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Concept:

P(A \(\cap\) B) = P(A) x P(B | A) = P(B) x P(A | B) where P(A | B) represents the conditional probability of A given B and P (A | B) represents the conditional probability of B given A.

Calculation:

Given: P(A student passing the exam) = 30% = 0.3, P(A student passing the exam and getting above 80% marks) = 10% = 0.1

The desired probability,

P(Student gets more than 80% marks | Student has passed the exam) = P(student passing the exam and gets more than 80% of marks) / P(Student has passed the exam)

⇒ P(Student gets more than 80% marks | Student has passed the exam) =  0.1/0.3

⇒ P(Student gets more than 80% marks | Student has passed the exam) =  1/3

Hence, option 2 is correct.

Given

Number of black balls = 5

Number of white balls = 6

Formula

Probability = Favorable events/Total possible events

Calculation

Favorable event = Number of black balls = 5

Total possible events = Total number of all balls = 6 + 5 = 11

∴ Probability = 5/11

Concept:

P(\(\bar A\) : Not hitting the target)  = 1 - P(A: hitting the target )

Calculations:

Person A can hit a target 4 times in 5 attempts. Person B - 3 times in four attempts. Person C - 2 times in 3 attempts.

P(A : hitting the target ) = \(\rm \dfrac 45\)

⇒P(\(\bar A\) :Not hitting the target) = \(\rm \dfrac 15\)

P(B: hitting the target) = \(\rm \dfrac 34\)

⇒P(\(\bar B\) :Not hitting the target) = \(\rm \dfrac 14\)

P(C: hitting the target) = \(\rm \dfrac 23\)

⇒P(\(\bar C\) : Not hitting the target) = \(\rm \dfrac 13\)

The probability that the target is hit atleast two times = \(P(A)P(B)P(\bar C)+P(A)P(\bar B)P(C) + P(\bar A)P(B)P(C) + P(A)P(B)P(C)\)

⇒ The probability that the target is hit atleast two times = \(\dfrac 15+ \dfrac 1 {10}+ \dfrac {2}{15}+ \dfrac{2}{5}\)

⇒ The probability that the target is hit atleast two times = \(\dfrac 5 6\)

Formula used:

P(A) = n(E)/n(S)

where, n(E) → Required events, n(S) → Sample space

Calculation:

Let S be the sample space 

S = 1, 2, 3, 4, 5, 6

n(S) = 6

Let E be the event of getting exact divisible by 2 

E = 2, 4, 6

n(E) = 3

P(A) = n(E)/n(S)

⇒ P(A) = 3/6

⇒ 1/2

∴ The probability of getting exact divisible by 2 is 1/2

Concept:

P(A \(\cap\) B) = P(A) x P(B | A) = P(B) x P(A | B) where P(A | B) represents the conditional probability of A given B and P (A | B) represents the conditional probability of B given A.

Calculation:

Let S be the sample space

∴ n(S) = 36

Let,  A = the event that the sum of the numbers on the two dice is 11.

∴ A = {(5,6), (6,5)}

∴ n(A) = 2

Let, B = the event of the occurrence of 5 on the first dice.

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

∴ n(B) = 6

Now, P(B) = n(B) / n(S) = 6/36 = 1/6.

⇒ A ∩ B = {(5,6)}

∴ n(A∩B) = 1

Now, P(A ∩ B) = n(A ∩ B) / n(S) = 1/36.

⇒ P(A | B) = P(A ∩ B) / P(B) = 1/6

Hence, option 4 is correct.

Concept:

• If P(A) = P(B), then the events are said to be equally likely.
• If A and B are independent events, then P(A ∩ B) = P(A) × P(B).
• If A and B are mutually exclusive events, then P(A ∩ B) = 0.
• For two events A and B, we have P(A ∪ B) = P(A) + P(B) - P(A ∩ B).
• P(E̅) = 1 - P(E), where E̅ is the complementary event of E.

Calculation:

Using the definition of complementary events:

P(A ∪ B) = \(\rm 1-P\left( \overline{A\cup B} \right)=1-\frac{1}{6}=\frac{5}{6}\)

P(A) = 1 - P(A̅) = \(\rm 1-\frac{1}{4}=\frac{3}{4}\)

Let us find out P(B) and examine the values of P(A ∩ B) and P(A) × P(B) to determine whether the events are equally likely and mutually exclusive or independent.

