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Chapter 4: Problem 71

Consider a roulette wheel consisting of 38 numbers 1 through \(36,0,\) and double \(0 .\) If Smith always bets that the outcome will be one of the numbers 1 through \(12,\) what is the probability that (a) Smith will lose his first 5 bets; (b) his first win will occur on his fourth bet?

Short Answer

Expert verified
(a) The probability that Smith will lose his first 5 bets is approximately 0.175. (b) The probability that Smith's first win will occur on his fourth bet is approximately 0.157.

Step by step solution

01

Recall the probability formula

The probability of an event happening is the number of successful outcomes divided by the total number of possible outcomes. In this case, there are 12 successful outcomes (numbers 1 through 12), and 38 possible outcomes (numbers 1 through 36, 0, and double 0).
02

Find the probability of a single bet

To find the probability of Smith winning a single bet, we can use the probability formula: \(P(win) = \frac{successful\ outcomes}{total\ outcomes}\), so the probability of winning a single bet is \(P(win) = \frac{12}{38}\). Similarly, the probability of losing a single bet is \(P(lose) = 1 - P(win) = \frac{26}{38}\). (a) Smith will lose his first 5 bets:
03

Calculate the probability of losing 5 bets in a row

Since the outcome of each bet is independent, we can find the probability of losing 5 bets in a row by multiplying the probability of losing a single bet 5 times: \(P(lose\ 5\ in\ a\ row) = P(lose)^5 = \left(\frac{26}{38}\right)^5\).
04

Simplify the expression

To find the probability, we can simplify the expression: \(P(lose\ 5\ in\ a\ row) = \left(\frac{26}{38}\right)^5 \approx 0.175\). Therefore, the probability that Smith will lose his first 5 bets is approximately 0.175. (b) First win will occur on his fourth bet:
05

Use the geometric distribution formula

In this case, we can use the geometric distribution formula to find the probability that Smith's first win occurs on the fourth bet: \(P(first\ win\ on\ 4th\ bet) = P(lose)^{3} * P(win)\).
06

Plug in the values and calculate

Plugging in the values, we get: \(P(first\ win\ on\ 4th\ bet) = \left(\frac{26}{38}\right)^{3} * \frac{12}{38}\).
07

Simplify the expression

To find the probability, we can simplify the expression: \(P(first\ win\ on\ 4th\ bet) = \left(\frac{26}{38}\right)^{3} * \frac{12}{38} \approx 0.157\). Therefore, the probability that Smith's first win will occur on his fourth bet is approximately 0.157.

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