Three dice are rolled. If the probability of getting different numbers on the three dice is $$\frac{p}{q}$$, where $$p$$ and $$q$$ are co-prime, then $$q-p$$ is equal to :

Total number of outcomes $=6 \times 6 \times 6=216$ Number of outcomes getting different numbers on the three dice are ${ }^6 P_3=\frac{6 !}{3 !}=120$ $\therefore$ Required probability $=\frac{120}{216}=\frac{5}{9}$ $\therefore p=5$ and $q=9$ $\therefore q-p=9-5=4$

Probability — Page 13 | JEE Mathematics

Q121

Let a die be rolled

n

times. Let the probability of getting odd numbers seven times be equal to the probability of getting odd numbers nine times. If the probability of getting even numbers twice is

\dfrac{k}{2^{15}}

, then

\mathrm{k}

is equal to :

A 15

B 60

C 30

D 90

Correct Answer

Option B

Solution

Given that, $\mathrm{P}($ odd number seven times $)=\mathrm{P}($ Odd number nine times)

\begin{aligned} & \Rightarrow{ }^n \mathrm{C}_7\left(\frac{1}{2}\right)^7\left(\frac{1}{2}\right)^{n-7}={ }^n \mathrm{C}_9\left(\frac{1}{2}\right)^9\left(\frac{1}{2}\right)^{n-9} \\\\ & \Rightarrow{ }^n \mathrm{C}_7={ }^n \mathrm{C}_9 \\\\ & \Rightarrow n=7+9=16 \end{aligned}

\begin{aligned} & \text{ Hence, } \mathrm{P}(\text{ Even number twice })=\frac{k}{2^{15}} \\\\ & \Rightarrow{ }^{16} \mathrm{C}_2\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^{16-2}=\frac{k}{2^{15}} \\\\ & \Rightarrow \frac{16 \times 15}{2} \times \frac{1}{2^{16}}=\frac{k}{2^{15}} \Rightarrow k=60 \end{aligned}

Q122

Let N denote the sum of the numbers obtained when two dice are rolled. If the probability that $${2^N}

A 12

B 6

C 8

D 10

Correct Answer

Option C

Solution

$N$ denote the sum of the numbers obtained when two dice are rolled. Such that $2^N < N$ !

\text{ i.e., } 4 \leq N \leq 12 \text{ i.e., } N \in\{4,5,6, \ldots 12\}

Now, $P(N=2)+P(N=3)=\dfrac{1}{36}+\dfrac{2}{36}=\dfrac{3}{36}=\dfrac{1}{12}$ So, required probability $=1-\dfrac{1}{12}=\dfrac{11}{12}=\dfrac{m}{n}$

4 m-3 n=4 \times 11-3 \times 12=44-36=8

Q123

If the probability that the random variable

\mathrm{X}

takes values

x

is given by

\mathrm{P}(\mathrm{X}=x)=\mathrm{k}(x+1) 3^{-x}, x=0,1,2,3, \ldots

, where

\mathrm{k}

is a constant, then

\mathrm{P}(\mathrm{X} \geq 2)

is equal to :

A

\dfrac{7}{18}

B

\dfrac{20}{27}

C

\dfrac{7}{27}

D

\dfrac{11}{18}

Correct Answer

Option C

Solution

As, we know that sum of all the probabilities $=1$

\begin{aligned} & \text{ So, } \sum_{x=1}^{\infty} \mathrm{P}(\mathrm{X}=x)=1 \\\\ & \Rightarrow k\left[1+2 \cdot 3^{-1}+3 \cdot 3^{-2}+\ldots . \infty\right]=1 \end{aligned}

\begin{aligned} & \text{ Let } S=1+\frac{2}{3}+\frac{3}{3^2}+\ldots .+\infty \\\\ & \Rightarrow \frac{S}{3}=0+\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\ldots .+\infty \end{aligned}

