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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!

 Feb 4, 2020
 #1
avatar+118587 
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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

\(k=\frac{1}{21}\)

 

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\(P(throwing 7) \\ = 2[P(1,6)+P(2,5)+P(3,4)]\\ =2[  k*6k  + 2k*5k + 3k*4k]\\ =2k^2 (6+10+12)\\ =56k^2\\ =\frac{56}{441}\\ =\frac{8}{63}\)

 

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   frown   wink

 Feb 4, 2020
edited by Melody  Feb 4, 2020
 #2
avatar+296 
+1

I double checked. Your answer is correct! Thanks for the answer.

 Feb 4, 2020

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