Question

A survey from the private bank found that 40% of adult client received their dividend from their saving. If 6 adult clients are selected at random, find the probability that at least 4 of them will received the dividend.

Answer 1

P(Adult client received their dividend from their saving) = 0.4

Number sample(n) = 6

We need to calculate probability that at least 4 of them will received the dividend, i.e. $P(X\geq4) = ?$

Here we will use binomial distribution probability which can be calculated as

$P(X=n ; N,p) = {N\choose n}*(p^n)*(1-p)^{(N-n)}$

$P(X\geq4) = P(X=4)+ P(X=5) + P(X=6) \\ = {6\choose4}*(0.4^4)*(1-0.4)^{(6-4)} + {6\choose5}*(0.4^5)*(1-0.4)^{(6-5)} \\+ {6\choose6}*(0.4^6)*(1-0.4)^{(6-6)} \\ = 0.1382 + 0.0368 + 0.0040 = 0.179$

So there is 17.9% probability that at least 4 of them will receive the dividend.