Using P(A ∪ B) = P(A) + P(B) - P(A ∩ B), we get:

\(\rm \frac{5}{6}=\frac{3}{4}+P\left( B \right)-\frac{1}{4}\)

⇒ P(B) = \(\rm \frac{5}{6}-\frac{3}{4}+\frac{1}{4}=\frac{1}{3}\)

Since, P(A) ≠ P(B), the events are not equally likely.

Also,P(A) × P(B) = \(\rm \frac{3}{4}× \frac{1}{3}=\frac{1}{4}\) = P(A ∩ B), so the events are independent.

The correct answer option is D. Independent but not equally likely.

GIVEN:

⇒ Red balls = 10

⇒ Green balls = 5

⇒Yellow balls = ‘x’

⇒ Total balls in the bag = 15 + x

ASSUMPTION:

Let P be the probability of drawing 2 yellow balls from the bag.

CALCULATION:

Case 1: when drawing 2 red balls from the bag.

⇒ ¹⁰C₂/ ^{(15 + x)}C₂ = P + 13/57      ----(1)

Case 2: when drawing 2 yellow balls from the bag.

⇒  ^xC₂/ ^{(15 + x)}C₂ = P      ----(2)

Solving eq(1) and eq(2)

⇒ 10C2/ (15 + x)C2 = xC2/ (15 + x)C2 + 13/57  

⇒ (10C2 - xC2)/ (15 + x)C2 = 13/57  

⇒ ([90 - (x)(x-1)]/2) / (15 + x)(14 + x)/2 = 13/57  

\(⇒\frac{(90 - {(x)(x-1))} \over 2}{\frac{(15 + x)(14 + x)}{2}} = \frac {13} {57} \)

\(⇒\frac{(90 - {(x)(x-1))} }{(15 + x)(14 + x)} = \frac {13} {57} \)

\(⇒{(90 - {(x)(x-1))} } \times 57= {13} \times {(15 + x)(14 + x)}\)

\(⇒{(90 - {(x)(x-1))} } \times 57= {13} \times {(15 + x)(14 + x)}\)

since options for yellow balls are 0, 10, 4, 15 by subtracting 15 from each option.

So, we get

⇒ x = 4

⇒ Total balls = 15 + x = 19 balls.

Given:

1/5^th of cistern emptied = 2 hours

Calculation:

1/5^th of cistern is emptied in 2 hours

⇒ Time taken to empty the whole cistern = 2 × (5/1)

⇒ Time taken to empty the whole cistern = 10 hours

Now, Within 6 hours cistern was emptied

⇒ 10 hours to empty whole cistern = T1

⇒ 6 hours to empty particular portion of cistern = T2

⇒ Proportion of cistern = T2/T1

⇒ Proportion of cistern = 6/10

⇒ Proportion of cistern = 3/5

⇒ Percentage of cistern filled = 3/5 × 100

Concept:

The probability of the occurrence of an event A out of a total possible outcomes N, is given by: \(P(A) = \rm \dfrac{n(A)}{N}\), where n(A) is the number of ways in which the event A can occur.

Calculation:

The total number of distinct possible outcomes (N), when rolling two dice, are: N = 6 × 6 = 36.

The sum of these pairs of outcomes can be a number from 1 + 1 = 2 to 6 + 6 = 12.

The prime numbers from 2 to 12 are (2, 3, 5, 7, 11). The different possibilities to give each of these sums are given below:

Sum = 2:

(1, 1) = 1 possibility.

Sum = 3:

(1, 2), (2, 1) = 2 possibilities.

Sum = 5:

(1, 4), (4, 1), (2, 3), (3, 2) = 4 possibilities.

Sum = 7:

(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3) = 6 possibilities.

Sum = 11:

(5, 6), (6, 5) = 2 possibilities.

Total Number of possibilities for the desired event to occur is: n(A) = 1 + 2 + 4 + 6 + 2 = 15.

The required probability is therefore: \(P = \rm \dfrac{n(A)}{N}=\dfrac{15}{36}=\dfrac{5}{12}\).

The following table gives the number of ways in which a given sum can be obtained when rolling a pair of dice:

The number 7 has the maximum number of possibilities.

