+296 Calamitous Clod rolls a pair of rigged dice. The probability of rolling k on a die is directly proportional to k. What is the probability that the sum of Calamitous Clod's dice is equal to 7?
Thanks!
+118687 Calamitous Clod rolls a pair of rigged dice. The probability of rolling k on a die is directly proportional to k.
What is the probability that the sum of Calamitous Clod's dice is equal to 7?
Mmmm. Am I supposed to assume that these are 6 sided dice?
Prob(rolling1) = k Where k is less than 1
Prob(rolling 2) = 2k
P(3) = 3k
..
P(6)=6k
k(1+2+3+4+5+6)=21k
21k=1 so
\(k=\frac{1}{21}\)
| 1 | 2 | 3 | 4 | 5 | 6 | |
|---|---|---|---|---|---|---|
| 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| 6 | 7 | 8 | 9 | 10 | 11 | 12 |
\(P(throwing 7) \\ = 2[P(1,6)+P(2,5)+P(3,4)]\\ =2[ k*6k + 2k*5k + 3k*4k]\\ =2k^2 (6+10+12)\\ =56k^2\\ =\frac{56}{441}\\ =\frac{8}{63}\)
You need to check what I have done.
It has been pointed out to me that k is the number on the die. I should have used a different letter for the proportionality constant.
If you change all my k's to a different letter, it would be a much better answer 
