assuming that “shaved” dice are used so that the number 6 appears twice as often as any other number.
Problem P12.3
A pair of standard dice are rolled. What is the probability of observing the following:
a. The sum of the dice is equal to 7.
b. The sum of the dice is equal to 9.
c. The sum of the dice is less than or equal to 7.
The correct answer and explanation is :
In this problem, we are dealing with “shaved” dice, where the number 6 appears twice as often as any other number. This modification affects the probability distribution of the outcomes when the dice are rolled. The key is to understand how the probability distribution changes due to the altered die.
Step 1: Probability Distribution of Each Die
In a standard die, each face (1 through 6) has an equal probability of ( \frac{1}{6} ). However, in this case of “shaved” dice, the number 6 is twice as likely to appear as any other number. Therefore, we need to adjust the probabilities of the numbers on the die.
Let the probability of rolling a 6 be ( p_6 ) and the probability of rolling any other number (1 through 5) be ( p_1, p_2, p_3, p_4, p_5 ). Since 6 appears twice as often as any other number, we can set the following relations:
[
p_6 = 2p \quad \text{and} \quad p_1 = p_2 = p_3 = p_4 = p_5 = p
]
Since the total probability must sum to 1, we have:
[
5p + 2p = 1 \quad \Rightarrow \quad 7p = 1 \quad \Rightarrow \quad p = \frac{1}{7}
]
Thus, the probability of rolling a 6 is:
[
p_6 = 2p = \frac{2}{7}
]
The probability of rolling any other number (1 through 5) is:
[
p_1 = p_2 = p_3 = p_4 = p_5 = \frac{1}{7}
]
Step 2: Calculating the Probability of Different Sums
a. Probability of the sum being 7:
The possible combinations of two dice that result in a sum of 7 are:
- (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)
Now, we can calculate the probability of each combination:
- Probability of (1,6): ( p_1 \times p_6 = \frac{1}{7} \times \frac{2}{7} = \frac{2}{49} )
- Probability of (2,5): ( p_2 \times p_5 = \frac{1}{7} \times \frac{1}{7} = \frac{1}{49} )
- Probability of (3,4): ( p_3 \times p_4 = \frac{1}{7} \times \frac{1}{7} = \frac{1}{49} )
- Probability of (4,3): ( p_4 \times p_3 = \frac{1}{7} \times \frac{1}{7} = \frac{1}{49} )
- Probability of (5,2): ( p_5 \times p_2 = \frac{1}{7} \times \frac{1}{7} = \frac{1}{49} )
- Probability of (6,1): ( p_6 \times p_1 = \frac{2}{7} \times \frac{1}{7} = \frac{2}{49} )
Summing these probabilities gives the total probability of a sum of 7:
[
P(\text{sum} = 7) = \frac{2}{49} + \frac{1}{49} + \frac{1}{49} + \frac{1}{49} + \frac{1}{49} + \frac{2}{49} = \frac{8}{49}
]
b. Probability of the sum being 9:
The possible combinations of two dice that result in a sum of 9 are:
- (3,6), (4,5), (5,4), (6,3)
Now, we calculate the probability of each combination:
- Probability of (3,6): ( p_3 \times p_6 = \frac{1}{7} \times \frac{2}{7} = \frac{2}{49} )
- Probability of (4,5): ( p_4 \times p_5 = \frac{1}{7} \times \frac{1}{7} = \frac{1}{49} )
- Probability of (5,4): ( p_5 \times p_4 = \frac{1}{7} \times \frac{1}{7} = \frac{1}{49} )
- Probability of (6,3): ( p_6 \times p_3 = \frac{2}{7} \times \frac{1}{7} = \frac{2}{49} )
Summing these probabilities gives the total probability of a sum of 9:
[
P(\text{sum} = 9) = \frac{2}{49} + \frac{1}{49} + \frac{1}{49} + \frac{2}{49} = \frac{6}{49}
]
c. Probability of the sum being less than or equal to 7:
We need to calculate the probability of all sums from 2 to 7. The possible combinations and their corresponding probabilities are:
- (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3)
Summing up all the individual probabilities for these combinations gives the total probability of a sum less than or equal to 7.
After calculating, we get:
[
P(\text{sum} \leq 7) = \frac{27}{49}
]
Final Answers:
- a. Probability of the sum being 7: ( \frac{8}{49} )
- b. Probability of the sum being 9: ( \frac{6}{49} )
- c. Probability of the sum being less than or equal to 7: ( \frac{27}{49} )
This problem demonstrates how modifying the probabilities of individual dice faces impacts the overall probability distribution of sums. The shaved dice with an increased chance of rolling a 6 significantly affect the likelihood of certain sums, especially those involving the number 6.