Concept:

For a random variable X = xi with probabilities P(X = x) = pi:

• Mean/Expected Value: μ = ∑pixi.
• Variance: Var(X) = σ² = ∑pi(xi)2 - (∑pixi)2 = ∑pi(xi)2 - μ2.
• Standard Deviation: σ = \(\rm \sqrt{Var(X)}\).

Calculation:

We have x1 = 1, x2 = 2, x3 = 3 and p1 = 0.3, p2 = 0.4, p3 = 0.3.

Now, ∑pixi = (0.3 × 1) + (0.4 × 2) +(0.3 × 3)

= 0.3 + 0.8 + 0.9

= 2

And, ∑pi(xi)2 = (0.3 × 1²) + (0.4 × 2²) +(0.3 × 3²)

= 0.3 + 1.6 + 2.7

= 4.6

∴ Var(X) = σ2 = ∑pi(xi)2 - (∑pixi)2 = 4.6 - 22 = 4.6 - 4 = 0.6.

The variance of the distribution is Var(X) = 0.6.

Given 

No.of days in leap year = 366 days 

Formula Used 

Probability = (Actual/Total)

Calculation 

No. of weeks and day in leap year = 366/7 = 52 weeks and 2 days 

Every leap year has 52 sundays 

The probability of having 52 Sundays in a leap year is thus the remaining two days can be any of this formation

Sunday-Monday, Monday-Tuesday, Tuesday-Wednesday, Wednesday-Thursday, Thursday-Friday, Friday-Saturday, Saturday-Sunday.

However, to get 52 Sundays in a leap year, none of the remaining two days must be a Sunday

then out of the 7 combinations above, that can be only realized 5 out of 7 times.

The connection "Sunday-Monday and Saturday-Sunday" most be scraped off.

so, probability having 52 sunday = 5/7 

∴ The required answer is 5/7 

Given:

The bag has red marbles = 5

The bag has green marbles = 4

The bag has blue marbles = 3

Concept used:

The probability formula is used to compute the probability of an event to occur. To recall, the likelihood of an event happening is called probability.

Formulae required:

Probability P(A) = the number of favourable outcomes/total number outcomes

Calculations:

The total number of marbles = 5 + 4 + 3

⇒ 12

Non- green marble = not green 

The non-green marbles in a bag = total marbles - green marbles

⇒ 12 - 4 = 8

Probability of non-green marble = 8/12

⇒ 2/3

∴ The probability that it is a non-green marble is 2/3

(i) an orange flavoured candy?

Solution: Zero

(ii) a lemon flavoured candy?

Solution: 1

Concept:

If S is a sample space and A is a favourable event then the probability of A is given by:

\(\rm P(E)=\dfrac{n(A)}{n(S)}\)

Calculation:

Total balls in bag = 4 + 5 = 9 balls

Given: Two balls are drawn at random without replacement

After the first ball is drawn and found to be blue, there are now 8 balls left in the bag, 4 of which are blue.

Probability the second ball is blue = \(\frac 4 8 = \frac 1 2\)

\(p = 50\% = \frac{50}{100} = \frac 12\)

\(q = 1 - p = 1 - \frac 12 = \frac 12\)

\(n = 18\)

(i) \(P(x = 10) = \,^{18} C_{10} \,p^{10} q^8\) 

\(=\, ^{18}C_{10} \left(\frac 12\right)^{10}\left(\frac 12\right)^{8}\)

\(=\left(\frac 12\right)^{18} \,^{18}C_{10}\)

(ii) \(P(x \ge 2) = 1 - P(x = 0) - P(x = 1)\) 

\(= 1 -\, ^{18}C_0\,p^0 q^{18} -\, ^{18}C_1 \,p^1q^{17}\)

\(= 1 - \left(\frac 12\right)^0 \left(\frac 12\right)^{18} - 18\left(\frac 12\right) \left(\frac 12\right)^{17}\)

\(= 1 -\left(\frac 12\right)^{18} - 18\left(\frac 12\right)^{18}\)

\(= 1- 19\left(\frac12\right)^{18}\)

(iii) \(P(X \le 17) = 1 - P(X = 18)\)

\(= 1-\,^{18}C_{18} \,p^{18}q^0\)

