Question

In the Japanese game show Sushi Roulette, the contestant spins a large wheel that’s divided into 12 equal sections. Nine of the sections have a sushi roll, and three have a “wasabi bomb.” When the wheel stops, the contestant must eat whatever food is on that section. To win the game, the contestant must eat one wasabi bomb. Find the probability that it takes 3 or more spins for the contestant to get a wasabi bomb. Show your method clearly.

Answer 1

Answer:

P = 0.545

Step-by-step explanation:

Here we can find the probability Q, in which you need one or two spins to win the game, and then:

P = 1 - Q

Will be the probability that it takes 3 or more spins to win.

Case where you need only one spin:

There are 3 out of 12 possible outcomes, then the probability of winning with only one spin is:

p = 3/12.

Case where you need two spins to win:

Here we should get a normal sushi roll in the first spin and a wasabi bomb in the second.

The probability of getting a sushi roll in the first spin is equal to the quotient between the number of sushi rolls (9) and the total number of possible outcomes (12), then:

p1 = 9/12

For the second spin we need to get a wasabi bomb, assuming that the susi roll is not replaced (so now there are 8 sections with sushi rolls and 3 with wasabi bombs, if you get the empty section you roll again or something like that)

Now the probability of getting a wasabi bomb is equal to the quotient between the number of wasabi bombs remaining (3) and the total number of outcomes remaining (11), this is:

p2 = 3/11

The joint probability is the product of the two individual probabilities:

p = p1*p2 = (9/12)*(3/11)

Now, the total probability of winning with one or two spins is equal to the sum of the probabilities for each case, then:

Q = (9/12)*(3/11)  + 3/12 = 0.455

Then the probability of needing 3 or more spins to win is:

P = 1 - Q = 1 - 0.455 = 0.545