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 April 16, 2024 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
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

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo

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