On subtracting, we get

\begin{aligned} & \frac{2 S}{3}=1+\frac{1}{3}+\frac{1}{3^2}+\ldots .+\infty \\\\ & \Rightarrow \frac{2 S}{3}=\frac{1}{1-\frac{1}{3}}=\frac{1}{\frac{2}{3}} \\\\ & \Rightarrow \frac{2 S}{3}=\frac{3}{2} \\\\ & \Rightarrow S=\frac{9}{4} \end{aligned}

\begin{aligned} & \text{ So, } k \times \frac{9}{4}=1 \Rightarrow k=\frac{4}{9} \\\\ & \text{ Now, } \mathrm{P}(\mathrm{X} \geq 2)=1-\mathrm{P}(\mathrm{X}<2) \\\\ & =1-\mathrm{P}(\mathrm{X}=0)-\mathrm{P}(\mathrm{X}=1) \\\\ & =1-\frac{4}{9}(1)-\frac{4}{9} \times \frac{2}{3} \\\\ & =1-\frac{4}{9}-\frac{8}{27}=\frac{27-12-8}{27}=\frac{7}{27} \end{aligned}

Q124

In a bolt factory, machines

A, B

and

C

manufacture respectively

20 \%, 30 \%

and

50 \%

of the total bolts. Of their output 3, 4 and 2 percent are respectively defective bolts. A bolt is drawn at random from the product. If the bolt drawn is found the defective, then the probability that it is manufactured by the machine

C

is :

A

\dfrac{2}{7}

B

\dfrac{9}{28}

C

\dfrac{5}{14}

D

\dfrac{3}{7}

Correct Answer

Option C

Solution

Given : $P(A)=\dfrac{20}{100}=\dfrac{2}{10}$

P(B)=\frac{30}{100}=\frac{3}{10}

P(C)=\frac{50}{100}=\frac{5}{10}

Let $\mathrm{E} \rightarrow$ Event that the bolt is defective.

\text{ So, } P(E / A)=\frac{3}{100},

P\left(\frac{E}{B}\right)=\frac{4}{100}, P\left(\frac{E}{C}\right)=\frac{2}{100}

So, $\mathrm{P}(\mathrm{C} / \mathrm{E})$

\begin{aligned} & =\frac{P\left(\frac{E}{C}\right) \times P(C)}{P\left(\frac{E}{A}\right) \times P(A)+P\left(\frac{E}{B}\right) \times P(B)+P\left(\frac{E}{C}\right) \times P(C)} \\\\ & =\frac{\frac{5}{10} \times \frac{2}{100}}{\frac{3}{100} \times \frac{2}{10}+\frac{4}{100} \times \frac{3}{10}+\frac{2}{100} \times \frac{5}{10}} \\\\ & =\frac{10}{6+12+10}=\frac{10}{28}=\frac{5}{14} \end{aligned}

Q125

A pair of dice is thrown 5 times. For each throw, a total of 5 is considered a success. If the probability of at least 4 successes is

\dfrac{k}{3^{11}}

, then

k

is equal to :

A 82

B 164

C 123

D 75

Correct Answer

Option C

Solution

Given, a pair of dice is thrown once, then number of outcomes $=6 \times 6=36$ A total of 5 are $\{(1,4),(2,3),(4,1),(3,2)\}$

\begin{aligned} & \therefore p =p(\text{ success })=\frac{4}{36}=\frac{1}{9} \\\\ & q =1-\frac{1}{9}=\frac{8}{9} \end{aligned}

Let $X$ represent the number of success

\begin{aligned} & \therefore P(\text{ at least } 4 \text{ success }) \\\\ &=P(X=4)+P(X=5) \\\\ &={ }^5 C_4 p^4 q+{ }^5 C_5 p^5 q^{5-5} \\\\ &=5\left(\frac{1}{9}\right)^4\left(\frac{8}{9}\right)+\left(\frac{1}{9}\right)^5 \\\\ &=\frac{40+1}{\left(3^2\right)^5}=\frac{41}{310} \\\\ &=\frac{41 \times 3}{3^{11}}=\frac{123}{3^{11}} \\\\ & \therefore k=123 \end{aligned}

Q126

Three dice are rolled. If the probability of getting different numbers on the three dice is

\dfrac{p}{q}

, where

p

and

q

are co-prime, then

q-p

is equal to :

