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Chapter 2: Q18E (page 65)

A wallet contains five \(10 bills, four \)5 bills, and six \(1 bills (nothing larger). If the bills are selected one by one in random order, what is the probability that at least two bills must be selected to obtain a first \)10 bill?

Short Answer

Expert verified

The probability that at least two bills must be selected to obtain a first $10 bill is 0.67.

Step by step solution

01

Given information

The number of $10 bills that a wallet contains is 5.

The number of $5 bills that a wallet contains is 4.

The number of $1 bills that a wallet contains is 6.

02

Compute the probability

From the provided information, the total number of bills is,

\(5 + 4 + 6 = 15\)

Let X represents the number of trials to select the first $10 bill.

The probability of selecting a $10 bill is \(\frac{5}{{15}}\).

The probability that at least two bills must be selected to obtain a first $10 bill is computed as,

\(\begin{aligned}P\left( {X \ge 2} \right) &= 1 - P\left( {X < 2} \right)\\ &= 1 - P\left( {X = 1} \right)\\ &= 1 - \left( {\frac{5}{{15}}} \right){\left( {1 - \frac{5}{{15}}} \right)^{1 - 1}}\\ &= 1 - \frac{1}{3}\\ &= 1 - 0.33\\ &= 0.67\end{aligned}\)

Therefore, the probability that at least two bills must be selected to obtain a first $10 bill is0.67.

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