Question

a certain game involves tossing 3 fair coins, and it pays 17¢ for 3 heads, 7¢ for 2 heads, and 5¢ for 1 head. is 7¢ a fair price to pay to play this game? that is, does the 7¢ cost to play make the game fair? the 7¢ cost to play is not a fair price to pay because the expected winnings are . (type an integer or a fraction. simplify your answer.)

Explanation:

Step1: Calculate probability of 3 heads

When tossing 3 fair coins, the total number of possible outcomes is $2\times2\times2 = 8$. The number of ways to get 3 heads is $C(3,3)=\frac{3!}{3!(3 - 3)!}=1$. So the probability $P(3H)=\frac{1}{8}$.

Step2: Calculate probability of 2 heads

The number of ways to get 2 heads out of 3 coins is $C(3,2)=\frac{3!}{2!(3 - 2)!}=3$. So the probability $P(2H)=\frac{3}{8}$.

Step3: Calculate probability of 1 head

The number of ways to get 1 head out of 3 coins is $C(3,1)=\frac{3!}{1!(3 - 1)!}=3$. So the probability $P(1H)=\frac{3}{8}$.

Step4: Calculate expected - value of winnings

The pay - off for 3 heads is 17 cents, for 2 heads is 7 cents, and for 1 head is 5 cents. The expected value $E(X)$ of the winnings is $E(X)=17\times\frac{1}{8}+7\times\frac{3}{8}+5\times\frac{3}{8}$.
\[

$$\begin{align*} E(X)&=\frac{17 + 21+15}{8}\\ &=\frac{53}{8}= 6.625 \end{align*}$$

\]

Answer:

$\frac{53}{8}$