Probability Ques 127
- Suppose the probability for $A$ to win a game against $B$ is 0.4 . If $A$ has an option of playing either a best of 3 games’ or a ‘best of 5 games" match against $B$, which option should choose so that the probability of his winning the match is higher? (no game ends in a draw).
$(1989,5$ M)
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Answer:
Correct Answer: 127.Best of 3 games
Solution:
Formula:
- Case I When $A$ plays 3 games against $B$.
In this case, we have $n=3, p=0.4$ and $q=0.6$
Let $X$ denote the number of wins. Then,
$$ P(X=r)={ }^{3} C _r(0.4)^{r}(0.6)^{3-r} ; r=0,1,2,3 $$
$\therefore \quad P _1=$ probability of winning the best of 3 games
$$ \begin{aligned} & =P(X \geq 2) \\ & =P(X=2)+P(X=3) \\ & ={ }^{3} C _2(0.4)^{2}(0.6)^{1}+{ }^{3} C _3(0.4)^{3}(0.6)^{0} \\ & =0.288+0.064=0.352 \end{aligned} $$
Case II When $A$ plays 5 games against $B$.
In this case, we have
$$ n=5, p=0.4 \quad \text { and } \quad q=0.6 $$
Let $X$ denotes the number of wins in 5 games.
Then,
$P(X=r)={ }^{5} C _r(0.4)^{r}(0.6)^{5}-r$, where $r=0,1,2 \ldots, 5$
$\therefore \quad P _2=$ probability of winning the best of 5 games
$=P(X \geq 3)$
$=P(X=3)+P(X=4)+P(X=5)$
$={ }^{5} C _3(0.4)^{3}(0.6)^{2}+{ }^{5} C _4(0.4)^{4}(0.6)+{ }^{5} C _5(0.4)^{5}(0.6)^{0}$
$=0.2304+0.0768+0.1024=0.31744$ Clearly, $P _1>P _2$. Therefore, first option i.e. ‘best of 3 games’ has higher probability of winning the match.
BETA
