Probability of rolling a sum of 7 with loaded dice

  • Thread starter Thread starter Panphobia
  • Start date Start date
  • Tags Tags
    Dice Probability
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
2 replies · 3K views
Panphobia
Messages
435
Reaction score
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:
Physics news on Phys.org
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 *.