Probability of Rolling 7 with Loaded Dice: A Homework Solution | 11/56 Odds

In summary: Thank you for catching that! In summary, a pair of loaded dice have a probability of 2/7 for 4 appearing on the first die and a probability of 3/8 for 3 appearing on the second die, with the rest of the numbers being equally likely events. The probability of rolling a 7 as the sum of the numbers on the two dice is 5/28.
  • #1
Panphobia
435
13

Homework Statement


A pair of dice is loaded. The probability of 4 appearing on the first die is 2/7.
and the probability of a 3 appearing on the second die is thrice as much as the other numbers
on the die. If the rest of the numbers are equally likely events in both dice, what is the
probability of 7 appearing as the sum of the numbers when the two dice are rolled.

The Attempt at a Solution



So I was wondering if the answer I got is correct by chance. I took all the cases.
1 6
6 1
2 5
5 2
3 4
4 3

Got the probabilities and added them up. But I am not sure I got it right because it says the rest of the numbers are equally likely to appear, does that mean the probability of the rest in both dice is exactly the same. This is because if the probability of rolling a 4 on the first die is 2/7 then the rest are
2/7 + 5x = 1
x = 1/7

and then that same logic for the second die

3n + 5n = 1
n = 1/8

so the probability is

2/7*3/8 + 5*1/7*1/8
= 11/56
 
Last edited:
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  • #2
Panphobia said:

Homework Statement


A pair of dice is loaded. The probability of 4 appearing on the first die is 2/7.
and the probability of a 3 appearing on the second die is thrice as much as the other numbers
on the die. If the rest of the numbers are equally likely events in both dice, what is the
probability of 7 appearing as the sum of the numbers when the two dice are rolled.

The Attempt at a Solution



So I was wondering if the answer I got is correct by chance. I took all the cases.
1 6
6 1
2 5
5 2
3 4
4 3

Got the probabilities and added them up. But I am not sure I got it right because it says the rest of the numbers are equally likely to appear, does that mean the probability of the rest in both dice is exactly the same. This is because if the probability of rolling a 4 on the first die is 2/7 then the rest are
2/7 + 5x = 1
x = 1/7

and then that same logic for the second die

3n + 5n = 1
n = 1/8

so the probability is

2/7*3/8 + 5*1/7*1/8
= 10/56 = 5/28

It is all correct. Except, sadly, 2*3 = 6 (not 5)!
 
  • #3
I am so sorry, that was a big fail. Haha mixed up the + and *.
 

FAQ: Probability of Rolling 7 with Loaded Dice: A Homework Solution | 11/56 Odds

1. What is the probability of rolling a 7 with loaded dice?

The probability of rolling a 7 with loaded dice is 11/56, or approximately 0.1964. This means that for every 56 rolls, we can expect to get a 7 about 11 times.

2. How were the odds of 11/56 determined for this homework solution?

The odds of 11/56 were determined by first calculating the total number of possible outcomes when rolling two dice, which is 36. Then, we determined the number of ways to get a 7 with two loaded dice, which is 11. Therefore, the probability of rolling a 7 is 11/36, or approximately 0.1964.

3. How do loaded dice affect the probability of rolling a 7?

Loaded dice have a higher probability of landing on certain numbers than regular dice. In this case, the loaded dice have a higher chance of landing on numbers that add up to 7, such as 3 and 4, which increases the overall probability of rolling a 7.

4. What other factors could affect the probability of rolling a 7 with loaded dice?

Besides the loaded dice, other factors that could affect the probability of rolling a 7 include the surface the dice are rolled on, the force and angle at which they are thrown, and any imperfections on the dice themselves.

5. Are there any real-life applications for understanding the probability of rolling a 7 with loaded dice?

Yes, understanding probability is important in many fields, including statistics, finance, and gambling. Knowing the odds of certain outcomes can help in making informed decisions and predictions.

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