Question
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?
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 $$$$$$$$$$P\left( S \right) = 1/6;;,P\left( F \right){\text{ }} = {\text{ }}5/6
Now, A can get success & failures in the following way
A got six in 1st trial, of which probability will be -P\left( S \right) = 1/6;;
A fail to get six then B fails to get six & then A got six, probability of this will be –
$P(FFS) = \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{1}{6}$
A fails to get six then B fails to get six , again A fails to get six then B fails to get six & then finally A got six , ,of which probability will be -P\left( {FFFFS} \right) = \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{1}{6}Soon,LetP(A)betheprobabilityofA’swinningAsobtainingasixbyAandnotasixbyBareanindependentevent.So,P\left( {A{\text{ }}} \right){\text{ }} = {\text{ }}P\left( S \right){\text{ }} + P\left( {FFS} \right) + P\left( {FFFFS} \right) + \ldots \ldots .$$
=61+61(65)2+61(65)4+........
=61×1−(65)21 =61×1−(65)21 [ [∵a+ar+ar2+....nterm=1−r,a]
by Geometric progression formula
⇒P(A)=1−362561
⇒P(A)=61×1136
⇒P(A)=116
Hence, the probability of A’s winning is 116.
Note: Remember formula of G.P to be applied, i.e., [∵a+ar+ar2+....nterm=1−r,a]& calculations should be done attentively.