The mean and the variance of a binomial distribution are 4 and 2 respectively. Find the probability of atleast 6 successes.

$\frac{37}{256}$

The correct answer is Option (4) → $\frac{37}{256}$

According to given, we have

mean $= 4 ⇒ np = 4$   ...(i)

and variance $= 2 ⇒ npq = 2$   ...(ii)

From (i) and (ii), we get

$4.q = 2⇒q = \frac{1}{2}$, so $p = 1-q=1-\frac{1}{2}=\frac{1}{2}$

From (i), we get $n.\frac{1}{2}= 4⇒n=8$.

If $r$ denotes the number of successes, then

$P(r) = {^nC}_rp^r q^{n-r} = {^8C}_r (\frac{1}{2})^r (\frac{1}{2})^{8-r} = {^8C}_r (\frac{1}{2})^8$.

∴ The probability of atleast 6 successes

$= P(6) + P(7) + P(8)$

$= {^8C}_6(\frac{1}{2})^8+{^8C}_7(\frac{1}{2})^8+{^8C}_8(\frac{1}{2})^8$

$=(\frac{1}{2})^8({^8C}_2+{^8C}_1+{^8C}_0)$

$=\frac{1}{256}(28+8+1)=\frac{37}{256}$