Answer to Question #99703 in Statistics and Probability for Nishit Tuli

Question

Lok Sabha in India is to discuss a key constitutional amendment, and there will be four
rounds of voting. In each round of voting, assume that the ruling party has a 60% chance
of winning, i.e. getting the resolution passed by majority vote. Further assuming that the
voting rounds are independent of each other, what is the probability that:
a. The ruling party will win 0 rounds, 1 round, 2 rounds, 3 rounds or all 4 rounds of
voting? b.The ruling party will win at least 1 round?

Answer 1

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

$$P(X\geq1)=1-P(X=0)=$$ $$=1-\binom{4}{0}0.6^0(1-0.6)^{4-0}=$$ $$=1-0.4^4 =1-0.0256=0.9744$$