Question

A and B throw a coin alternately till one of them gets a ‘head’ and wins the game. Find their respective probabilities of winning . -Maths 9th

Answer 1

Let A : Event of A getting a head ⇒ \(\bar{A}\) : Event of A not getting a head ∴ P(A) = \(rac{1}{2}\) and P(\(\bar{A}\)) = 1 - \(rac{1}{2}\) = \(rac{1}{2}\)Similarly, B : Event of B getting a head \(\bar{B}\) : Event of B not getting a head.So P(B) = \(rac{1}{2}\) and P(\(\bar{B}\)) = 1 - \(rac{1}{2}\) = \(rac{1}{2}\).Let A start the game. A's wins if he throws a head in the 1st throw or 3rd throw or 5th throw or throw and so on.Probability of A's winning in 1st throw = P(A) = \(rac{1}{2}\)Probability of A's winning in 3rd throw = P(not A). P(not B). P(A) = P(\(\bar{A}\)) x P(\(\bar{B}\)) x P(A)= \(rac{1}{2}\)x \(rac{1}{2}\)x \(rac{1}{2}\) = \(\big(rac{1}{2}\big)^3\)Similarly, probability of A's winning in 5th throw =  P(\(\bar{A}\)) x P(\(\bar{B}\)) x  P(\(\bar{A}\)) x P(\(\bar{B}\)) x P(A)= \(rac{1}{2}\)x \(rac{1}{2}\)x \(rac{1}{2}\)x \(rac{1}{2}\)x \(rac{1}{2}\) = \(\big(rac{1}{2}\big)^5\) and so on. Since all these events are mutually exclusive,p(A winning the game first) = \(rac{1}{2}\) + \(\big(rac{1}{2}\big)^3\) + \(\big(rac{1}{2}\big)^5\) + \(\big(rac{1}{2}\big)^7\) ... = \(rac{1}{2}\)\(\bigg[\) 1 + \(\big(rac{1}{2}\big)^2\) + \(\big(rac{1}{2}\big)^4\) + \(\big(rac{1}{2}\big)^6\) + ...\(\bigg]\)= \(rac{1}{2}\)\(\bigg[rac{1}{1-\big(rac{1}{2}\big)^2}\bigg]\) = \(rac{1}{2}\) x \(rac{1}{rac{3}{4}}\) = \(rac{1}{2}\) x \(rac{4}{3}\) = \(rac{2}{3}\).(∵ For an infinite GP, S∞ = \(rac{ ext{First term}}{ ext{1-common ratio}}\))Since A and B are mutually exclusive events, as either of them will win, P(B winning the game first) = 1 - \(rac{2}{3}\) = \(rac{1}{3}\).