A pair of dice is rolled, what is the probability of each of the following events.

Question

a.     A sum of at least 7 shows up

b.     A sum of most 8 appears

c.     One number is three or the other number is even

d.     Both numbers are even and greater than or equal to 4,

Answer 1

Dice numbers are low enough to work out the cases:

a) 3-4, 5-2,6-1 and 4-3,5-2,6-1 out of 6² possible rolls. P(2 dice adding to 7) = 6/36

b) sum of at most 8: 36 - N(9) - N(10) - N(11) - N(12) is the number of rolls that give you at most 8

N(9) is 4, N(10) = 3 (doubles only count once: 64,46, 55) N(11) = 2 N(12) = 1 so N(<= 8) = 36 - 10 = 26

P(<=8) = 26/36

c) Again count: 1st die 3 and 2nd die even 2, 4, or 6 is 3 cases. 2nd die 3 and first even is another 3

P = 6/36

d) Both even >= 4 so 1st die 4 or 6 and 2nd die 4 or 6. 4 cases, so P is 4/36

Please consider a tutor. Take care.