Let "X=" the number of winnned rounds of voting: "X\\sim B(n, p)"
"P(X=x)=\\binom{n}{x}p^x(1-p)^{n-x}, x=0,1,2,...,n"Given that "n=4, p=0.6"
a. The probability that the ruling party will win 0 rounds is
"P(X=0)=\\binom{4}{0}0.6^0(1-0.6)^{4-0}=0.4^4=""=0.0256" The probability that the ruling party will win 1 round is
"P(X=1)=\\binom{4}{1}0.6^1(1-0.6)^{4-1}=4\\cdot0.6\\cdot0.4^3=""=0.1536" The probability that the ruling party will win 2 rounds is
"P(X=2)=\\binom{4}{2}0.6^2(1-0.6)^{4-2}=6\\cdot0.6^2\\cdot0.4^2=""=0.3456" The probability that the ruling party will win 3 rounds is
"P(X=3)=\\binom{4}{3}0.6^3(1-0.6)^{4-3}=4\\cdot0.6^3\\cdot0.4=""=0.3456" The probability that the ruling party will win 4 rounds is
"P(X=4)=\\binom{4}{4}0.6^4(1-0.6)^{4-4}=0.6^4=""=0.1296" b.The probability that the ruling party will win at least 1 round is
"P(X\\geq1)=1-P(X=0)=""=1-\\binom{4}{0}0.6^0(1-0.6)^{4-0}=""=1-0.4^4 =1-0.0256=0.9744"
Comments
Leave a comment