You toss two number cubes. If a sum of 7 or 11 comes up, you get 7 points; if not, you lose 2 points.
The probabilities for each of the sums are:
[tex]
\begin{array}{llll}
P(2)=1 / 36 & P(3)=1 / 18 & P(4)=1 / 12 & P(5)=1 / 9 \\
P(6)=5 / 36 & P(7)=1 / 6 & P(8)=5 / 36 & P(9)=1 / 9 \\
P(10)=1 / 12 & P(11)=1 / 18 & P(12)=1 / 36
\end{array}
[/tex]
What is the probability of a sum of 7 or 11?
Answer (1)
Calculate the probability of a sum of 7 or 11: P ( 7 \t or 11 ) = 6 1 + 18 1 = 9 2 .
Calculate the probability of not getting a sum of 7 or 11: P ( \t n o t ( 7 \t or 11 )) = 1 − 9 2 = 9 7 .
Calculate the expected value: E ( X ) = ( 7 ) ∗ 9 2 + ( − 2 ) ∗ 9 7 .
The expected value is: E ( X ) = 0 .
Explanation
Understand the problem We are given the probabilities of each sum when tossing two number cubes. If the sum is 7 or 11, we gain 7 points; otherwise, we lose 2 points. We need to calculate the expected value of the points gained or lost in a single toss.
Calculate the probability of a sum of 7 or 11 First, we need to calculate the probability of getting a sum of 7 or 11. We are given that P ( 7 ) = 6 1 and P ( 11 ) = 18 1 . Therefore, the probability of getting a sum of 7 or 11 is: P ( 7 \t or 11 ) = P ( 7 ) + P ( 11 ) = 6 1 + 18 1 To add these fractions, we need a common denominator, which is 18. So, we have: P ( 7 \t or 11 ) = 18 3 + 18 1 = 18 4 = 9 2
Calculate the probability of not getting a sum of 7 or 11 Next, we need to calculate the probability of not getting a sum of 7 or 11. This is the complement of the probability we just calculated: P ( \t n o t ( 7 \t or 11 )) = 1 − P ( 7 \t or 11 ) = 1 − 9 2 = 9 9 − 9 2 = 9 7
Calculate the expected value Now, we can calculate the expected value. The expected value is the sum of each outcome multiplied by its probability. In this case, we have two outcomes: gaining 7 points (if the sum is 7 or 11) and losing 2 points (if the sum is not 7 or 11). So, the expected value is: E ( X ) = ( 7 ) ∗ P ( 7 \t or 11 ) + ( − 2 ) ∗ P ( \t n o t ( 7 \t or 11 )) = ( 7 ) ∗ 9 2 + ( − 2 ) ∗ 9 7 E ( X ) = 9 14 − 9 14 = 0
State the final answer Therefore, the expected value of the points gained or lost in a single toss of the two number cubes is 0.
Examples
This type of probability calculation is used in various games of chance, such as card games or dice games, to determine the fairness and expected return of the game. For example, casinos use expected value calculations to ensure that, on average, they make a profit from the games they offer. Understanding expected value helps players make informed decisions about whether to play a game and how much to wager.
