Probability — Page 15 | JEE Mathematics

Q141

A company has two plants

A

and

B

to manufacture motorcycles.

60 \%

motorcycles are manufactured at plant

A

and the remaining are manufactured at plant

B .80 \%

of the motorcycles manufactured at plant

A

are rated of the standard quality, while

90 \%

of the motorcycles manufactured at plant

B

are rated of the standard quality. A motorcycle picked up randomly from the total production is found to be of the standard quality. If

p

is the probability that it was manufactured at plant

B

, then

126 p

is

A 54

B 66

C 56

D 64

Correct Answer

Option A

Solution

\begin{aligned} & P(\text{ standard automobile from } A)=\frac{6}{10} \times \frac{8}{10}=\frac{12}{25} \\ & P(\text{ standard automobile from } B)=\frac{4}{10} \times \frac{9}{10}=\frac{9}{25} \end{aligned}

\begin{aligned} & \text{ Required Probability } \frac{\frac{9}{25}}{\frac{12}{25}+\frac{9}{25}} \\ & P=\frac{9}{21}=\frac{3}{7} \end{aligned}

So,

126 P=126 \times \frac{3}{7}=54

Q142

Bag 1 contains 4 white balls and 5 black balls, and Bag 2 contains n white balls and 3 black balls. One ball is drawn randomly from Bag 1 and transferred to Bag 2. A ball is then drawn randomly from Bag 2. If the probability, that the ball drawn is white, is

\dfrac{29}{45}

, then n is equal to:

A 5

B 6

C 4

D 3

Correct Answer

Option B

Solution

\begin{aligned} & \text{ Bag } 1=\{4 \mathrm{~W}, 5 \mathrm{~B}\} \\ & \text{ Bag } \mathbf{2}=\{\mathbf{n W}, \mathbf{3 B}\} \\ & \mathrm{P}\left(\frac{\mathrm{~W}}{\mathrm{Bag} 2}\right)=\frac{29}{45} \\ & \Rightarrow \mathrm{P}\left(\frac{\mathrm{~W}}{\mathrm{~B}_1}\right) \times \mathrm{P}\left(\frac{\mathrm{~W}}{\mathrm{~B}_2}\right)+\mathrm{P}\left(\frac{\mathrm{~B}}{\mathrm{~B}_1}\right) \times \mathrm{P}\left(\frac{\mathrm{~W}}{\mathrm{~B}_2}\right)=\frac{29}{45} \\ & \frac{4}{9} \times \frac{\mathrm{n}+1}{\mathrm{n}+4}+\frac{5}{9} \times \frac{\mathrm{n}}{\mathrm{n}+4}=\frac{29}{45} \end{aligned}

$\mathrm{n = 6}$

Q143

A coin is tossed three times. Let

X

denote the number of times a tail follows a head. If

\mu

and

\sigma^2

denote the mean and variance of

X

, then the value of

64\left(\mu+\sigma^2\right)

is:

A 64

B 32

C 51

D 48

Correct Answer

Option D

Solution

Outcome $x_i$ $p_i$ HHH 0 $\dfrac{1}{8}$ TTT 0 $\dfrac{1}{8}$ HHT 1 $\dfrac{1}{8}$ HTH 1 $\dfrac{1}{8}$ THH 0 $\dfrac{1}{8}$ TTH 0 $\dfrac{1}{8}$ THT 1 $\dfrac{1}{8}$ HTT 1 $\dfrac{1}{8}$ $\begin{aligned} & \mu=\sum x_i P_i=\dfrac{1}{2} \\ & \sigma^2=\sum x_i^2 P_i-\mu^2 \\ & =\dfrac{1}{2}-\dfrac{1}{4}=\dfrac{1}{4} \\ & 64\left(\mu+\sigma^2\right)=64\left[\dfrac{1}{2}+\dfrac{1}{4}\right] \\ & =64 \times \dfrac{3}{4}=48\end{aligned}$

Q144

Two balls are selected at random one by one without replacement from a bag containing 4 white and 6 black balls. If the probability that the first selected ball is black, given that the second selected ball is also black, is

\dfrac{m}{n}

, where

\operatorname{gcd}(m, n)=1

, then

m+n

is equal to :

A 4

B 14

C 11

D 13

Correct Answer

Option B

Solution

In a bag containing 4 white and 6 black balls, two balls are selected one by one without replacement.

We need to find the probability that the first ball is black, given that the second ball is black.

Let’s define the events: A: The first ball selected is black.

B: The second ball selected is black.

The probability $P(A|B)$ is given by the formula: $P(A|B) = \dfrac{P(A \cap B)}{P(B)}$ Let's compute each part separately: Probability of both balls being black $P(A \cap B)$ : The chance that the first ball is black is $\dfrac{6}{10}$ .

Given that one black ball is removed, the probability that the second is also black is $\dfrac{5}{9}$ .