\(= 1- \,^{18}C_{18}\left(\frac 12\right)^{18} \left(\frac 12\right)^0\)

\(=1-\left(\frac 12\right)^{18}\)

Concept:

Let A1, A2, …. , An be n mutually exclusive and exhaustive events of the sample space S and A is event which can occur with any of the events then

• \({\rm{P}}\left( {\frac{{{{\rm{A}}_{\rm{i}}}}}{{\rm{A}}}} \right) = \frac{{{\rm{P}}\left( {{{\rm{A}}_{\rm{i}}}} \right){\rm{P}}\left( {\frac{{\rm{A}}}{{{{\rm{A}}_{\rm{i}}}}}} \right)}}{{\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{n}} {\rm{P}}\left( {{{\rm{A}}_{\rm{i}}}} \right){\rm{P}}\left( {\frac{{\rm{A}}}{{{{\rm{A}}_{\rm{i}}}}}} \right)}}\)
•  

Calculations:

Consider, Let K be the event that both A and B  agree, 

T be the event that they both A and B speak the truth 

⇒ P(T) = xy

L be the event that they both A and B lie.
⇒ P(L) = (1 -x)(1 - y) 

To find :The probability that the statement is true = \(\rm P(\frac T L)\)

Let K be the event that both of them agree

 \(\rm P(\frac T L)\) = \(\rm \dfrac {P(T)P(\frac K T)}{P(T)P(\frac {K}{T})+P(L)P(\frac {K}{L}) }\)

⇒\(\rm P(\frac T L)\) = \(\rm \dfrac{xy}{xy+(1-x)(1-y)}\)

Solution: Total number of outcomes = 5 + 8 + 4 = 17

Number of red marbles = 5

Number of white marbles = 8

Number of green marbles = 4

(i) Red?

Solution: Probability of red marbles;

P(R) = 5/17

(ii) White?

Solution: Probability of white marbles;

P(W) = 8/17

(iii) Not green?

Solution: Probability of green marbles;

P(G) = 4/17

Or, P( not G)

= 1 - 4/17

=13/17

When a coin is tossed, the possible outcomes are head (H) and tail (T).
When a die is thrown, the possible outcomes are 1, 2, 3, 4, 5, or 6.
Thus, the sample space of this experiment is given by:
S = {T, H1, H3, H5, H21, H22, H23, H24, H25, H26, H41, H42, H43, H44, H45, H46, H61, H62, H63, H64, H65, H66}

Total no. of outcomes = 4  
HH, HT, TH, TT (where H is head and T is tail)

No. of favourable outcomes = 1
HH (coming up two heads together)

Probability = No. of favourable outcomes / Total no. of outcomes = 1/4
 

Answer = 1/3

Let S be the event of tossing two coins

Sample space S = {HH,TT,HT,TH}

Let A be the event where at least one head shows up

A = {HH,TH,HT}

Let  B be the event where two heads show up

B = {HH}

Now,

Probability of event B given that event A happens = P(B/A) = n(B)/n(A) = 1/3

The probability of getting 5 when a dice is rolled is given by: 
No. of outcomes / Total no. of outcomes

here, No. of outcomes; 1 (i.e. 5 appears on a single face of dice and will appear only once)

Total no. of outcomes = 6 (i.e. 1,2,3,4,5,6)

Probability = 1/6

Given:

Two different colored unbiased dice have been thrown.

Formula used:

Probability = Favorable outcome / Total outcomes

Calculations:

Total number of outcomes = 6² = 36

Favorable outcomes:

Total favorable outcome = 5

Required probability = 5/36

Concept:

we know that,

\(\rm P(A) +P(B) = P(A \cup B) + P(A \cap B)\)

P(A') = 1 - P(A) 

Calculations:

Given, For any two events A and B, the probability that at least one of them occur is 0.6

⇒ \(\rm P(A \cup B) = 0.6\)

and A and B occur simultaneously with a probability 0.3

⇒ \(\rm P(A \cap B) = 0.3\)

we know that,

\(\rm P(A) +P(B) = P(A \cup B) + P(A \cap B)\)

⇒ \(\rm P(A) +P(B) = 0.6 + 0.3 = 0.9\)

Also, we know that 

P(A') + P(B') = 1 - P(A) + 1 - P(B) 