A 3

B 4

C 1

D 2

Correct Answer

Option B

Solution

Total number of outcomes $=6 \times 6 \times 6=216$ Number of outcomes getting different numbers on the three dice are ${ }^6 P_3=\dfrac{6 !}{3 !}=120$ $\therefore$ Required probability $=\dfrac{120}{216}=\dfrac{5}{9}$ $\therefore p=5$ and $q=9$ $\therefore q-p=9-5=4$

Q127

Let Ajay will not appear in JEE exam with probability

\mathrm{p}=\dfrac{2}{7}

, while both Ajay and Vijay will appear in the exam with probability

\mathrm{q}=\dfrac{1}{5}

. Then the probability, that Ajay will appear in the exam and Vijay will not appear is :

A

\dfrac{9}{35}

B

\dfrac{3}{35}

C

\dfrac{24}{35}

D

\dfrac{18}{35}

Correct Answer

Option D

Solution

We are given that the probability of Ajay not appearing in the JEE exam is $\mathrm{p}=\dfrac{2}{7}$ , and the probability that both Ajay and Vijay will appear in the exam is $\mathrm{q}=\dfrac{1}{5}$ .

We are asked to find the probability that Ajay will appear in the exam and Vijay will not.

Let's denote this probability as $\mathrm{r}$ .

To find $\mathrm{r}$ , we need to use the concept of complementary events.

The probability that Ajay will appear in the exam is the complement of the probability that he will not appear.

So,

P(\text{Ajay appears}) = 1 - P(\text{Ajay does not appear}) = 1 - \mathrm{p} = 1 - \frac{2}{7} = \frac{5}{7}.

The event that both Ajay and Vijay appear in the exam is independent of the event that only Ajay appears (and Vijay does not).

Therefore, we can express the probability that only Ajay will appear (and Vijay will not) as the difference of Ajay appearing minus both Ajay and Vijay appearing, because the probability of both appearing ( $\mathrm{q}$ ) is included in the probability of Ajay appearing:

\mathrm{r} = P(\text{Ajay appears}) - P(\text{Both Ajay and Vijay appear}) = \frac{5}{7} - \frac{1}{5}.

To subtract these two fractions, we need a common denominator, which would be $35$ in this case. So,

\mathrm{r} = \frac{5}{7} \cdot \frac{5}{5} - \frac{1}{5} \cdot \frac{7}{7} = \frac{25}{35} - \frac{7}{35} = \frac{25 - 7}{35} = \frac{18}{35}.

Therefore, the probability that Ajay will appear in the exam and Vijay will not appear is $\mathrm{r} = \dfrac{18}{35}$ , which corresponds to Option D.

Q128

A bag contains 8 balls, whose colours are either white or black. 4 balls are drawn at random without replacement and it was found that 2 balls are white and other 2 balls are black. The probability that the bag contains equal number of white and black balls is :

A

\dfrac{2}{5}

B

\dfrac{2}{7}

C

\dfrac{1}{7}

D

\dfrac{1}{5}

Correct Answer

Option B

Solution

$\begin{aligned} & \mathrm{P}(4 \mathrm{~W} 4 \mathrm{~B} / 2 \mathrm{~W} 2 \mathrm{~B})= \\\\ & \dfrac{P(4 W 4 B) \times P(2 W 2 B / 4 W 4 B)}{P(2 W 6 B) \times P(2 W 2 B / 2 W 6 B)+P(3 W 5 B) \times P(2 W 2 B / 3 W 5 B)} \\ & +\ldots \ldots \ldots \ldots+P(6 W 2 B) \times P(2 W 2 B / 6 W 2 B)\end{aligned}$ $\begin{aligned} & =\dfrac{\dfrac{1}{5} \times \dfrac{{ }^4 \mathrm{C}_2 \times{ }^4 \mathrm{C}_2}{{ }^8 \mathrm{C}_4}}{\dfrac{1}{5} \times \dfrac{{ }^2 \mathrm{C}_2 \times{ }^6 \mathrm{C}_2}{{ }^8 \mathrm{C}_4}+\dfrac{1}{5} \times \dfrac{{ }^3 \mathrm{C}_2 \times{ }^5 \mathrm{C}_2}{{ }^8 \mathrm{C}_4}+\ldots+\dfrac{1}{5} \times \dfrac{{ }^6 \mathrm{C}_2 \times{ }^2 \mathrm{C}_2}{{ }^8 \mathrm{C}_4}} \\\\ & =\dfrac{2}{7}\end{aligned}$

