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Question: A & B roll a die successively till one of them gets a six. What are the chances of A’s winning if he...

A & B roll a die successively till one of them gets a six. What are the chances of A’s winning if he starts the game?

Explanation

Solution

We will have to consider no. of trials & events in them to get the sum of probabilities of all those in such a way so that we can get probability of A’s winning. While adding all the probabilities apply the formula of G.P.

Complete step-by-step answer: Let’s assume S & F be the events of success (getting 6 on die) & failure (not getting 6 on die) of A respectively. Where $$$$$$$$$$P\left( S \right) = 1/6;;,, P\left( F \right){\text{ }} = {\text{ }}5/6 Now, A can get success & failures in the following way A got six in 1st trial, of which probability will be -P\left( S \right) = 1/6;; A fail to get six then B fails to get six & then A got six, probability of this will be – $P(FFS) = \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{1}{6}$ A fails to get six then B fails to get six , again A fails to get six then B fails to get six & then finally A got six , ,of which probability will be -P\left( {FFFFS} \right) = \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{1}{6}Soon,Let So on, LetP(A)betheprobabilityofAswinningAsobtainingasixbyAandnotasixbyBareanindependentevent.So,be the probability of A’s winning As obtaining a six by A and not a six by B are an independent event. So,P\left( {A{\text{ }}} \right){\text{ }} = {\text{ }}P\left( S \right){\text{ }} + P\left( {FFS} \right) + P\left( {FFFFS} \right) + \ldots \ldots .$$
=16+16(56)2+16(56)4+........= \dfrac{1}{6} + \dfrac{1}{6}{\left( {\dfrac{5}{6}} \right)^2} + \dfrac{1}{6}{\left( {\dfrac{5}{6}} \right)^4} + ........
=16×[11(56)2]= \dfrac{1}{6} \times \left[ {\dfrac{1}{{1 - {{\left( {\dfrac{5}{6}} \right)}^2}}}} \right] =16×[11(56)2] = \dfrac{1}{6} \times \left[ {\dfrac{1}{{1 - {{\left( {\dfrac{5}{6}} \right)}^2}}}} \right] [ [a+ar+ar2+....nterm=a1r,]\left[ {\because a + ar + a{r^2} + ....nterm = \dfrac{a}{{1 - r,}}} \right]
by Geometric progression formula
P(A)=1612536\Rightarrow P(A) = \dfrac{{\dfrac{1}{6}}}{{1 - \dfrac{{25}}{{36}}}}
P(A)=16×3611\Rightarrow P(A) = \dfrac{1}{6} \times \dfrac{{36}}{{11}}
P(A)=611\Rightarrow P(A) = \dfrac{6}{{11}}
Hence, the probability of A’s winning is 611\dfrac{6}{{11}}.

Note: Remember formula of G.P to be applied, i.e., [a+ar+ar2+....nterm=a1r,]\left[ {\because a + ar + a{r^2} + ....nterm = \dfrac{a}{{1 - r,}}} \right]& calculations should be done attentively.