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 Smith will lose his first 5 bets; his first win will occur on his fourth bet?

Answer 1

Answer:

0.15;

0.101

Step-by-step explanation:

Since there are 38 total options and Smith is betting on 12 possible outcomes per bet, the probability of him winning (W) and losing (L) each bet are:

$W=\frac{12}{38} \\L= 1 - W =\frac{26}{38}$

The probability that Smith will lose his first 5 bets (L5) is:

$L_{5} = (\frac{26}{38})^{5}  \\L_{5} = 0 .15$

The probability that his first win will occur on his fourth bet (W4) is the probability of him losing the first three bets multiplied by the probability of winning the fourth:

$W_{4}=(\frac{26}{38})^3*\frac{12}{38}\\W_{4}=0.101$