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Question: A best of 9 series is to be played between two teams \({T_1}\) and \({T_2}\), that is, the first tea...

A best of 9 series is to be played between two teams T1{T_1} and T2{T_2}, that is, the first team to win 5 games is the winner. The team T1{T_1} has a chance of 23\dfrac{2}{3} of winning any game. Let P=abP = \dfrac{a}{b} be the probability (expressed as lowest rational) that exactly 7 games will need to be played to determine a winner, find (a+b)\left( {a + b} \right).

Explanation

Solution

Team 1 wins with a probability of 23\dfrac{2}{3}. Thus, the probability that team 1 loses a match will be equal to 13\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 aa and bb hence the sum.

Complete step by step solution:
Given:
Probability of winning game by team 1=23 = \dfrac{2}{3}
Now, we will find the probability of losing the game by team 1.
Thus, Probability of losing game by team 1=1−23=13 = 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=13 = \dfrac{1}{3}
We know the winner will be declared in the 7thgame.
Thus, the probability of winner being declared in 7th game=probability of team 1 winning 4 games out of 6 and winning 7th games +  probability of team 2 winning 4 games out of 6 and winning 7th games.................(1)  = {\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, probability of team 1 winning 4 games out of 6 and winning 7th games=6C4×(23)4×(13)2×(23){\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;
nCr=n!(n−r)!r!{}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}
Therefore,
probability of team 1 winning 4 games out of 6 and winning 7th games=6!(6−4)!4!×(23)4×(13)2×(23){\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
probability of team 1 winning 4 games out of 6 and winning 7th games=160729{\text{probability of team 1 winning 4 games out of 6 and winning }}{{\text{7}}^{{\text{th}}}}{\text{ games}} = \dfrac{{160}}{{729}}
Here, probability of team 2 winning 4 games out of 6 and winning 7th games=6C4×(13)4×(23)2×(13){\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;
nCr=n!(n−r)!r!{}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}
Therefore,
probability of team 2 winning 4 games out of 6 and winning 7th games=6!(6−4)!4!×(13)4×(23)2×(13){\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
probability of team 2 winning 4 games out of 6 and winning 7th games=20729{\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=160729+20729=180729=2081 = \dfrac{{160}}{{729}} + \dfrac{{20}}{{729}} = \dfrac{{180}}{{729}} = \dfrac{{20}}{{81}}
According to question,
The probability of winner being declared in 7th game=ab = \dfrac{a}{b}
Therefore,
a=20 b=81  a = 20 \\\ b = 81 \\\
Thus, the value of (a+b)=20+81=101\left( {a + b} \right) = 20 + 81 = 101
The required value of (a+b)\left( {a + b} \right) is 101.

Note: Here we have calculated the value of terms likenCm=n!m!×(n−m)!{}^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