Question 31 (OR 1st Question) - CBSE Class 12 Sample Paper for 2020 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at March 23, 2023 by Teachoo

Two numbers are selected at random (without replacement) from first 7 natural numbers. If X denotes the smaller of the two numbers obtained, find the probability distribution of X. Also, find mean of the distribution.

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Question 31 (OR 1st Question) Two numbers are selected at random (without replacement) from first 7 natural numbers. If X denotes the smaller of the two numbers obtained, find the probability distribution of X. Also, find mean of the distribution.
Let X : be the smaller of two numbers obtained
The possible outcomes are
Sample space = S = {(1, 2),(1, 3),(1, 4),(1, 5),(1, 6),(1, 7)(2, 1),(2, 3),(2, 4),(2, 5),(2, 6),(2, 7) ) ((3, 1),(3, 2),(3, 4),(3, 5),(3, 6),(3, 7)(4, 1),(4, 2),(4, 3),(4, 5),(4, 6),(4, 7) )((5, 1),(5, 2),(5, 3),(5, 4),(5, 6),(5, 7)(6, 1),(6, 2),(6, 3),(6, 4),(6, 5),(6, 7)(7, 1),(7, 2), (7, 3), (7, 4), (7, 5), (7, 6)))}
Total number of possible outcomes = 42
The smaller number can be: 1, 2, 3, 4, 5 or 6
So, the values of X can be : 1, 2, 3, 4, 5 or 6
{(1, 2),(1, 3),(1, 4),(1, 5),(1, 6),(1, 7),@(2, 1),(3, 1),(4, 1),(5, 1),(6, 1),(7, 1) )}
{(2, 3),(2, 4),(2, 5),(2, 6),(2, 7),@(3, 2), (4, 2),(5, 2),(6, 2),(7, 2))}
{(3, 4),(3, 5),(3, 6),(3, 7),@(4, 3),(5, 3),(6, 3),(7, 3) )}
{(4, 5),(4, 6),(4, 7),@(5, 4),(6, 4),(7, 4) )}
{(5, 6),(5, 7),@(6, 5), (7, 5) )}
{(6, 7),@(7, 6))}
12
12/42 =6/21
10
Thus, the probability distribution is
The mean Expected value is given by
𝝁=𝑬(𝑿)=∑_(𝑖=1)^𝑛 𝑥𝑖𝑃𝑖
= 1 × 6/21+"2 × " 5/21+ 3 × 4/21+ 4 × 3/21+ 5 × 2/21 + 6 × 1/21
= (6 + 10 + 12 + 12 + 10 + 6)/21
= 56/21
= 𝟖/𝟑

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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