Question

Alice, Bob, and Carol repeatedly take turns tossing a die. Alice begins; Bob always follows Alice, Carol always follows Bob, and Alice always follows Carol. Find the probability that Carol will be the first one to toss a six. (The probability of obtaining a six on any toss is $\dfrac{1}{6}$, independent of the outcome of any other toss.)
A. $\dfrac{1}{3}$
B. $\dfrac{2}{9}$
C. $\dfrac{5}{{18}}$
D. $\dfrac{{25}}{{91}}$

Answer 1

Hint: Find the probability that Carol wins in the first round. After that, find the probability that Carol wins in the second round and so on. After that sum all the probabilities since she can win in any round. Then take-out the common part, the remaining part will form GP. Apply the formula of the sum of GP of infinite terms, $S = \dfrac{a}{{1 - r}}$. After that do a calculation to get the desired result.

Complete step-by-step solution:
Since the probability of obtaining a six on a toss is $\dfrac{1}{6}$.
Then, the probability of not obtaining a six on a toss is,
$ \Rightarrow 1 - \dfrac{1}{6}$
Take LCM of the terms,
$ \Rightarrow \dfrac{{6 - 1}}{6}$
Subtract the terms in the numerator,
$ \Rightarrow \dfrac{5}{6}$
As it is given that first Alice will toss, then Bob will toss and then Carol will toss.
If Carol wins in the first round, then she must have rolled a six after two non-sixes have occurred by Alice and Bob. The probability will be,
\[ \Rightarrow \left( {\dfrac{5}{6}} \right)\left( {\dfrac{5}{6}} \right)\left( {\dfrac{1}{6}} \right) = {\left( {\dfrac{5}{6}} \right)^2}\left( {\dfrac{1}{6}} \right)\]
If Carol wins in the second round, then she must have rolled a six after five non-sixes have occurred. The probability will be,
\[ \Rightarrow \left( {\dfrac{5}{6}} \right)\left( {\dfrac{5}{6}} \right)\left( {\dfrac{5}{6}} \right)\left( {\dfrac{5}{6}} \right)\left( {\dfrac{5}{6}} \right)\left( {\dfrac{1}{6}} \right) = {\left( {\dfrac{5}{6}} \right)^5}\left( {\dfrac{1}{6}} \right)\]
Similarly, the probability of Carol winning in the third round is \[{\left( {\dfrac{5}{6}} \right)^8}\left( {\dfrac{1}{6}} \right)\] and wining in the fourth round is \[{\left( {\dfrac{5}{6}} \right)^{11}}\left( {\dfrac{1}{6}} \right)\] and so on.
It is possible that the game could continue forever. The probability that Carol wins the game is equal to the sum of the probabilities that she wins in any given round,
$ \Rightarrow P = {\left( {\dfrac{5}{6}} \right)^2}\left( {\dfrac{1}{6}} \right) + {\left( {\dfrac{5}{6}} \right)^5}\left( {\dfrac{1}{6}} \right) + {\left( {\dfrac{5}{6}} \right)^8}\left( {\dfrac{1}{6}} \right) + {\left( {\dfrac{5}{6}} \right)^{11}}\left( {\dfrac{1}{6}} \right) + \ldots $
Take \[{\left( {\dfrac{5}{6}} \right)^2}\left( {\dfrac{1}{6}} \right)\] common on the right side,
$ \Rightarrow P = {\left( {\dfrac{5}{6}} \right)^2}\left( {\dfrac{1}{6}} \right)\left( {1 + {{\left( {\dfrac{5}{6}} \right)}^3} + {{\left( {\dfrac{5}{6}} \right)}^6} + {{\left( {\dfrac{5}{6}} \right)}^9} + \ldots } \right)$
Since, the terms inside the bracket form a GP, whose first term is 1 and the common ratio is ${\left( {\dfrac{5}{6}} \right)^3}$.
The formula of the sum of GP of infinite terms is,
$S = \dfrac{a}{{1 - r}}$
Use this formula to calculate the sum,
$ \Rightarrow P = {\left( {\dfrac{5}{6}} \right)^2}\left( {\dfrac{1}{6}} \right) \times \dfrac{1}{{1 - {{\left( {\dfrac{5}{6}} \right)}^3}}}$
Take LCM in the denominator,
$ \Rightarrow P = \dfrac{{{5^2}}}{{{6^3}}} \times \dfrac{1}{{\dfrac{{{6^3} - {5^3}}}{{{6^3}}}}}$
Move ${6^3}$ in the numerator,
$ \Rightarrow P = \dfrac{{{5^2}}}{{{6^3}}} \times \dfrac{{{6^3}}}{{{6^3} - {5^3}}}$
Simplify the terms,
$ \Rightarrow P = 25 \times \dfrac{1}{{216 - 125}}$
Subtract the value in the denominator,
$ \Rightarrow P = \dfrac{{25}}{{91}}$

Hence, the option(D) is the correct answer.

Note: Probability is a measure of the likelihood of an event to occur. Many events cannot be predicted with total certainty. We can predict only the chance of an event to occur i.e. how likely they are to happen, using it.
The probability formula is defined as the possibility of an event to happen is equal to the ratio of the number of favorable outcomes and the total number of outcomes.
$P\left( E \right) = \dfrac{{n\left( E \right)}}{{n\left( S \right)}}$
A probability of 0 means that an event is impossible.
A probability of 1 means that an event is certain.
An event with a higher probability is more likely to occur.
Probabilities are always between 0 and 1.