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

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

Short answer

1. The probability to lose first 5 bets by smith is$0.15$
2. The probability to occur first win on his fourth bet is$0.101$

Step-by-step solution

Given Information (Part-a)

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 $12$.

We have to find what is the probability that Smith will lose his first$5$bets.

Solution (Part-a)

We obtain that the probability that Smith loses a single bet is $\frac{26}{38}$.

Therefore, the probability that he loses all his five betting is:

Simplify ${\left(\frac{26}{38}\right)}^5$

Apply the exponent rule,

We need to factor

${26}^5:2^5\cdot {13}^5$

${38}^5:2^5\cdot {19}^5$

Therefore, $=\frac{2^5\cdot {13}^5}{2^5\cdot {19}^5}$

Cancel the common factor $2^5$

Now,

$=\frac{{13}^5}{{19}^5}$

$=0.15$

Final Answer (Part-a)

The probability to lose first 5 bets by smith is$0.15$

Given Information(Part-b)

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 $12$

We have to find what is the probability that his first win will occur on his fourth bet?

Solution (Part-b)

Define random variable $Y$that marks the first time that he wins a bet.

We know that$Y~\mathrm{Geom}\left(\frac{12}{38}\right)$

Therefore, the required probability is simply ,

$P(Y=4)={\left(\frac{26}{38}\right)}^3·\frac{12}{38}=0.101$

:Final Answer (Part-b) 

The required probability is simply $P\left(Y=4\right)=$$0.101$