Q. 6.7 In the Japanese game show Sushi ... [FREE SOLUTION]

91影视

The Practice of Statistics for AP Examination

Daren Starnes, Josh Tabor

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 837 pages

ISBN: 9781319113339

Chapter 6: Q. 6.7 (page 435)

In the Japanese game show Sushi Roulette, the contestant spins a large wheel that鈥檚 divided into 12 equal sections. Nine of the sections show a sushi roll, and three have a 鈥渨asabi bomb.鈥� 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 鈮� 3) 鈮� 0.6836

Step by step solution

Given information

The Probability of success is,

\begin{array}{rcl} p & = & \frac{3}{12} \\ = & \frac{1}{4} \\ = & 0.25 \end{array}

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:

$\begin{array}{rcl} P (X & = & k) = q^{k - 1} 脳 p \\ = & (1 - p)^{k - 1} 脳 p \end{array}$

To calculate the likelihood, apply the complement rule.

$\begin{array}{rcl} P (X & 鈮� & 3) = P (X = 1) + P (X = 2) + P (X = 3) \\ = & (1 - 0.25)^{1 - 1} (0.25) + (1 - 0.25)^{2 - 1} (0.25) + (1 - 0.25)^{3 - 1} (0.25) \\ = & 0.25 + 0.75 (0.25) + (0.75)^{2} (0.25) \\ 鈮� & 0.6836 \\ = & 68.36 % \end{array}$