The mean and variance of a binomial variable   X are 2 and 1 respectively. The probability that X takes…

The mean and variance of a binomial variable X are 2 and 1 respectively. The probability that X takes values greater than 1, is
  1. 516
  2. 816
  3. 1116
  4. 116

Solution

Given, mean np=2               ...(i)

And variance npq=1             ...(ii)

From Eqs.(i) and (ii), we get

q=12

∴ p=1-q=12

From Eq.(i), n×12=2

∴  n=4

The binomial distribution is 12+124

Now, PX>1=PX=2+PX=3+P(X=4)

= 4C2122122+4C3123121+4C4124

=6+4+116

=1116

Asked in: AP EAMCET 2021 (20 Aug Shift 1)

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