Algebra

Probability

Hard

common

Two cards are drawn simultaneously from a well shuffled pack of 52 cards. Then variance of the number of kings is

$\frac{680}{(221)^2}$

$\frac{6080}{(221)^2}$

$\frac{680}{221}$

$\frac{6800}{(221)^2}$

✅ Correct Option: 4

Let X be the number of kings. $P(X=0) = \binom{48}{2}/\binom{52}{2} = 188/221$ , $P(X=1) = 32/221$ , $P(X=2) = 1/221$ . Then $E(X) = 34/221$ and $E(X^2) = 36/221$ . Variance $= E(X^2) - [E(X)]^2 = 36/221 - 1156/(221)^2 = (7956 - 1156)/(221)^2 = 6800/(221)^2$ .

2026: 29 May Shift 2

An insurance company insured 200 scooter drivers, 400 car drivers and 600 Truck drivers. The probability of an accident involving scooter driver, car driver and truck driver are 0.01, 0.03 and 0.15 respectively. One of the insured person meets an accident, then the probability that he is a truck driver is:-

2025: 22 May Shift 1

Let X denotes the number of doublets obtained in 3 throws of a pair of dice.

Match List-I with List-II

List-I	List-II
(A) $P(X = 0)$	(I) $\frac{1}{216}$
(B) $P(X = 1)$	(II) $\frac{15}{216}$
(C) $P(X = 2)$	(III) $\frac{75}{216}$
(D) $P(X = 3)$	(IV) $\frac{125}{216}$

Choose the correct answer from the options given below:

2025: 15 May Shift 2

Two events A and B will be independent, then

2022: 23 Aug Shift 1

The conditional probability that his answer is correct when it is given that he knew it :

2026: 30 May Shift 1

If E and F are two events such that P(E) = 0.8, P(F) = 0.7 and P(E ∩ F) = 0.6 then $P(\bar{E}/\bar{F})$ is:

2026: 19 May Shift 1

The random variable X has the following probability distribution:

X	1	2	3	4	5
P(X)	$\frac{1}{10}$	$\frac{1}{10}$	$\frac{1}{3}$	k	$\frac{1}{3}$

The correct value of k is:

Two cards are drawn simultaneously from a well shuffled pack of 52 cards. Then variance of the number of kings is

Solution

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