Question

a player plays a roulette game in a casino by betting on a single number each time. because the wheel has 38 numbers, the probability that the player will win in a single play is $\frac{1}{38}$. note that each play of the game is independent of all previous plays. the gambler claims that because he has 1 chance in 38 of winning each time he plays, he is certain to win at least once if he plays 38 times. does this sound reasonable to you? find the probability that he will win at least once in 38 plays. round your answer to 4 decimal places.

Explanation:

Step1: Find the probability of losing a single play

The probability of losing a single play is $1 - \frac{1}{38}=\frac{37}{38}$.

Step2: Find the probability of losing all 38 plays

Since the plays are independent, the probability of losing all 38 plays is $(\frac{37}{38})^{38}$.

Step3: Find the probability of winning at least once in 38 plays

The probability of winning at least once in 38 plays is $P = 1-(\frac{37}{38})^{38}$.
Calculate $(\frac{37}{38})^{38}\approx0.3660$. Then $P = 1 - 0.3660=0.6340$.

Answer:

$0.6340$