Question

A pair of dice are rolled find the following? the odds in favor of a sum of 4 -the odds in favor a sum of 7 or 11 - the probability of a product that is even and greater than 5

Answer 1

(a)

Odds in favor of a sum of 4 = Number of favorable cases of sum of 4/ Number of unfavorable cases of sum of 4

Cases of sum of 4:

(1,3), (3,1), (2,2): 3 Nos.

So,

Number of favorable cases of sum of 4 = 3

Total Cases = 6 X 6 = 36

So,

Number of unfavorable cases of sum of 4 = 36 - 3 = 33

So,

Substituting, we get:

Odds in favor of a sum of 4 = 3/33 = 1/11

(b)

Odds in favor of a sum of 7 or 11 = Number of favorable cases of sum of 7 or 11/ Number of unfavorable cases of sum of 7 or 11

Cases of sum of 7:

(1,6), (6,1), (2,5), (5,2), (3,4), (4,3): 6 Nos.

Cases of sum of 11:

(5,6), (6,5): 2 Nos

So,

Number of favorable cases of sum of 7 or 11 = 6 + 2 = 8

Total Cases = 6 X 6 = 36

So,

Number of unfavorable cases of sum of 4 = 36 - 8 = 28

So,

Substituting, we get:

Odds in favor of a sum of 7 or 11 = 8/28 = 2/7

(c)

Odds in favor of a Product that is even and greater than 5 = Number of favorable cases of product that is even and greater than 5 / Number of unfavorable cases of product that is even and greater than 5.

Cases of product that is even and greater than 5

(1,6), (6,1), (2,3), (3,2), (2,4), (4,2), (2,5), (5,2), (2,6), (6,2), (3,4), (4,3), (3,6), (6,3), (4,4), (4,5), (5,4), (4,6), (6,4), (5,6), (6,5), (6,6): 22 Nos.

So,

Number of favorable cases of Product that is even and greater than 5 = 22

Total Cases = 6 X 6 = 36

So,

Number of unfavorable cases of Product that is even and greater than 5 = 36 - 22 = 14

So,

Substituting, we get:

Odds in favor of a Product that is even and greater than 5 = 22/14 = 11/7