Question

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?

Answer 1

Hint: 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 - \[P\left( {FFFFS} \right) = \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{1}{6}\]
So on,
Let \[P(A)\] be the probability of A’s winning
As obtaining a six by A and not a six by B are an independent event.
So, \[P\left( {A{\text{ }}} \right){\text{ }} = {\text{ }}P\left( S \right){\text{ }} + P\left( {FFS} \right) + P\left( {FFFFS} \right) + \ldots \ldots .\]
$ = \dfrac{1}{6} + \dfrac{1}{6}{\left( {\dfrac{5}{6}} \right)^2} + \dfrac{1}{6}{\left( {\dfrac{5}{6}} \right)^4} + ........$
$ = \dfrac{1}{6} \times \left[ {\dfrac{1}{{1 - {{\left( {\dfrac{5}{6}} \right)}^2}}}} \right]$ $ = \dfrac{1}{6} \times \left[ {\dfrac{1}{{1 - {{\left( {\dfrac{5}{6}} \right)}^2}}}} \right]$ [ $\left[ {\because a + ar + a{r^2} + ....nterm = \dfrac{a}{{1 - r,}}} \right]$
  by Geometric progression formula
$ \Rightarrow P(A) = \dfrac{{\dfrac{1}{6}}}{{1 - \dfrac{{25}}{{36}}}}$
$ \Rightarrow P(A) = \dfrac{1}{6} \times \dfrac{{36}}{{11}}$
$ \Rightarrow P(A) = \dfrac{6}{{11}}$
Hence, the probability of A’s winning is $\dfrac{6}{{11}}$.

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