In a Binomial distribution consisting of 5 independent trials, probabilities of exactly 1 and 2 successes are 0.4096 and 0.2048 respectively, then the probability, of getting exactly 4 successes, is

Step-by-step solution

Let P ( X =1) be probability of one success and P ( X =2) be probability of two success array ll & P(X=1)= ^5 C ₁ p ^1 q ^4=0.4096 ...(i) & P ( X =2)= ^5 C ₂ p ^2 q ^3=0.2048...(ii) array Where p= probability of success q = probability of failure Dividing (i) by (ii), we get aligned & ^5 C ₁ pq ^4 ^5 C ₂ p ^2 q ^3 = 0.4096 0.2048 & 5 q 10 p =2 & q =4 p & Also, p + q =1 & p+4 p=1 aligned array ll & p= 1 5 & & q= 4 5 array Now, Probability of getting 4 successes aligned & = P ( X =4) & = ^5 C ₄ p ^4 q & =5 ( 1 5 )^4