Probability — Page 11 | JEE Mathematics

Q101

An urn contains 6 white and 9 black balls. Two successive draws of 4 balls are made without replacement. The probability, that the first draw gives all white balls and the second draw gives all black balls, is :

A

\dfrac{3}{256}

B

\dfrac{5}{256}

C

\dfrac{3}{715}

D

\dfrac{5}{715}

Correct Answer

Option C

Solution

\frac{{ }^6 \mathrm{C}_4}{{ }^{15} \mathrm{C}_4} \times \frac{{ }^9 \mathrm{C}_4}{{ }^{11} \mathrm{C}_4}=\frac{3}{715}

Hence option (3) is correct.

Q102

Let

\mathrm{A}

and

\mathrm{B}

be two events such that

P(B \mid A)=\dfrac{2}{5}, P(A \mid B)=\dfrac{1}{7}

and

P(A \cap B)=\dfrac{1}{9} \cdot

Consider (S1)

P\left(A^{\prime} \cup B\right)=\dfrac{5}{6}

, (S2)

P\left(A^{\prime} \cap B^{\prime}\right)=\dfrac{1}{18}

Then :

A Both (S1) and (S2) are true

B Both (S1) and (S2) are false

C Only (S1) is true

D Only (S2) is true

Correct Answer

Option A

Solution

P(A/B) = {1 \over 7} \Rightarrow {{P(A \cap B)} \over {P(B)}} = {1 \over 7}

\Rightarrow P(B) = {7 \over 9}

P(B/A) = {2 \over 5} \Rightarrow {{P(A \cap B)} \over {P(A)}} = {2 \over 5}

P(A) = {5 \over 2}\,.\,{1 \over 9} = {5 \over {18}}

S2:P(A' \cap B') = {1 \over {18}}

S1:

and

P(A' \cup B) = {1 \over 9} + {6 \over 9} + {1 \over {18}} = {5 \over 6}.

Q103

Let

S=\{1,2,3, \ldots, 2022\}

. Then the probability, that a randomly chosen number n from the set S such that

\mathrm{HCF}\,(\mathrm{n}, 2022)=1

, is :

A

\dfrac{128}{1011}

B

\dfrac{166}{1011}

C

\dfrac{127}{337}

D

\dfrac{112}{337}

Correct Answer

Option D

Solution

S = {1, 2, 3, ..........

2022} HCF (n, 2022) = 1 $\Rightarrow$ n and 2022 have no common factor Total elements = 2022 2022 = 2 $\times$ 3 $\times$ 337 M : numbers divisible by 2.

{2, 4, 6, ........, 2022}

\,\,\,\,

n(M) = 1011 N : numbers divisible by 3. {3, 6, 9, ........, 2022}

\,\,\,\,

n(N) = 674 L : numbers divisible by 6. {6, 12, 18, ........, 2022}

\,\,\,\,

n(L) = 337 n(M $\cup$ N) = n(M) + n(N) $-$ n(L) = 1011 + 674 $-$ 337 = 1348 0 = Number divisible by 337 but not in M $\cup$ N {337, 1685} Number divisible by 2, 3 or 337 = 1348 + 2 = 1350 Required probability

= {{2022 - 1350} \over {2022}}

= {{672} \over {2022}}

= {{112} \over {337}}

Q104

Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be black in colour. Then the probability, that the transferred ball is red, is :

A

\dfrac{4}{9}

B

\dfrac{5}{18}

C

\dfrac{1}{6}

D

\dfrac{3}{10}

Correct Answer

Option B

Solution

Let $E \rightarrow$ Ball drawn from Bag II is black.

$E_{R} \rightarrow$ Bag I to Bag II red ball transferred.

$E_{B} \rightarrow$ Bag I to Bag II black ball transferred.

$E_{w} \rightarrow$ Bag I to Bag II white ball transferred.

