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.
"n=5, p=60\\%=0.6, q=0.4\n\\\\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)\n\\\\=1-[P(X=0)+P(X=1)+P(X=2)]\n\\\\=\\ ^5C_0(0.6)^0(0.4)^5+\\ ^5C_1(0.6)^1(0.4)^4+\\ ^5C_2(0.6)^2(0.4)^3\n\\\\=0.31744"
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