Algebra

Probability

Medium

core

A and B throw a die alternatively till one of them gets a number more than 4 and wins the game. Then the probability of winning the game by B, if A starts first :

$\frac{2}{5}$

$\frac{3}{5}$

$\frac{1}{5}$

$\frac{4}{5}$

✅ Correct Option: 1

$p = \frac{2}{6} = \frac{1}{3}$ (success), $q = \frac{2}{3}$ .

B wins on 2nd, 4th, 6th... throws.

$P(B) = qp + q^3 p + ... = \frac{qp}{1-q^2} = \frac{(2/3)(1/3)}{5/9} = \frac{2}{5}$ .

2025: 21 May Shift 1

A can solve 90% problems and B can solve 70% problems of the book. A problem is selected at random from the book. The probability that the problem is solved, is equal to

2026: 22 May Shift 1

Match List-I with List-II

List-I	List-II
(A) For a random variable X, $E(X)=3$ and $E(X^2)=11$ , then $Var(X)$ is	(I) 25
(B) For a random variable X, $E(X)=3$ and $E(X^2)=13$ , then $Var\left(\frac{X}{2}\right)$ is	(II) 2
(C) In a binomial distribution, if $n=20, q=0.75$ , then its mean is	(III) 1
(D) If in a binomial distribution, mean is 5 and variance is 4, then the number of trials is	(IV) 5

Choose the correct answer from the options given below:

2026: 14 May Shift 2

A pair of coins thrown then

Match the LIST-I with LIST-II

	LIST-I		LIST-II
A.	Probability of occurance of head on both coins is	I.	$\dfrac{3}{4}$
B.	Probability of occurance of at least one tail is	II.	$\dfrac{1}{2}$
C.	Probability of occurance of one tail only is	III.	$1$
D.	Probability of a tail appears on one coin given that only one coin shows head.	IV.	$\dfrac{1}{4}$

Choose the correct answer from the options given below:

2025: 13 May Shift 1

The probability distribution of a ramdom variable $X$ is:

$P(X=x)=\begin{cases}kx^2,& x=1,2,3\\2kx,&x=4,5,6\\0,&\text{otherwise}\end{cases}$ Where $K$ is a constant

Match List-I with List-II

List-I	List-II
(A) $k$	(I) 7/22
(B) $P(X \geq 4)$	(II) 1/44
(C) $P(X < 4)$	(III) 95/22
(D) $E[X]$	(IV) 15/22

Choose the correct answer from the options given below:

2026: 21 May Shift 1

A pair of dice is thrown. If the two numbers appearing on them are different then the probability that the sum of the number is 6, is.

2025: 22 May Shift 1

Probability that a man speaks truth is $\frac{3}{4}$ . He throws a die and reports that it is a six. The probability that it is actually a six is

A and B throw a die alternatively till one of them gets a number more than 4 and wins the game. Then the probability of winning the game by B, if A starts first :

Solution

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