$P\left(E_{R} / E\right)=\dfrac{P\left(E / E_{R}\right) \cdot P\left(E_{R}\right)}{P\left(E / E_{R}\right) P\left(E_{R}\right)+P\left(E / E_{B}\right) P\left(E_{B}\right)+P\left(E / E_{W}\right) P\left(E_{W}\right)}$ Here, $P\left(E_{R}\right)=3 / 10, \quad P\left(E_{B}\right)=4 / 10, \quad P\left(E_{W}\right)=3 / 10$ and

\begin{aligned} & P\left(E / E_{R}\right)=5 / 10, \quad P\left(E / E_{B}\right)=6 / 10, \quad P\left(E / E_{W}\right)=5 / 10 \\\\ & \therefore \quad P\left(E_{R} / E\right)=\frac{15 / 100}{15 / 100+24 / 100+15 / 100} \\\\ & =\frac{15}{54}=\frac{5}{18} \end{aligned}

Q105

Two dice are thrown independently. Let

\mathrm{A}

be the event that the number appeared on the

1^{\text{st }}

die is less than the number appeared on the

2^{\text{nd }}

die,

\mathrm{B}

be the event that the number appeared on the

1^{\text{st }}

die is even and that on the second die is odd, and

\mathrm{C}

be the event that the number appeared on the

1^{\text{st }}

die is odd and that on the

2^{\text{nd }}

is even. Then :

A A and B are mutually exclusive

B the number of favourable cases of the events A, B and C are 15, 6 and 6 respectively

C B and C are independent

D the number of favourable cases of the event

(\mathrm{A\cup B)\cap C}

is 6

Correct Answer

Option D

Solution

$\begin{aligned} & A=\{(1,2),(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6),(4,5),(4,6),(5,6)\} \\\\ & n(A)=15 \\\\ & B=\{(2,1),(2,3),(2,5),(4,1),(4,3),(4,5),(6,1),(6,3),(6,5)\} \\\\ & n(B)=9 \\\\ & C=\{(1,2),(1,4),(1,6),(3,2),(3,4),(3,6),(5,2),(5,4),(5,6)\} \\\\ & n(C)=9\end{aligned}$

(4,5) \in A \text{ and }(4,5) \in B

$\therefore A$ and $B$ are not exclusive events

\begin{aligned} & n((A \cup B) \cap C)=n(A \cap C)+n(B \cap C)-n(A \cap B \cap C) \\\\ = & 3+3-0 \\\\ = & 6 \end{aligned}

Option (D) is correct.

\begin{aligned} & n(B)=\frac{9}{36}, n(C)=\frac{9}{36}, n(B \cap C)=0 \\\\ & \Rightarrow n(B) \cdot n(C) \neq n(B \cap C) \\\\ & \therefore B \text{ and } C \text{ are not independent. } \end{aligned}

Q106

In a binomial distribution

B(n,p)

, the sum and the product of the mean and the variance are 5 and 6 respectively, then

6(n+p-q)

is equal to :

A 52

B 50

C 51

D 53

Correct Answer

Option A

Solution

\begin{aligned} & \text{ Given } \\\\ & \mathrm{np}+\mathrm{npq}=5 \\\\ & \Rightarrow \mathrm{np}(1+\mathrm{q})=5 ........(i) \\\\ & \text{ and (np) (npq) }=6 \\\\ & \Rightarrow \mathrm{n}^2 \mathrm{p}^2 \mathrm{q}=6 ........(ii) \\\\ & (\mathrm{i})^2 \div(\mathrm{ii}) \\\\ & \frac{(1+q)^2}{9}=\frac{25}{6} \\\\ & \Rightarrow 6 \mathrm{q}^2-13 \mathrm{q}+6=0 \\\\ & \Rightarrow \mathrm{q}=\frac{2}{3}, \frac{3}{2} \text{ (rejected) } \\\\ & \mathrm{p}=1-\frac{2}{3}=\frac{1}{3} \\\\ & \frac{n}{3}\left(1+\frac{2}{3}\right)=5 \\\\ & \Rightarrow \mathrm{n}=9 \\\\ & 6(\mathrm{n}+\mathrm{p}-\mathrm{q})=52 \end{aligned}

Q107

A bag contains 6 balls. Two balls are drawn from it at random and both are found to be black. The probability that the bag contains at least 5 black balls is :

