Probability 5 Question 2

2. Let a random variable X have a binomial distribution with mean 8 and variance 4 . If P(X2)=k216, then k is equal to

(2019 Main, 12 April I)

(a) 17

(b) 121

(c) 1

(d) 137

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Answer:

Correct Answer: 2. (d)

Solution:

  1. Let for the given random variable ’ X ’ the binomial probability distribution have n-number of independent trials and probability of success and failure are p and q respectively. According to the question, Mean =np=8 and variance =npq=4

q=12p=1q=12 Now, n×12=8n=16P(X=r)=16Cr1216P(X2)=P(X=0)+P(X=1)+P(X=2)16C01216+16C11216+16C21216=1+16+120216=137216=k216k=137

(given)



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