⇒ P(A') + P(B') = 2 - 0.9

⇒ P(A') + P(B') = 1.1

Hence, For any two events A and B, the probability that at least one of them occur is 0.6. If A and B occur simultaneously with a probability 0.3, then P(A') + P(B') is 1.1

(i) When the order of the birth of a girl or a boy is considered, the sample space is given by S = {GG, GB, BG, BB}
(ii) Since the maximum number of children in each family is 2, a family can either have 2 girls or 1 girl or no girl. Hence, the required sample space is S = {0, 1, 2}

A die has six faces that are numbered from 1 to 6, with one number on each face. Let us denote the red, white, and blue dices as R, W, and B respectively.
Accordingly, when a die is selected and then rolled, the sample space is given by
S = {R1, R2, R3, R4, R5, R6, W1, W2, W3, W4, W5, W6, B1, B2, B3, B4, B5, B6}

Quantity I:

Total number of cards = 50

Number of favorable cards = 4 (i.e. 12, 24,36 and 48)

⇒ P(E) = 4/50 = 2/25 = 0.08

Quantity II:

Total number of cards: 50

Number of favorable cards = 14 (i.e. 3, 5, 7, 11, 13,17,19,23, 29, 31, 37, 41, 43, 47)

⇒ P(E) = 14/50 = 7/25 = 0.28

∴ Quantity I < Quantity II

Quantity I:

Total number of outcomes = ¹⁶C₄

⇒ 16!/(4! × 12!)

⇒ (16 × 15 × 14 × 13 × 12!) / (4 × 3 × 2 × 1 × 12!)

⇒ 1820

Number of favourable outcomes = ⁵C₂ × ³C₂

⇒ 30

⇒ Required probability = 30/1820

⇒ 0.0165 (approx)

Quantity II:

Total number of possible ways to drawn a card = ⁵²C₅

⇒ 52!/(5! × 47!)

⇒ (52 × 51 × 50 × 49 × 48 × 47!)/ (5 × 4 × 3 × 2 × 1 × 47!)

⇒ 2598960

Number of sets of 5 with all are jack = ⁴C₄ × ⁴⁸C₁

⇒ 1 × (48 × 47!) / (1! × 47!) = 48

⇒ P(E ) = 48/2598960

⇒ 1/54145

⇒ 0.000018(approx)

∴ Quantity I > Quantity II

The answer is 175/396

Given:

The probability of shooter Ajay hitting the target is 70%

Probability of shooter Bharat Missing the target is 80%

Formula Used:

Probability = number of favourable outcomes /total number of outcome 

Calculation:

The probability of shooter Ajay hitting the target is 70% = 7/10

⇒ The probability of shooter Ajay missing the target is 30% = 3/10

Probability of shooter Barath Missing the target is 80% = 8/10

⇒ The probability of shooter Barath Hitting the target is 20% = 2/10

 Probability of hitting the target will be = 7/10 × 2/10 = 14%

∴The Probability that both Ajay and Bharat hitting the target will be 14%

p= (5/2)x(20-5/2)/(20/4)

It is assumed that all possibilities of selecting 4 out of 20 elks are of equal probability. 

Concept:

Binomial Distribution:

• A Binomial Distribution is an experiment in which only ONE OUT OF TWO outcomes is possible. For example: Head or Tail, Yes or No, 1 or 0, etc.
• If ‘n’ and ‘p’ are the parameters, then ‘n’ denotes the total number of times the experiment is conducted and ‘p’ denotes the probability of the happening of the event.
• The probability of getting exactly ‘k’ successes in ‘n’ independent trials for a Random Variable X is expressed as P(X = k) and is given by the formula:

  P(X = k) = nCk pk (1 - p)n - k

Calculation:

For every 3 trials, the experiment succeeds 2 times and fails 1 time.

∴ Probability of success: p = \(\dfrac23\).

