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.
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) \\\\\n\n= {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)} \\\\\n\n= 0.1382 + 0.0368 + 0.0040 = 0.179"
So there is 17.9% probability that at least 4 of them will receive the dividend.
Comments
Leave a comment