Question

60% of people those who purchase life insurance are women. If the owners of 5 life

insurance are randomly selected, then find the probability that there are

(i) no women

(ii) exactly 3 women

(iii)more than 2 women.

Answer 1

Solution:

$n=5, p=60\%=0.6, q=0.4 \\X\sim Bin(n,p)$

(i)

$P(X=0)=\ ^5C_0(0.6)^0(0.4)^5=0.01024$

(ii)

$P(X=3)=\ ^5C_3(0.6)^3(0.4)^2=0.3456$

(iii)

$P(X>2)=1-P(X\le 2) \\=1-[P(X=0)+P(X=1)+P(X=2)] \\=\ ^5C_0(0.6)^0(0.4)^5+\ ^5C_1(0.6)^1(0.4)^4+\ ^5C_2(0.6)^2(0.4)^3 \\=0.31744$