Algebra

Probability

Medium

applied

A random variable X has the following probability distribution:

X 0 1 2 3 4 5 6 7
P(X) 0 k 2k 2k 3k $k^2$ $2k^2$ $7k^2+k$

Then, which of the following are correct ?

A. $k = 0.1$

B. $P(X<3) = 0.3$

C. $P(0<X<3) = 0.8$

D. $P(X>6) = 0.17$

Choose the correct answer from the options given below:

X	0	1	2	3	4	5	6	7
P(X)	0	k	2k	2k	3k	$k^2$	$2k^2$	$7k^2+k$

A, B and D only

A, C and D only

B, C and D only

B and C only

✅ Correct Option: 1

Summing all probabilities gives $10k^2+9k = 1$ , so $k = 0.1$ and A holds. Then $P(X<3) = k+2k = 0.3$ , so B holds while C fails since $P(0<X<3)$ is also $0.3$ . Finally $P(X>6) = 7k^2+k = 0.17$ , so D holds.

2025: 30 May Shift 2

A random variable X has the following probability distribution:

X	0	1	2	3
P(X)	0.2	0.1	0.3	0.4

The variance of X will be

2025: 27 May Shift 1

It is given that 3% of items manufactured by an industry are defective. The probability that a packet of 250 items contains one defective item is: [Given: $e^{-7.5} \approx 0.000553$ ]

2022: 6 Aug Shift 2

There are three identical boxes I, II and III, each containing two balls. In box I, both balls are red, In box II, both balls are blue and box III contains one blue ball and one red ball. A boy randomly chooses a box and takes out a ball at random from it. If the ball is red, then the probability that the other ball in the box is also red colour is:

2025: 14 May Shift 1

If A is any event associated with sample space and If $E_1, E_2, E_3$ are mutually exclusive and exhaustive events. Then which of the following are true?

(A) $P(A) = P(E_1)P(E_2|A) + P(E_2)P(E_2|A) + P(E_3)P(E_3|A)$

(B) $P(A) = P(A|E_1)P(E_1) + P(A|E_2)P(E_2) + P(A|E_3)P(E_3)$

(D) $P(A|E_i) = \frac{P(E_i|A)P(E_i)}{\sum_{i=1}^3 P(E_i|A)P(E_i)}$ , $i = 1,2,3$

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) $\frac{1}{6}$

2025: 2 June Shift 1

The probability distribution of a random variable $x$ is, $P(x) = \frac{k}{2^x}, x = 0, 1, 2, 3$ . Then

Match List-I with List-II

List-I	List-II
(A) $k$	(I) $\frac{2}{15}$
(B) $P(x = 1)$	(II) $\frac{1}{5}$
(C) $P(1 < x < 3)$	(III) $\frac{8}{15}$
(D) $P(x \geq 2)$	(IV) $\frac{4}{15}$

Choose the correct answer from the options given below:

Solution

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