Q129

A coin is biased so that a head is twice as likely to occur as a tail. If the coin is tossed 3 times, then the probability of getting two tails and one head is

A

\dfrac{1}{9}

B

\dfrac{2}{9}

C

\dfrac{1}{27}

D

\dfrac{2}{27}

Correct Answer

Option B

Solution

To solve this problem, we need to first determine the probability of getting a head (H) and the probability of getting a tail (T).

Since a head is twice as likely to occur as a tail, we can denote the probability of getting a tail as

P(T) = p

and the probability of getting a head as

P(H) = 2p

. These probabilities must sum to 1 because those are the only two possible outcomes for each coin toss :

P(H) + P(T) = 1

2p + p = 1

3p = 1

p = \frac{1}{3}

Therefore, the probability of getting a tail (T) is

P(T) = \frac{1}{3}

and the probability of getting a head (H) is

P(H) = 2 \times \frac{1}{3} = \frac{2}{3}

.

Now to find the probability of getting two tails and one head, we need to consider the different sequences in which this can occur.

There are three unique sequences: TTH, THT, and HTT.

The probability of each sequence is found by multiplying the probabilities of each individual event since each coin toss is independent:

P(TTH) = P(T) \times P(T) \times P(H) = \left(\frac{1}{3}\right)^2 \times \frac{2}{3} = \frac{1}{9} \times \frac{2}{3} = \frac{2}{27}

P(THT) = P(T) \times P(H) \times P(T) = \frac{1}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{2}{27}

P(HTT) = P(H) \times P(T) \times P(T) = \frac{2}{3} \times \left(\frac{1}{3}\right)^2 = \frac{2}{27}

The overall probability of getting two tails and one head in any order is the sum of these individual probabilities :

P(2T1H) = P(TTH) + P(THT) + P(HTT) = \frac{2}{27} + \frac{2}{27} + \frac{2}{27} = \frac{6}{27}

Simplifying this expression gives us:

P(2T1H) = \frac{6}{27} = \frac{2}{9}

Therefore, the correct answer is : Option B :

\frac{2}{9}

Q130

Three rotten apples are accidently mixed with fifteen good apples. Assuming the random variable

x

to be the number of rotten apples in a draw of two apples, the variance of

x

is

A

\dfrac{57}{153}

B

\dfrac{40}{153}

C

\dfrac{37}{153}

D

\dfrac{47}{153}

Correct Answer

Option B

Solution

3 bad apples, 15 good apples. Let

\mathrm{X}

be no of bad apples

\begin{aligned} & \text{ Then } \mathrm{P}(\mathrm{X}=0)=\frac{{ }^{15} \mathrm{C}_2}{{ }^{18} \mathrm{C}_2}=\frac{105}{153} \\ & \mathrm{P}(\mathrm{X}=1)=\frac{{ }^3 \mathrm{C}_1 \times{ }^{15} \mathrm{C}_1}{{ }^{18} \mathrm{C}_2}=\frac{45}{153} \\ & \mathrm{P}(\mathrm{X}=2)=\frac{{ }^3 \mathrm{C}_2}{{ }^{18} \mathrm{C}_2}=\frac{3}{153} \\ & \mathrm{E}(\mathrm{X})=0 \times \frac{105}{153}+1 \times \frac{45}{153}+2 \times \frac{3}{153}=\frac{51}{153} \\ & =\frac{1}{3} \\ & \operatorname{Var}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^2\right)-(\mathrm{E}(\mathrm{X}))^2 \\ & =0 \times \frac{105}{153}+1 \times \frac{45}{153}+4 \times \frac{3}{153}-\left(\frac{1}{3}\right)^2 \\ & =\frac{57}{153}-\frac{1}{9}=\frac{40}{153} \end{aligned}