A

\dfrac{3}{7}

B

\dfrac{5}{6}

C

\dfrac{5}{7}

D

\dfrac{2}{7}

Correct Answer

Option C

Solution

Let $E_{i} \rightarrow$ Bag have at least $i$ black balls $E \rightarrow 2$ balls are drawn & both black

\begin{aligned} & \therefore P\left(\frac{E_{5} \text{ or } E_{6}}{E}\right)=\frac{P\left(\frac{E}{E_{5}}\right)+P\left(\frac{E}{E_{6}}\right)}{\sum\limits_{i=1}^{6} P\left(\frac{E}{E_{i}}\right)} \\\\ & =\frac{\frac{{ }^{5} C_{2}}{{ }^{6} C_{2}}+\frac{{ }^{6} C_{2}}{{ }^{6} C_{2}}}{0+\frac{{ }^{2} C_{2}}{{ }^{6} C_{2}}+\frac{{ }^{3} C_{2}}{{ }^{6} C_{2}}+\frac{{ }^{4} C_{2}}{{ }^{6} C_{2}}+\frac{{ }^{5} C_{2}}{{ }^{6} C_{2}}+\frac{{ }^{6} C_{2}}{{ }^{6} C_{2}}} \\\\ & =\frac{10+15}{1+3+6+10+15}=\frac{25}{35}=\frac{5}{7} \end{aligned}

Q108

If an unbiased die, marked with

-2,-1,0,1,2,3

on its faces, is thrown five times, then the probability that the product of the outcomes is positive, is :

A

\dfrac{27}{288}

B

\dfrac{521}{2592}

C

\dfrac{440}{2592}

D

\dfrac{881}{2592}

Correct Answer

Option B

Solution

{}^5{C_0} \times {3^5} = 243

{}^5{C_2} \times {2^2} \times {3^3} = 1080

{}^5{C_4} \times {2^4}\,.\,3 = 240

$\therefore$ required probability

= {{243 + 1080 + 240} \over {6 \times 6 \times 6 \times 6 \times 6}} = {{521} \over {2592}}

Q109

Let

\mathrm{S} = \{ {w_1},{w_2},......\}

be the sample space associated to a random experiment. Let

P({w_n}) = {{P({w_{n - 1}})} \over 2},n \ge 2

. Let

A = \{ 2k + 3l:k,l \in N\}

and

B = \{ {w_n}:n \in A\}

. Then P(B) is equal to :

A

\dfrac{3}{32}

B

\dfrac{1}{32}

C

\dfrac{1}{16}

D

\dfrac{3}{64}

Correct Answer

Option D

Solution

P({w_1}) + {{P({w_1})} \over 2} + {{P({w_1})} \over {{2^2}}}\, + \,..... = 1

$\therefore$

P({w_1}) = {1 \over 2}

Hence,

P({w_n}) = {1 \over {{2^n}}}

Every number except 1, 2, 3, 4, 6 is representable in the form

2k + 3l

where

k,l \in N

. $\therefore$

P(B) = 1 - P({w_1}) - P({w_2}) - P({w_3}) - P({w_4}) - P({w_6})

= {3 \over {64}}

Q110

Fifteen football players of a club-team are given 15 T-shirts with their names written on the backside. If the players pick up the T-shirts randomly, then the probability that at least 3 players pick the correct T-shirt is :

A

\dfrac{1}{6}

B

\dfrac{2}{15}

C

\dfrac{5}{24}

D 0.08

Correct Answer

Option D

Solution

Required probability $=1-\dfrac{D_{(15)}+{ }^{15} C_1 \cdot D_{(14)}+{ }^{15} C_2 D_{(13)}}{15 !}$ Taking $\mathrm{D}_{(15)}$ as $\dfrac{15 !}{e}$ $\mathrm{D}_{(14)}$ as $\dfrac{14 \text{ ! }}{e}$ $\mathrm{D}_{(13)}$ as $\dfrac{13 \text{ ! }}{e}$

\text{ We get, } \text{ Required probability } = 1-\left(\frac{\frac{15 !}{e}+15 \cdot \frac{14 !}{e}+\frac{15 \times 14}{2} \times \frac{13 !}{e}}{15 !}\right)

=1-\left(\frac{1}{e}+\frac{1}{e}+\frac{1}{2 e}\right)=1-\frac{5}{2 e} \approx .08