$P(A \cap B) = \dfrac{6}{10} \times \dfrac{5}{9} = \dfrac{30}{90} = \dfrac{1}{3}$ Probability that the second ball is black $P(B)$ : This can occur if either the first ball was white or black: First ball white, second black: $\dfrac{4}{10} \times \dfrac{6}{9}$ Both balls black: $\dfrac{6}{10} \times \dfrac{5}{9}$ $P(B) = \dfrac{4}{10} \times \dfrac{6}{9} + \dfrac{6}{10} \times \dfrac{5}{9} = \dfrac{24}{90} + \dfrac{30}{90} = \dfrac{54}{90} = \dfrac{3}{5}$ Now, using these probabilities: $P(A|B) = \dfrac{\dfrac{1}{3}}{\dfrac{3}{5}} = \dfrac{1}{3} \times \dfrac{5}{3} = \dfrac{5}{9}$ Here, $\dfrac{5}{9}$ is in its simplest form ( $\operatorname{gcd}(5, 9) = 1$ ), so $m = 5$ and $n = 9$ .

Therefore, $m + n = 5 + 9 = 14$ .

Q145

If A and B are two events such that

P(A) = 0.7

,

P(B) = 0.4

and

P(A \cap \overline{B}) = 0.5

, where

\overline{B}

denotes the complement of B, then

P\left(B \mid (A \cup \overline{B})\right)

is equal to

A

\dfrac{1}{3}

B

\dfrac{1}{2}

C

\dfrac{1}{4}

D

\dfrac{1}{6}

Correct Answer

Option C

Solution

\begin{aligned} & \mathrm{P}(\mathrm{~A})=\frac{7}{10}, \mathrm{P}(\mathrm{~B})=\frac{4}{10} \\ & \mathrm{P}(\mathrm{~A} \cup \overline{\mathrm{~B}})=\frac{5}{10} \\ & \mathrm{P}\left(\frac{\mathrm{~B}}{\mathrm{~A} \cup \overline{\mathrm{~B}}}\right)=\frac{\mathrm{P}(\mathrm{~B} \cap(\mathrm{~A} \cup \overline{\mathrm{~B}}))}{\mathrm{P}(\mathrm{~A} \cup \overline{\mathrm{~B}})} \\ & =\frac{\mathrm{P}((\mathrm{~B} \cap \overline{\mathrm{~B}}) \cup(\mathrm{B} \cap \mathrm{~A}))}{\mathrm{P}(\mathrm{~A} \cup \overline{\mathrm{~B}})}=\frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{P}(\mathrm{~A} \cup \overline{\mathrm{~B}})} \end{aligned}

\begin{aligned} & =\frac{\mathrm{P}(\mathrm{~A})-\mathrm{P}(\mathrm{~A} \cap \overline{\mathrm{~B}})}{\mathrm{P}(\mathrm{~A})+\mathrm{P}(\overline{\mathrm{~B}})-\mathrm{P}(\mathrm{~A} \cap \overline{\mathrm{~B}})}=\frac{\frac{7}{10}-\frac{5}{10}}{\frac{7}{10}+\left(1-\frac{4}{10}\right)-\frac{5}{10}} \\ & =\frac{2}{8}=\frac{1}{4} \end{aligned}

Q146

If

A

and

B

are two events such that

P(A \cap B)=0.1

, and

P(A \mid B)

and

P(B \mid A)

are the roots of the equation

12 x^2-7 x+1=0

, then the value of

\dfrac{P(\bar{A} \cup \bar{B})}{P(\bar{A} \cap \bar{B})}

is :

A

\dfrac{4}{3}

B

\dfrac{7}{4}

C

\dfrac{9}{4}

D

\dfrac{5}{3}

Correct Answer

Option C

Solution

To solve this problem, start by considering the equation given for the probabilities: $12x^2 - 7x + 1 = 0$ The roots of this equation are: $x = \dfrac{1}{3}, \dfrac{1}{4}$ Assume $P(A \mid B) = \dfrac{1}{3}$ and $P(B \mid A) = \dfrac{1}{4}$ .

From the definitions of conditional probability, we have: $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{1}{3}$ $P(B \mid A) = \dfrac{P(A \cap B)}{P(A)} = \dfrac{1}{4}$ Given that $P(A \cap B) = 0.1$ , we can use these equations to find $P(B)$ and $P(A)$ : From $\dfrac{0.1}{P(B)} = \dfrac{1}{3}$ , we find $P(B) = 0.3$ .

From $\dfrac{0.1}{P(A)} = \dfrac{1}{4}$ , we find $P(A) = 0.4$ .