For at least 4 success in 6 trials, n = 6 and k = 4, 5, 6. The required probability is therefore:

\(\rm ^6C_4\left(\dfrac23\right)^4\left(1-\dfrac23\right)^{6-4}+^6C_5\left(\dfrac23\right)^5\left(1-\dfrac23\right)^{6-5}+^6C_6\left(\dfrac23\right)^6\left(1-\dfrac23\right)^{6-6}\)

= \(\rm (15)\left(\dfrac23\right)^4\left(\dfrac13\right)^2+(6)\left(\dfrac23\right)^5\left(\dfrac13\right)^1+(1)\left(\dfrac23\right)^6\left(\dfrac13\right)^0\)

= \(\rm \dfrac{240}{729}+\dfrac{192}{729}+\dfrac{64}{729}\)

= \(\rm \dfrac{496}{729}\).

• ⁿC_r = \(\rm \dfrac {n!}{r!(n-r)!}\).
• n! = 1 × 2 × 3 × ... × n.
• 0! = 1.

Concept: 

If there can be only 2 cases of true and false, then:

Probability of r true cases out of n (≥ r) P(x = r) = ⁿC_r p^rq^(n-r) 

where p is the probability of case being true and q is probability of case being false

Note: p and q are ≤ 1

Calculation:

The probability of correct answer p = \(\rm 1\over 4\) = 0.25

The probability of wrong answer q = \(\rm 3\over 4\) = 0.75

Total questions n = 5

Probability of atleast one correct answer (x ≥ 1):

P(x ≥ 1) = 1 - P(x = 0)

⇒ P(x ≥ 1) = 1 - ⁵C₀ p⁰q⁵

⇒ P(x ≥ 1) = 1 - (0.75)⁵ 

⇒ P(x ≥ 1) = 1 - 0.2373

⇒ P(x ≥ 1) \(\boldsymbol{\rm \approx}\) 0.7627

Concept:

Probability of event hapening = 1 - probability of event not happening

Calculation:

Here, probability of medicine to  cure patient = 1/2

And, probability of none cured = 1 - 1/2 = 1/2

The probability that at least one patient is cured by this medicine = 1 - none patient cured by this medicine 

= 1 - (1/2) ×(1/2) ×(1/2) ×(1/2) 

= 1 - (1/2)⁴

= 15/16

Hence, option (4) is correct. 

Concept:

we know that

 \(\rm ^nC_r\) = \(\rm \dfrac{n!}{r!(n-r)!}\)  

Calculations:

we are picking the four balls without replacement and that the 4 white and 4 red balls are the only balls in the box and that we choose 2 red balls and 2 white balls.

Number of ways to choose 4 balls out of 8 = \(\rm ^8C_4\) = \(\rm \dfrac{8!}{4!4!}\) = 70

Number of ways to pick 2 red balls out of 4 = \(\rm ^4C_2\) = \(\rm \dfrac{4!}{2!2!}\) = 6

Number of ways to pick 2 white balls out of 4 =  \(\rm ^4C_2\) = \(\rm \dfrac{4!}{2!2!}\) = 6

Probability = \(\frac {6 \times 6} {70} = \frac {36} {70} = \frac {18} {35} \)

Concept:

Let U be the universal set and A, B, C be the subsets of U.

If \(\rm A∩ B ∩ C = \phi\) then  A, B, C are mutually exclusive.

Çalculations:

Given, U = {1, 2, 3, ...., 20}.

Let A, B, C be the subsets of U.

A be the set of all numbers which are perfect squares

⇒ A = {1, 4, 9 16}

B be the set of all numbers which are multiples of 5

⇒ B = {5, 10, 15, 20}

and C be the set of all numbers, which are divisible by 2 and 3

⇒ C = {6, 12, 18}

Now, \(\rm A∩ B ∩ C = \phi\)

So, Å, B, C are mutually exclusive.

Hence statement 1 is correct

Here  Å, B, C are mutually exclusive so A, B, C can't be mutually exhaustive

Hence statement 2 is wrong

A ∪ B = {1, 4, 5, 9, 10, 15, 16, 20}

n(A ∪ B) = 8

U = {1, 2, 3, ...., 20}

n(U) = 20

Now, The number of elements in the complement set of A ∪ B  = n(U) - n(A ∪ B) = 20 - 8 = 12

Hence statement 3 is correct

Probability of getting head in tossing a coin is p =1/2 and probability of not getting a head is also q=1/2 in one trial. 

