Algebra

Probability

Medium

common

Two cards are drawn simultaneously at random from a well shuffled pack of 52 Cards. Let X be the random variable which denotes number of kings in the draw. Then the probability distribution of X is

X	0	1	2
P(X)	$\dfrac{^{48}C_2}{^{52}C_2}$	$\dfrac{^4C_1 \times ^{48}C_1}{^{52}C_2}$	$\dfrac{^4C_2}{^{52}C_2}$

X	0	1	2
P(X)	$\dfrac{^4C_0}{^{52}C_2}$	$\dfrac{^4C_1}{^{52}C_2}$	$\dfrac{^4C_2}{^{52}C_2}$

X	0	1	2
P(X)	$\dfrac{^4C_2}{^{52}C_2}$	$\dfrac{^4C_1 \times ^{48}C_1}{^{52}C_2}$	$\dfrac{^{48}C_2}{^{52}C_2}$

X	0	1	2
P(X)	$\dfrac{^4C_0}{^{52}C_2}$	$\dfrac{^{48}C_1}{^{52}C_2}$	$\dfrac{^4C_2}{^{52}C_2}$

✅ Correct Option: 1

Two cards are drawn simultaneously at random from a well shuffled pack of 52 cards. X denotes the number of kings in the draw.

Total cards = 52

Kings in deck = 4

Non-kings = 48

Cards drawn = 2

X (number of kings) can be:

X = 0: No kings (both cards are non-kings)

X = 1: Exactly one king (one king and one non-king)

X = 2: Two kings (both cards are kings)

Total ways to select 2 cards from 52:

$^{52}C_2$

For X = 0 (no kings), both cards must be non-kings.

Ways to choose 2 cards from 48 non-kings = $^{48}C_2$

$P(X = 0) = \frac{^{48}C_2}{^{52}C_2}$

For X = 1 (exactly one king), one card is a king and one is a non-king.

Ways to choose 1 king from 4 kings = $^4C_1$

Ways to choose 1 non-king from 48 non-kings = $^{48}C_1$

Total ways = $^4C_1 \times ^{48}C_1$

$P(X = 1) = \frac{^4C_1 \times ^{48}C_1}{^{52}C_2}$

For X = 2 (two kings), both cards must be kings.

Ways to choose 2 kings from 4 kings = $^4C_2$

$P(X = 2) = \frac{^4C_2}{^{52}C_2}$

The probability distribution of X is:

X	0	1	2
P(X)	$\frac{^{48}C_2}{^{52}C_2}$	$\frac{^4C_1 \times ^{48}C_1}{^{52}C_2}$	$\frac{^4C_2}{^{52}C_2}$

2025: 16 May Shift 1

If the random variable X is normally distributed with mean 16 and standard deviation 4, then the standard normal variable Z corresponding to X = 17 is

2022: 7 Aug Shift 2

A shopkeeper sells three types of flower seeds $A_1, A_2, A_3$ . They are sold as a mixture where the proportions are 4 : 4 : 2 respectively. The germination rates of the three types of seeds are 45%, 60% and 35% respectively. Calculate the probability in the following cases.

The probability of a randomly chosen seed to germinate is :

2026: 30 May Shift 1

A company manufactured a product through 4 units. A, B, C and D. The probability that a product is manufactured are $\frac{3}{10},\frac{1}{5},\frac{1}{10}$ and $\frac{2}{5}$ by units A, B, C and D respectively. The probability that it will be defective are $\frac{1}{4},\frac{1}{3}$ and $\frac{1}{12}$ if it is produced by A, B, C respectively and there is no defective from unit D. A product is chosen, it is found to be defective.

Match the LIST-I with LIST-II

LIST-I	LIST-II
A. The Probability that the defective product is from A	I. 4/9
B. The Probability that the defective product is from B	II. 0
C. Probability of defective product	III. 1/2
D. Probability that the defective product is from D	IV. 3/20

Choose the correct answer from the options given below:

2024: 16 May Shift 1

If the random variable $X$ has the following distribution:

$X$	0	1	2	otherwise
$P(X)$	$k$	$2k$	$3k$	$0$

Match List-I with List-II:

List-I	List-II
(A) $k$	(I) $\frac{5}{6}$
(B) $P(X < 2)$	(II) $\frac{4}{3}$
(C) $E(X)$	(III) $\frac{1}{2}$
(D) $P(1 \leq X \leq 2)$	(IV) $\frac{1}{6}$

2022: 4 Aug Shift 1

Two cards are drawn successively with replacement from a well shuffled deck of 52 cards. The probability distribution of the number of kings will be:

2025: 15 May Shift 1

Match List-I with List-II

An urn contains 4 white and 3 red balls. In a random draw of three balls, the probability of

List-I	List-II
(A) No red ball is	(I) $\frac{12}{35}$
(B) Only 1 red ball is	(II) $\frac{1}{35}$
(C) Exactly 2 red balls is	(III) $\frac{4}{35}$
(D) no white ball is	(IV) $\frac{18}{35}$

Choose the correct answer from the options given below:

Two cards are drawn simultaneously at random from a well shuffled pack of 52 Cards. Let X be the random variable which denotes number of kings in the draw. Then the probability distribution of X is

Solution

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