Question

An individual uses the following gambling system at Las Vegas. He bets $ 1 that the roulette wheel will come up red. If he wins, he quits. If he loses then he makes the same bet a second time only this time he bets $ 2; and then regardless of the outcome, quits. Assuming that he has a probability of 12 of winning each bet, what is the probability that he goes home a winner? Why is this system not used by everyone?

Answer 1

Answer:

3/4

Step-by-step explanation:

Given that an individual uses the following gambling system at Las Vegas. He bets $ 1 that the roulette wheel will come up red

Prob of winning in one trial = 1/2

For him to go home as winner either he has to win in the I case or he loses I one and wins the second one.

Hence probability = prob he wins the Ist +prob he loses first and wins second

=$\frac{1}{2} +\frac{1}{2} \frac{1}{2} \\=\frac{3}{4}$

This system will not be used because normally prob of winning will not be 1/2 but much more less than that.  Hence prob of losing would be more so everyone may not use this.