Example on Random Variable (Sheldon Ross)

Three balls are to be randomly selected without replacement from an urn containing 20 balls numbered 1 through 20. If we bet that at least one of the balls that are drawn has a number as large as or larger than 17, what is the probability that we win the bet?

Answer

Let X denote the largest number selected. Then X is a random variable taking on one of the values 3, 4, . . . , 20. Furthermore, if we suppose that each of the \(\binom{20}{3}\) possible selections are equally likely to occur, then 

\(P(X = i) = \frac{ \binom{i-1}{2} }{ \binom{20}{3}} \ \ \ \ , i= 3, .... 20\)

Therefore,

\(P(X = 20) = \frac{ \binom{19}{2} }{ \binom{20}{3}} = \frac{3}{20} = .150 \\ P(X = 19) = \frac{ \binom{18}{2} }{ \binom{20}{3}} = \frac{51}{380} \approx .134 \\ P(X = 18) = \frac{ \binom{17}{2} }{ \binom{20}{3}} = \frac{34}{285} \approx .199 \\ P(X = 17) = \frac{ \binom{16}{2} }{ \binom{20}{3}} = \frac{2}{19} \approx .105\)

Hence, since the event \(\{ X \geq 17 \} \) is the union of the disjoint events \(\{X = i \} , \ \ \ \ i = 17, 18 , 19, 20\) 

It follows that the probability of our winning the bet is given by:

\(P\{ X \geq 17\} \approx .105 + .119 + .134 + .150 = .508\)

 

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