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?

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

=

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.

answer: answer is c!

ok i got to organize the

3/15 and 12/55

3*

aha

3*4 equals 12 but not

a: no

b:

8/24 12/35

no

gautie

it's c

5/18 = 25/90

5*5=25

18*5 = 90

hope that you!

one more happy customer!

< 3

step-by-step explanation:

grab rope, tie a loop-de-loop, put it around ur neck, now jump while its tied to a high place. now your shoes are cool.

step-by-step explanation: