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Chapter 6: Q. 6.7 (page 435)

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

Short Answer

Expert verified

Thus, the required Probability,

P(X≤3)≈0.6836

Step by step solution

01

Given information

The Probability of success is,

p=312=14=0.25
02

Calculation

The distribution will be geometric because the variable is the number of spins required till first success.

The definition of geometric probability is as follows:

P(X=k)=qk-1×p=(1-p)k-1×p

To calculate the likelihood, apply the complement rule.

P(X≤3)=P(X=1)+P(X=2)+P(X=3)=(1-0.25)1-1(0.25)+(1-0.25)2-1(0.25)+(1-0.25)3-1(0.25)=0.25+0.75(0.25)+(0.75)2(0.25)≈0.6836=68.36%

As a result, there is a68.36 percent chance that the contestant will get a wasabi bomb in three or fewer spins.

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Most popular questions from this chapter

Life insurance A life insurance company sells a term insurance policy to 21-year-old males that pays \(100,000 if the insured dies within the next 5 years. The probability that a randomly chosen male will die each year can be found in mortality tables. The company collects a premium of \)250 each year as payment for the insurance. The amount Y that the company earns on a randomly selected policy of this type is \(250 per year, less the \)100,000 that it must pay if the insured dies. Here is the probability distribution of Y:

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