Question

Five fair dice are rolled simultaneously

a.) If we add all the numbers that show up, find the probability that we have at least a sum of seven. (Please show your work for this part :)!)

Answer 1

Five fair dice are rolled

so if we say that their sum is S so S has a minimum value of 5.

so we have too find that

Pr(S >= 7) = 1 - Pr(S = 5) - Pr(S = 6)

Pr(S = 5) = Pr(All dices show up the number 1) = (1/6)⁵ = 1/7776

P(S= 6) = Pr(Four dices shows number 1 and one shows number 2) = ⁴C₁ (1/6)⁵ = 1/1944

Pr(S >= 7) = 1 - 5/7776 = 7771/7776

Answer 2

SOLUTION :

5 dices are rolled simultaneously. 

P( Sum ≥ 7) 

= 1 - P(Sum < 7)

Now possible outcomes for sum < 7 are :

(1,1,1,1,1), (1, 1,1, 1, 2) ; (1,1,1,2,1), (1, 1, 2, 1, 1), (1,2, 1, 1, 1) and (2, 1,1,1,1)

There are 6 outcomes.

P (each outcome) = (1/6)^5 

P (sum < 7) 

= 6 * (1/6)^5

= 1/6^4

= 1/1296

Hence, 

P( sum ≥ 7) 1 - 1/1296 = 1295/1296 = 0.999228395 (ANSWER).