Probabilities of Picking Correct Cases

Example

There are five matches on each weekend of a basketball season. Ben takes part in a competition in which he earns one point if he picks more than half of the winning teams for a weekend, and zero points otherwise. The probability that Ben correctly picks the team that wins any given match is \( 0.7 \).

Part 1

Find the probability that Ben earns one point for a given weekend.

\( \displaystyle \begin{align} \Pr(\text{correct pick}) &= 0.7 \\ \Pr(\text{incorrect pick}) &= 0.3 \\ \Pr(\text{One point}) &= \Pr(\text{3, 4 or 5 winning teams are picked}) \\ &= \Pr(\text{3 correct picks}) + \Pr(\text{4 correct picks}) + \Pr(\text{5 correct picks}) \\ &= {5 \choose 3} 0.3^2 \times 0.7^3 + {5 \choose 4} 0.3^1 \times 0.7^4 + {5 \choose 5} 0.3^0 \times 0.7^5 \\ &= 0.83692 \end{align} \)

Part 2

Hence, find the probability that Ben earns one point for a given fifteen-week season. Given your answer correct to one significant figure.

\( \displaystyle \begin{align} \Pr(\text{One point for 15 weeks}) &= 0.83692^{15} \\ &= 0.069224 \cdots \\ &= 0.07 \ \text{(one significant figure)} \end{align} \)

Part 3

Find the probability that ben earns at most 13 points during the fifteen-week season. Given your answer correct to two significant figures.

\( \displaystyle \begin{align} \Pr(\text{at most 13 points}) &= \Pr(\text{0 or 1 or 2 or } \cdots \text{ or 13 points}) \\ &= 1 – \Pr(\text{14 points}) – \Pr(\text{15 points}) \\ &= 1 – {15 \choose 14} 0.16308^1 \times 0.83692^{14} – {15 \choose 15} 0.16308^0 \times 0.83692^{15} \\ &= 0.72844 \cdots \\ &= 0.73 \ \text{(two significant figures)} \end{align} \)

 

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