Let x denote the number of doublets in three throws of a pair of dice with the following probability distribution., x, 0, 1, 2, 3, P(x), 25/72k, 15/72k, 3/72k, 1/360k, If value of k is equal to m/n * (m,n) = 1, then m + n is equal to Option 1: 8 Option 2: 19 Option 3: 16 Option 4: 18

Answer: 8. Solution: Since this is a valid probability distribution, all probabilities must sum to 1. 25k/72 + 15k/72 + 3k/72 + k/360 = 1 Taking k common: k(25/72 + 15/72 + 3/72 + 1/360) = 1 LCM of 72 and 360 is 360: 25/72 = 125/360, 15/72 = 75/360, 3/72 = 15/360, 1/360 = 1/360 k x (125 + 75 + 15 + 1)/360 = 1 k x 216/360 = 1 k x 3/5 = 1 k = 5/3 Here m = 5 and n = 3, with (5, 3) = 1. m + n = 5 + 3 = 8

Algebra

Probability

Medium

common

Let x denote the number of doublets in three throws of a pair of dice with the following probability distribution.

x 0 1 2 3
P(x) $\frac{25}{72}k$ $\frac{15}{72}k$ $\frac{3}{72}k$ $\frac{1}{360}k$

If value of k is equal to $\frac{m}{n} \cdot \gcd(m,n) = 1$ , then $m + n$ is equal to

x	0	1	2	3
P(x)	$\frac{25}{72}k$	$\frac{15}{72}k$	$\frac{3}{72}k$	$\frac{1}{360}k$

✅ Correct Option: 1

Since this is a valid probability distribution, all probabilities must sum to 1.

$\frac{25k}{72} + \frac{15k}{72} + \frac{3k}{72} + \frac{k}{360} = 1$

Taking $k$ common:

$k\left(\frac{25}{72} + \frac{15}{72} + \frac{3}{72} + \frac{1}{360}\right) = 1$

LCM of 72 and 360 is 360:

$\frac{25}{72} = \frac{125}{360}, \quad \frac{15}{72} = \frac{75}{360}, \quad \frac{3}{72} = \frac{15}{360}, \quad \frac{1}{360} = \frac{1}{360}$

$k \times \frac{125 + 75 + 15 + 1}{360} = 1$

$k \times \frac{216}{360} = 1$

$k \times \frac{3}{5} = 1$

$k = \dfrac{5}{3}$

Here $m = 5$ and $n = 3$ , with $\gcd(5, 3) = 1$ .

$m + n = 5 + 3 = 8$

2025: 15 May Shift 2

If a random variable X has the following probability distribution:

X	0	1	2	3
P(X)	K	K/2	K/4	K/8

then,

Match List-I with List-II

List-I	List-II
(A) The value of K is	(I) 2/15
(B) P(0 < X < 2) is	(II) 1/15
(C) P(1 < X < 3) is	(III) 8/15
(D) P(X > 2) is	(IV) 4/15

Choose the correct answer from the options given below:

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The probability that seed is of type $A_2$ given that seed germinate.

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A random variable X has the following probability distribution:

X	2	3	4	5
P(X)	5/k	7/k	9/k	11/k

Then the value of $\frac{k}{4}$ is:

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In a game, a man wins Rs. 5 for getting a number greater than 4 and loses Rs. 1 otherwise, when a fair dice is thrown. The man decided to throw a die thrice but to quit as and when he gets a number greater than 4. The expected value of the amount(in Rs.) he wins or loses is:

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The probability distribution of number of doublets in three throws of a pair of dice is

Solution

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