Question

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?

Answer 1

Answer:

The probability that  Smith will lose his first 5 bets is 0.15

The probability that  his first win will occur on his fourth bet is 0.1012

Step-by-step explanation:

Consider the provided information.

A roulette wheel consisting of 38 numbers 1 through 36, 0, and double 0. Smith always bets that the outcome will be one of the numbers 1 through 12,

It is given that Smith always bets on the numbers 1 through 12.

There are 12 numbers from 1 to 12.

Thus, the probability of success (winning) is=  $\frac{12}{38}$

The probability of not success (loses) is=  $1-\frac{12}{38}=\frac{26}{38}$

Part (A) Smith will lose his first 5 bets.

The probability  that Smith loses his first 5 bets is,

$\frac{26}{38}\times\frac{26}{38}\times\frac{26}{38}\times\frac{26}{38}\times\frac{26}{38}=(\frac{26}{38})^5\approx0.15$

Hence, the probability that  Smith will lose his first 5 bets is 0.15

Part (B)  His first win will occur on his fourth bet?

Smith’s first win occurring on the fourth bet means that he loses the first 3 bets and wins on the fourth bet. That is,

$\frac{26}{38}\times\frac{26}{38}\times\frac{26}{38}\times\frac{12}{38}=(\frac{26}{38})^3\times\frac{12}{38}\approx0.1012$

Hence, the probability that  his first win will occur on his fourth bet is 0.1012