Algebra

Probability

Medium

core

A box contains 2 black and 4 red balls and another box contains 4 black and 3 red balls. If a ball drawn at random from one of the two boxes, then the probability of getting a black ball is

$\frac{2}{6}$

$\frac{4}{7}$

$\frac{19}{42}$

$\frac{6}{13}$

✅ Correct Option: 3

There are two boxes:

Box 1: 2 black + 4 red = 6 total balls

Box 2: 4 black + 3 red = 7 total balls

A ball is drawn at random from one of the two boxes. This means one box is randomly selected first, then a ball is drawn from that chosen box.

Probability of selecting Box 1 = $\frac{1}{2}$

Probability of getting black from Box 1 = $\frac{2}{6}$

Probability of selecting Box 1 and getting black = $\frac{1}{2} \times \frac{2}{6}$

$= \frac{2}{12}$

$= \frac{1}{6}$

Probability of selecting Box 2 = $\frac{1}{2}$

Probability of getting black from Box 2 = $\frac{4}{7}$

Probability of selecting Box 2 and getting black = $\frac{1}{2} \times \frac{4}{7}$

$= \frac{4}{14}$

$= \frac{2}{7}$

Total probability of getting a black ball = $\frac{1}{6} + \frac{2}{7}$

LCM of 6 and 7 = 42

$\frac{1}{6} = \frac{7}{42}$

$\frac{2}{7} = \frac{12}{42}$

Total probability = $\frac{7}{42} + \frac{12}{42}$

$= \frac{19}{42}$

Therefore, the probability of getting a black ball is $\frac{19}{42}$ .

2026: 25 May Shift 1

Bag A contains 3 red and 4 black balls, and Bag B contains 5 red and 'n' black balls. One ball is drawn at random from one of the bags and it is found to be red. If the probability that it was from bag B is 35/68, then the value of 'n' is:

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: 22 May Shift 2

Two numbers are selected at random from the integers 1 through 19. If the sum is even, then the probability that both are odd numbers is:

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

2025: 22 May Shift 2

If A and B are independent events and $P(A) = \frac{1}{2}$ $P(B) = \frac{1}{3}$ , then

Match List-I with List-II

List-I	List-II
(A) $P(A \cap B)$	(I) $\frac{1}{2}$
(B) $P(\bar{A})P(B) + P(A)P(\bar{B})$	(II) $\frac{1}{3}$
(C) $P(A \mid B) + P(B \mid A)$	(III) $\frac{1}{6}$
(D) $P(A \cap \bar{B})$	(IV) $\frac{5}{6}$

Choose the correct answer from the options given below:

2025: 19 May Shift 1

Bag A contains 2 unbiased and 3 biased coins whereas Bag B contains 3 unbiased and 2 biased coins. A bag is selected at random and 2 coins are taken out simultaneously. The probability, that both coins are unbiased is:

A box contains 2 black and 4 red balls and another box contains 4 black and 3 red balls. If a ball drawn at random from one of the two boxes, then the probability of getting a black ball is

Solution

Related questions:

Keyboard Shortcuts

CUET Mathematics Past Year Questions (Topic-Wise)