CBSE Class 12 Sample Paper for 2020 Boards

Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

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

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### Transcript

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