Now, calculate $P(A \cup B)$ using the formula for the union of two events: $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.4 + 0.3 - 0.1 = 0.6$ The task is to find the value of: $\dfrac{P(\overline{A} \cup \overline{B})}{P(\overline{A} \cap \overline{B})}$ Using De Morgan's laws and the complements: $P(\overline{A} \cup \overline{B}) = P(\overline{A \cap B}) = 1 - P(A \cap B) = 1 - 0.1 = 0.9$ $P(\overline{A} \cap \overline{B}) = 1 - P(A \cup B) = 1 - 0.6 = 0.4$ Finally, compute the ratio: $\dfrac{P(\overline{A} \cup \overline{B})}{P(\overline{A} \cap \overline{B})} = \dfrac{0.9}{0.4} = \dfrac{9}{4}$

Q147

Bag

B_1

contains 6 white and 4 blue balls, Bag

B_2

contains 4 white and 6 blue balls, and Bag

B_3

contains 5 white and 5 blue balls. One of the bags is selected at random and a ball is drawn from it. If the ball is white, then the probability that the ball is drawn from Bag

B_2

is:

A

\dfrac{2}{5}

B

\dfrac{4}{15}

C

\dfrac{1}{3}

D

\dfrac{2}{3}

Correct Answer

Option B

Solution

$E_1: B a g B_1$ is selected

\begin{array}{lll} B_1 & B_2 & B_3 \\ 6 \mathrm{~W} 4 \mathrm{~B} & 4 \mathrm{~W} 6 \mathrm{~B} & 5 \mathrm{~W} 5 \mathrm{~B} \end{array}

$E_2:$ bag $B_2$ is selected $E_3: B a g B_3$ is selected A : Drawn ball is white We have to find $P\left(\dfrac{E_2}{A}\right)$

\begin{aligned} & P\left(\frac{E_2}{A}\right)=\frac{P\left(E_2\right) P\left(\frac{A}{E_2}\right)}{P\left(E_1\right) P\left(\frac{A}{E_1}\right)+P\left(E_2\right) P\left(\frac{A}{E_2}\right)+P\left(E_3\right) P\left(\frac{A}{E_3}\right)} \\ & =\frac{\frac{1}{3} \times \frac{4}{10}}{\frac{1}{3} \times \frac{6}{10}+\frac{1}{3} \times \frac{4}{10}+\frac{1}{3} \times \frac{5}{10}} \\ & =\frac{4}{15} \end{aligned}

Q148

One die has two faces marked 1 , two faces marked 2 , one face marked 3 and one face marked 4 . Another die has one face marked 1 , two faces marked 2 , two faces marked 3 and one face marked 4. The probability of getting the sum of numbers to be 4 or 5 , when both the dice are thrown together, is

A

\dfrac{2}{3}

B

\dfrac{3}{5}

C

\dfrac{4}{9}

D

\dfrac{1}{2}

Correct Answer

Option D

Solution

$\mathrm{a}=$ number or dice 1 $\mathrm{b}=$ number on dice 2

(a, b)=(1,3),(3,1),(2,2),(2,3),(3,2),(1,4),(4,1)

Required probability

\begin{aligned} & =\frac{2}{6} \times \frac{2}{6}+\frac{1}{6} \times \frac{1}{6}+\frac{2}{6} \times \frac{2}{6}+\frac{2}{6} \times \frac{2}{6}+\frac{1}{6} \times \frac{2}{6}+\frac{2}{6} \times \frac{1}{6}+\frac{1}{6} \times \frac{2}{6} \\ & =\frac{18}{36}=\frac{1}{2} \end{aligned}

Q149

A

and

B

alternately throw a pair of dice. A wins if he throws a sum of 5 before

B

throws a sum of 8 , and

B

wins if he throws a sum of 8 before

A

throws a sum of 5 . The probability, that A wins if A makes the first throw, is

A

\dfrac{8}{19}

B

\dfrac{9}{19}

C

\dfrac{8}{17}

D

\dfrac{9}{17}

Correct Answer

Option B

Solution

\begin{aligned} & \mathrm{p}\left(\mathrm{~S}_5\right)=\frac{1}{9} \\ & \mathrm{p}\left(\mathrm{~S}_5\right)=\frac{5}{36} \\ & \text{ required prob }=\frac{1}{9}+\frac{8}{9} \cdot \frac{31}{36} \cdot \frac{1}{9}+\left(\frac{8}{9} \cdot \frac{31}{36}\right)^2 \cdot \frac{1}{9}+\ldots \infty \\ & =\frac{\frac{1}{9}}{1-\frac{62}{81}}=\frac{9}{19} \end{aligned}

Q150

Let

\mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right]

be a square matrix of order 2 with entries either 0 or 1 . Let E be the event that A is an invertible matrix. Then the probability

\mathrm{P}(\mathrm{E})

is :

A

\dfrac{3}{8}

B

\dfrac{1}{8}

C

\dfrac{3}{16}

D

\dfrac{5}{8}

Correct Answer

Option A

Solution

C-I $\left|\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right| \rightarrow 4$ ways C-II $\left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right| \&\left|\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right| \rightarrow 2$ ways

\mathrm{P}=\frac{\text{ favourable }}{\text{ total }}=\frac{6}{16}=\frac{3}{8}