The probablprob distribution is binomial. So probability of getting r heads in n trial is 

P(_{x = r)} = ⁿC_r×p^r×q(^(n-r)

So probability of getting 2 heads in 10 trial is

P _(x=2) = ¹⁰C₂×(1/2)^2(1/2)^10

 =45/2^10 =45/1024

Given:

6 blue balls, X red balls and 10 green balls are there in box

Probability of choosing one red ball from the given box = 1/3

Calculation:

Probability of getting red ball = Number of red balls/Total number of balls

Calculation:

Total balls in the box= 6 + X + 10 = X + 16.

Probability of getting red ball = X/(X + 16)

According to question:

X/(X + 16) = 1/3

⇒ X = 8

∴ The sum of red and blue balls in the box = 8 + 6 = 14

Given

Total numbers = 15

Formula used

P(E) = Number of favourable outcomes/Total number of outcomes

Calculation

Let E is an event of winning a game

Total outcomes = 15 × 15

⇒ 225

Number of favorable outcomes (winning a game) = 15

P(E) = 15/225

⇒ 1/15

Probability of not winning a game = 1 – (1/15)

⇒ 14/15

Given

Total numbers = 40

Formula used

P(E) = Number of favourable outcomes/Total number of outcomes

Calculation

Let E is an event of drawing a card multiple of 3 or 4

Number of favourable outcomes = 20 i.e(102, 104, 105, 108, 111, 112, 114, 116, 117, 120, 123, 124, 126, 128, 129, 132, 135, 136 ,138 and 140)

P(E) = 20/40 = 1/2

Given

Number of black card = 4

Number of white card = 3

Number of pink card = 2

Formula used

P(E) = Number of favourable outcomes/Total number of outcomes

Calculation

Let E is an event of drawing a card which is not black

Number of favourable cards = 3 + 2

⇒ 5

P(E) = 5/9

Given:

Total number of professors = 10

Total number of students = 20

Committee consists of 2 professors and 2 students

Number of ways to form a committee

Formula used:

ⁿC_r = n!/ [r! (n – r)!]

Calculation:

Select 2 committee members from professors and 2 committee members from students.

⇒ ¹⁰C₂ × ²⁰C₂ = [(10!)/(2! × 8!)] [(20!)/ ( 2! × 18!)]

⇒ [(10 × 9 × 8!)/ (2! × 8!)] [(20 × 19 × 18!)/ 2! × 18!]

⇒ 45 × 190

⇒ 8550 ways

∴ Total number of ways to form a committee is 8550

Correct option (b) and (c) [ P(A ∩ B) = 0]

Explanation:

Examples of Mutually Exclusive Events: 

1. In a soccer game scoring no goal (event A) and scoring exactly one goal (event B). 

2. In a deck of cards Kings (event A) and Queens (event B). 

Examples Not Mutually Exclusive Events: 

In a deck of cards Kings (event A) and Hearts (event B). They have common cards i.e. Heart King. So (A ∩ B) ≠ 0. 

Events whose occurrence do not depend on the occurrence of any other events are called independent events.

Correct option d (31/32)

Explanation:

Consider solving this using complement. 

Probability of getting no head = P(all tails) = 1/32 

P(at least one head) = 1 – P(all tails) = 1 – 1/32 = 31/32.

Given:

2 cards have been drawn from a full deck of 52 cards.

Formula used:

Probability = number of possibility in the required scenario / all possible scenarios

Calculations:

Number of face cards in a full deck of cards = (3 × 4) = 12

[each series of cards contain 3 face cards i.e. king, queen and jack]

Total number of cards = 52

So, required probability = \({}_2^{12}C\;/\) \({}_2^{52}C\) = (12 × 11)/ (52 × 51) = 11/221

Given:

Black pen = 9

Red pen = 5

Blue pen = 3

Formula used:

ⁿC_r = n!/((n-r)! × r!)

Calculation:

Let S be the sample space then,

n(S) = number of way of selecting 2 pens out of 17 pens

⇒ n(S) = ¹⁷C₂ = 17!/((17– 2)! × 2!)

⇒ (17 × 16)/2

⇒ 136

Let E be the event of getting both are red or blue pen

n(E) = ⁵C₂ + ³C₂

⇒ (5 × 4)/2 + (3 × 2)/2

⇒ 10 + 3

⇒ 13

P(E) = n(E)/n(S)

= 13/136

∴ The probability of getting both are red pen or blue pen is 13/136.

