Four boys participated in a competition in which their respective chances of winning prizes are \(\frac{1}{5}, \frac{1}{4}, \frac{1}{3}\) and \(\frac{1}{2}\...
Question 1 Report
Four boys participated in a competition in which their respective chances of winning prizes are \(\frac{1}{5}, \frac{1}{4}, \frac{1}{3}\) and \(\frac{1}{2}\). What is the probability that at most two of them win prizes?
Answer Details
Let the winning probabilities be \(p_1 = \tfrac{1}{5},\ p_2 = \tfrac{1}{4},\ p_3 = \tfrac{1}{3},\ p_4 = \tfrac{1}{2}\), with losing probabilities \(q_1 = \tfrac{4}{5},\ q_2 = \tfrac{3}{4},\ q_3 = \tfrac{2}{3},\ q_4 = \tfrac{1}{2}\).
"At most two win" is easier via the complement: \(1 - P(3\text{ win}) - P(4\text{ win})\).
All four win:
\[P(4) = \tfrac{1}{5}\cdot\tfrac{1}{4}\cdot\tfrac{1}{3}\cdot\tfrac{1}{2} = \frac{1}{120}\]Exactly three win (one loses at a time, summed):
\[\tfrac{4}{5}\cdot\tfrac{1}{4}\cdot\tfrac{1}{3}\cdot\tfrac{1}{2} + \tfrac{1}{5}\cdot\tfrac{3}{4}\cdot\tfrac{1}{3}\cdot\tfrac{1}{2} + \tfrac{1}{5}\cdot\tfrac{1}{4}\cdot\tfrac{2}{3}\cdot\tfrac{1}{2} + \tfrac{1}{5}\cdot\tfrac{1}{4}\cdot\tfrac{1}{3}\cdot\tfrac{1}{2}\] \[= \frac{4 + 3 + 2 + 1}{120} = \frac{10}{120}\]So \(P(3\text{ or }4) = \dfrac{10}{120} + \dfrac{1}{120} = \dfrac{11}{120}\), and:
\[P(\text{at most 2}) = 1 - \frac{11}{120} = \frac{109}{120} \approx 0.908\]Download The App On Google Playstore
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