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A and B play for a prize; A is to throw a die first, and is to win if he throws 6. If he fails B is to throw, and to win if he throws 6 or 5. If he fails, A is to throw again and to win with 6 or 5 or 4 and so on. Find the chance of each player.

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Last updated date: 27th Mar 2024
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MVSAT 2024
Answer
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Hint: We will calculate the winning chance of player A and player B individually. We have to do the calculation throw by throw individually for both players. We will start by considering the case of player A first - he wins in the first throw, he wins after B fails or he wins in his next chance. The probabilities in each case can be multiplied to get the total probability of A winning. Same can be repeated for B.

Complete step-by-step answer:
It is given in the question that two players A and B play for a prize. Also, this game has a sequence or we can say that this game has a rule that both of them have to follow.

At first player A will throw the die and if he gets 6, he wins and if the outcome is different from 6 then player B will throw the die. Now, for player B, if he throws the die and he gets outcomes as 5 or 6, then he wins otherwise A will again throw the die and this time for player A the winning outcomes are 4, 5 or 6 and this game will continue in the same pattern.
If we simplify this, then we can say that if A gets 6 in the first throw, then he wins or if he fails to get 6, then B will throw and if B gets 5 or 6, then B wins. But if he fails, then again A will throw the die and this time if A gets 4, 5 or 6 then A will win and this will continue in the same sequence.

So, we have to find the chance of winning for each player. Now, we will calculate the winning probability of player A as:
Probability A will win in the first throw \[P{{\left( A \right)}_{1}}=\dfrac{1}{6}\].
But if A fails to get 6 then B will throw and if B fails then again A gets a second chance to win this game. The probability of A to win this game in second chance is \[P{{\left( A \right)}_{2}}=\dfrac{5}{6}\times \dfrac{4}{6}\times \dfrac{3}{6}\] .
Again if A fails in the second chance then B will throw the die and if B also fails then A will get the third chance to throw the die. So, probability of winning of player A in third chance \[P{{\left( A \right)}_{3}}=\dfrac{5}{6}.\dfrac{4}{6}.\dfrac{3}{6}.\dfrac{2}{6}.\dfrac{5}{6}\].
Or, we can say that probability of winning of player A in all three throw $=P{{\left( A \right)}_{1}}+P{{\left( A \right)}_{2}}+P{{\left( A \right)}_{3}}$. On adding the value of $P{{\left( A \right)}_{1}}+P{{\left( A \right)}_{2}}+P{{\left( A \right)}_{3}}$respectively, we get,
$\begin{align}
  & P\left( A \right)=\dfrac{1}{6}+\dfrac{5}{6}.\dfrac{4}{6}.\dfrac{3}{6}+\dfrac{5}{6}.\dfrac{4}{6}.\dfrac{3}{6}.\dfrac{2}{6}.\dfrac{5}{6} \\
 & P\left( A \right)=\dfrac{1}{6}+\dfrac{5}{18}+\dfrac{25}{324} \\
\end{align}$
Taking LCM of 6, 18 and 324, we get;
$\begin{align}
  & P\left( A \right)=\dfrac{54+90+25}{324} \\
 & P\left( A \right)=\dfrac{169}{324} \\
\end{align}$

Or, we get the probability of winning from player A is $\dfrac{169}{324}$.
Similarly, we will calculate the winning probability of player B. So, player B can get his first throw only if A fails to get 6 in his first throw. Or, the probability of winning of player B in first throw $P{{\left( B \right)}_{1}}=\dfrac{5}{6}.\dfrac{2}{6}$.
Again, if the player A loses twice then only player B will get a second chance. Or, the probability of winning of player B in second throw $P{{\left( B \right)}_{2}}=\dfrac{5}{6}.\dfrac{4}{6}.\dfrac{3}{6}.\dfrac{4}{6}$.
Similarly, if the player A loses thrice then only player B will get a third chance. Or, the probability of winning of player B in third throw $P{{\left( B \right)}_{3}}=\dfrac{5}{6}.\dfrac{4}{6}.\dfrac{3}{6}.\dfrac{2}{6}.\dfrac{1}{6}.1$.
Or, we can say that probability of winning of player B in all three throw is given as, $P{{\left( B \right)}_{1}}+P{{\left( B \right)}_{2}}+P{{\left( B \right)}_{3}}$.
Or, we get $P\left( B \right)=P{{\left( B \right)}_{1}}+P{{\left( B \right)}_{2}}+P{{\left( B \right)}_{3}}$. On adding the value of $P{{\left( B \right)}_{1}}+P{{\left( B \right)}_{2}}+P{{\left( B \right)}_{3}}$ we get,
$\begin{align}
  & P\left( B \right)=\dfrac{5}{6}.\dfrac{2}{6}+\dfrac{5}{6}.\dfrac{4}{6}.\dfrac{3}{6}.\dfrac{4}{6}+\dfrac{5}{6}.\dfrac{4}{6}.\dfrac{3}{6}.\dfrac{2}{6}.\dfrac{1}{6}.1 \\
 & P\left( B \right)=\dfrac{5}{18}+\dfrac{1}{27}+\dfrac{5}{324} \\
\end{align}$
Taking LCM of 18, 27 and 324, we get,
$\begin{align}
  & P\left( B \right)=\dfrac{90+12+5}{324} \\
 & P\left( B \right)=\dfrac{107}{324} \\
\end{align}$
Or, we get probability of winning player B is $\dfrac{107}{324}$.
Thus, we have probability of player A to win this game is $\dfrac{169}{324}$ whereas the probability of player B to win this game is $\dfrac{107}{324}$.

Note: Do all the calculations throw by throw to avoid mistakes in such problems. Please note that students often skip to consider the chances of losing the opposite players. But we have to consider both chances of winning and losing for both player A and B for each time to get the correct answer.