P(B)=(5)/(8) and P(AcupB)=(3)/(4)`, then `P((overline(A))/(overline(B)))` is equal toA. `(1)/(4)`B. `(1)/(9)`C. `(2)/(3)`D. `(3)/(4)` |
| Answer» Correct Answer - (c) | |
| 343. |
If A and B are independent events such that `P(A) gt0, P(B) gt 0`, thenA. A and B are mutually exclusiveB. A and `overline(B)` are dependentC. `overline(A)` and B are dependentD. `P(A//B)+P(overline(A)//B)=1`. |
| Answer» Correct Answer - D | |
| 344. |
Statement-1: if A and B are two events, such that `0 lt P(A),P(B) lt 1,` then `P((A)/(overline(B)))+P((overline(A))/(overline(B)))=(3)/(2)` Statement-2: If A and B are two events, such that `0 lt P(A), P(B) lt 1,` then `P(A//B)=(P(AcapB))/(P(B)) and P(overline(B))=P(A capoverline(B))+P(overline(A) cap overline(B))`A. Statement-1 is true, Statement-2 is true: Statement-2 is a correct explanation for Statement-1B. Statement-1 is true, Statement-2 is true: Statement-2 is not a correct explanation for Statement-1C. Statement-1 is true, Statement-2 is falseD. Statement-1 is false, Statement-2 is true |
| Answer» Correct Answer - d | |
| 345. |
Statement-1 If A and B be the event in a sample space, such that `P(A)=0.3 and P(B)=0.2,` then `P(Acapoverline(B))` cannot be found. Statement-2 `P(Acapoverline(B))=P(A)+P(overline(B))-P(Acapoverline(B))`A. Statement-1 is true, Statement-2 is true: Statement-2 is a correct explanation for Statement-1B. Statement-1 is true, Statement-2 is true: Statement-2 is not a correct explanation for Statement-1C. Statement-1 is true, Statement-2 is falseD. Statement-1 is false, Statement-2 is true |
| Answer» Correct Answer - a | |
| 346. |
If two events A and B are such that `P(overline(A))=0.3, P(B)=0.4 and P(Acapoverline(B))=0.5,` then `P((B)/(Acupoverline(B)))` is equal toA. `(1)/(4)`B. `(1)/(5)`C. `(2)/5)`D. `(3)/(5)` |
| Answer» Correct Answer - (a) | |
| 347. |
If A and B are two events such that `P(A)gt0 and P(B)!=1,` then `P((overline(A))/(B))` is equal toA. `1-P((A)/(B))`B. `1-P((A)/(overline(B)))`C. `P((overline(A))/(B))`D. |
| Answer» Correct Answer - (b) | |
| 348. |
probability distribution of a random variable X is If `a=P(Xge2) and b=P(Xlt3)` then A. `altb`B. `agtb`C. a=bD. None of these |
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Answer» Correct Answer - C `a=P(Xge2)=P(X=2)+P(X=3)+P(X=4)` `=4k+2k+k=7k` `b=P(Xlt3)=P(X=0)+(P(X=1)+P(X=2)` `=k+2k+4k=7k` `therefore" "a=7k=b` |
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| 349. |
The pdf of a curve X is `f(x)={{:(k/sqrtx","0ltxlt4),(0","xle0" or "xge4):}` Then , `P(Xge1)` is equal toA. 0.2B. 0.3C. 0.4D. 0.5 |
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Answer» Correct Answer - D `therefore" "oversetoounderset(-oo)intf(x)dx=1` `therefore" "overset4underset0intk/sqrtxdx=1` `rArr" "k[2sqrtx]_0^4=1` `rArr" "2k(sqrt4-sqrt0)=1` `rArr" "4k=1` `therefore" "k=1/4` Now, `P(Xge1)=P(1leXlt4)=overset4underset1intf(x)dx` `=overset4underset1intk/sqrtxdx=2k[sqrtx]_1^4` `=2(1/4)(2-1)=1/2=0.5` |
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| 350. |
4 five-rupee coins, 3 two-rupee coins and 2 one-rupee coins are stacked together in a column at random. The probability that the coins of the same denomination are consecutive isA. `13//9!`B. `1//210`C. `1//35`D. none of these |
| Answer» Correct Answer - B | |