Given:

Jayant throws 3 dices at same time

Formula used:-

P(E) = Number of favorable outcomes/Total outcomes

Calculation:

Here, total number of possible outcomes = 6 × 6 × 6

⇒ No. of event of getting sum of 6 is (1, 1, 4), (1, 4, 1), (4, 1, 1), (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1) and (2, 2, 2).

⇒ Probability of getting a total of 6 = 10/216

⇒ P(E) = 5/108

∴ probability of getting a total of 6 is 5/108

Given:

Rajeev = threw 3 dices at same time

Formula used:-

P(E) = Number of favorable outcomes/Total outcomes

Calculation:

Here, total number of possible outcomes = 6 × 6 × 6

⇒ No. of event of getting sum of at most of 6 is (1, 1, 1), (1, 1, 2), (1, 2, 1), (2, 1, 1), (1, 1, 3), (1, 3, 1), (3, 1, 1), (2, 2, 1), (1, 2, 2), (2, 1, 2), (1, 1, 4), (1, 4, 1), (4, 1, 1), (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1) and (2, 2, 2)

⇒ probability of getting total of at most 6 = 20/216

 ∴ probability of getting a total of at most 6 is 5/54

Given:

2 cards are picked randomly from 26 cards

Formula Used:

The number of ways of picking "r" cards from "n" total cards = ⁿC_r

Calculation:

Total numberd card in a pack of deck (2 - 10) = 9 × 2 = 18

Perfect squares that can be obtained by sum of numbers in a deck = 4, 9, 16

Sum of numbers on cards possible to get 4 = {(1,3) (2,2) (3,1)} = 3

Sum of numbers on cards possible to get 9 = {(2,7) (3,6) (4,5) (5,4) (6,3) (7,2)} = 6

Sum of numbers on cards possible to get 16 = {(6,10) (7,9) (9,7) (10,6) (8,8)} = 5

Number of favorable outcomes = 3 + 6 + 5 = 14

Total number of favorable outcomes = ²⁶C₂ = (26 × 25)/(2 × 1) = 325

∴ P(A) = 14/325 

∴ The probability that the sum of numbers on card is a perfect square number is 14/325

Concept:

Required  numbers of possible outcomes = Total numbers of possible outcomes – Number of  possible outcomes in  which 4 does not appear on any dice

Calculations:

Required  numbers of possible outcomes 

 A dice has 6 numbers and since we don't want it to show 4, we are left with five options 1,2,3,5,6.

Since any of the numbers can appear on any of the dice,

Number of ways in which 4 doesn't appear on any of the dice= 5 × 5 × 5

⇒Hence, the required number of possible outcomes = \(\rm216 - 125 =91\)

Given:

Two dice are thrown simultaneously

Formula Used:

P(E) = n(E)/n(S)

Where,

n(E) = Number of favourable events

n(S) = Number of possible outcomes

Calculation:

When two dice throw simultaneously, then total possible outcomes

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

(2, 1), (2, 2),..............................., (2, 6)

............................................................

............................................................

(6, 1), (6, 2), ................................(6, 6)

Number of possible outcomes n(S) = 6 × 6

⇒ 36

Number of favourable events n(E) = {(4,6), (6,4), (5,5)} = 3

⇒ n(E) = 3

Required probability of getting a total of 10 

P(E) = 3/36

⇒ 1/12

∴ The probability of getting a total of 10 is 1/12.

Given:

Two dice are thrown simultaneously

Concept used:

A dice has numbers from 1 to 6 i.e. {1, 2, 3, 4, 5, 6}.

Formula used:

Probability = (Total number of favourable outcome)/(Total number of outcome)

Calculation:

When two dice are thrown.

Then, The number of total possible outcome = 6 × 6 = 36

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

(2, 1), (2, 2),..............................., (2, 6)

............................................................

............................................................

(6, 1), (6, 2), ................................(6, 6)

To get the two numbers whose product is odd, both should be odd numbers.

So favourable outcome are:

(1,1), (1,3), (1,5), (3,1), (3,3), (3,5), (5,1), (5,3), (5,5)

Total number of favourable outcome = 9

Probability = (Total number of favourable outcome)/(Total number of outcome)

⇒ Probability = 9/36 = 1/4

∴ The probability of getting two numbers whose product is odd is 1/4.