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

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The probability of rolling a sum of 7 with a pair of loaded dice is calculated to be 11/56. The first die has a probability of 4 appearing as 2/7, while the second die has a probability of 3 appearing that is three times the probability of other numbers. The calculations involve determining the probabilities of all combinations that yield a sum of 7, leading to the final result of 11/56. The discussion highlights the importance of correctly applying multiplication and addition in probability calculations.

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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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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)!
 
I am so sorry, that was a big fail. Haha mixed up the + and *.
 

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