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 Oct. 24, 2019 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.

Note
: - This
is similar to
Ex 13.4, 12 of NCERT – Chapter 13 Class 12

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
๐=๐ฌ(๐ฟ)=โ_(๐=1)^๐ ๐ฅ๐๐๐
= 1 ร 6/21+"2 ร " 5/21+ 3 ร 4/21+ 4 ร 3/21+ 5 ร 2/21 + 6 ร 1/21
= (6 + 10 + 12 + 12 + 10 + 6)/21
= 56/21
= ๐/๐

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.