Chapter 4: Q.4.72 (page 168)

Two athletic teams play a series of games; the first team to win 4 games is declared the overall winner. Suppose that one of the teams is stronger than the other and wins each game with probability .6, independently of the outcomes of the other games. Find the probability, for i = 4, 5, 6, 7, that the stronger team wins the series in exactly i games. Compare the probability that the stronger team wins with the probability that it would win a 2-outof-3 series.

Short Answer

Expert verified

The probability that the stronger team will win in $2$ out of $3$ series is : $(\begin{array}{l} 2 \\ 1 \end{array}) 0 . 6^{2} \cdot 0.4 = 0.288$

Step by step solution

:Given Information

If $i = 4$ , the stronger team will win the series in exactly $i$ games if and only if they win all four games. Hence, the Probability is $0 . 6^{4} = 0.1296$ .

Suppose $i \in {5, 6, 7}$ . Say that the last $i^{t h}$ game has won by the stronger team. Chose $i - 4$ places out of first $i - 1$ places and say that in these games has won weaker team. So, the probability that the stronger team has won the series in exactly $i$ games is $(\begin{array}{l} i - 1 \\ i - 4 \end{array}) 0 . 6^{4} \cdot 0 . 4^{i - 4}$ .

Step 2:Explanation

$i = 5 = 0.207$

$i = 6 = 0.207$

$i = 7 = 0.1666$

therefore the probability that the stronger team wins in $2$ out of $3$ series is , $(\begin{array}{l} 2 \\ 1 \end{array}) 0 . 6^{2} \cdot 0.4 = 0.288$ .

Final Answer

The final answer is $(\begin{array}{l} 2 \\ 1 \end{array}) 0 . 6^{2} \cdot 0.4 = 0.288$

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Short Answer

Step by step solution

:Given Information

Step 2:Explanation

Final Answer

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