Question

A best of 9 series is to be played between two teams ${T_1}$ and ${T_2}$, that is, the first team to win 5 games is the winner. The team ${T_1}$ has a chance of $\dfrac{2}{3}$ of winning any game. Let $P = \dfrac{a}{b}$ be the probability (expressed as lowest rational) that exactly 7 games will need to be played to determine a winner, find $\left( {a + b} \right)$.

Answer 1

Hint: Team 1 wins with a probability of $\dfrac{2}{3}$. Thus, the probability that team 1 loses a match will be equal to $\dfrac{1}{3}$, this also represents the probability of winning the match by team 2. Since, the winner will be declared in the 7th match. Thus, the probability of the winner being declared in the 7th match will be equal to the sum of the probability of winning 5 matches by team 1 and the probability of winning 5 matches by team 2. From there, we will get the required value of $a$ and $b$ hence the sum.

Complete step by step solution:
Given:
Probability of winning game by team 1$ = \dfrac{2}{3}$
Now, we will find the probability of losing the game by team 1.
Thus, Probability of losing game by team 1$ = 1 - \dfrac{2}{3} = \dfrac{1}{3}$
Here, the probability of losing game by team 1 is equal to the Probability of winning game by team 2.
Therefore,
Probability of winning game by team 2$ = \dfrac{1}{3}$
We know the winner will be declared in the 7thgame.
Thus, the probability of winner being declared in 7th game$
= {\text{probability of team 1 winning 4 games out of 6 and winning }}{{\text{7}}^{{\text{th}}}}{\text{ games + }} \\
{\text{probability of team 2 winning 4 games out of 6 and winning }}{{\text{7}}^{{\text{th}}}}{\text{ games}}.................\left( 1 \right) \\
$
Here, ${\text{probability of team 1 winning 4 games out of 6 and winning }}{{\text{7}}^{{\text{th}}}}{\text{ games}} = {}^6{C_4} \times {\left( {\dfrac{2}{3}} \right)^4} \times {\left( {\dfrac{1}{3}} \right)^2} \times \left( {\dfrac{2}{3}} \right)$
We know the combination formula;
${}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$
Therefore,
${\text{probability of team 1 winning 4 games out of 6 and winning }}{{\text{7}}^{{\text{th}}}}{\text{ games}} = \dfrac{{6!}}{{\left( {6 - 4} \right)!4!}} \times {\left( {\dfrac{2}{3}} \right)^4} \times {\left( {\dfrac{1}{3}} \right)^2} \times \left( {\dfrac{2}{3}} \right)$
On further simplification, we get
${\text{probability of team 1 winning 4 games out of 6 and winning }}{{\text{7}}^{{\text{th}}}}{\text{ games}} = \dfrac{{160}}{{729}}$
Here, ${\text{probability of team 2 winning 4 games out of 6 and winning }}{{\text{7}}^{{\text{th}}}}{\text{ games}} = {}^6{C_4} \times {\left( {\dfrac{1}{3}} \right)^4} \times {\left( {\dfrac{2}{3}} \right)^2} \times \left( {\dfrac{1}{3}} \right)$
We know the combination formula;
${}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$
Therefore,
${\text{probability of team 2 winning 4 games out of 6 and winning }}{{\text{7}}^{{\text{th}}}}{\text{ games}} = \dfrac{{6!}}{{\left( {6 - 4} \right)!4!}} \times {\left( {\dfrac{1}{3}} \right)^4} \times {\left( {\dfrac{2}{3}} \right)^2} \times \left( {\dfrac{1}{3}} \right)$
On further simplification, we get
${\text{probability of team 2 winning 4 games out of 6 and winning }}{{\text{7}}^{{\text{th}}}}{\text{ games}} = \dfrac{{20}}{{729}}$
Putting values of probability obtained in equation 1, we get
Thus, the probability of winner being declared in 7th game$ = \dfrac{{160}}{{729}} + \dfrac{{20}}{{729}} = \dfrac{{180}}{{729}} = \dfrac{{20}}{{81}}$
According to question,
The probability of winner being declared in 7th game$ = \dfrac{a}{b}$
Therefore,
$
  a = 20 \\
  b = 81 \\
$
Thus, the value of $\left( {a + b} \right) = 20 + 81 = 101$
The required value of $\left( {a + b} \right)$ is 101.

Note: Here we have calculated the value of terms like${}^n{C_m} = \dfrac{{n!}}{{m! \times \left( {n - m} \right)!}}$, this is the ratio of factorial of the terms.
We need to know the meaning of the factorial for solving problems like that.
(i)Factorial of any positive integer is defined as the multiplication of all the positive integers less than or equal to the given positive integers.
(ii)Factorial of zero is one.
(iii)Factorials are commonly used in permutations and combinations problems.
(iv)Factorials of negative integers are not defined