Question
The mean and variance of a random variable X having binomial distribution are 4 and 2 respectively, then P ( X =1) is
BITSAT
Mathematics
Probability
2021
BITSAT 2021
BITSAT Mathematics Question (2021) — Solution
Options
- A. 1 4
- B. 1 32
- C. 1 16
- D. 1 8
Answer
B. 1 32
Step-by-step solution
aligned n p &=4 \\ n p q &=2 \\ q &= 1 2 , p= 1 2 , n=8 \\ P ( X &=1)= ^ 8 C _ 1 ( 1 2 ) ( 1 2 )^ 7 =8 1 2^ 8 = 1 2^ 5 = 1 32 aligned
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