A fair 6faced die is tossed twice Letting E and F represent

A fair 6-faced die is tossed twice. Letting E and F represent the outcomes of the two tosses, compute the following probabilities (a) The sum of E and Fis 11. (b) The sum of E and F is even. (c) The sum of E and F is odd and greater than 3. (d) E is even and less than 6, and F is odd and greater than 1, (e) E is greater than 2, and F is less than 4 (f) E is 4, and the sum of E and F is odd.

Solution

(a) Total number of possible outcomes = 6*6 = 36

Number of favorable outcomes = 2 (i.e. 5+6 or 6+5)

Probability that sum of E and F is 11 = Number of favorable outcomes / Total outcomes = 2/36 = 0.0556

(b) Number of favorable outcomes = 6*3 = 18   (i.e., 1+1, 1+3, 1+5, 2+2, 2+4, 2+6, 3+1, 3+3, 3+5, 4+2, 4+4, 4+6, 5+1, 5+3, 5+5, 6+2, 6+4, 6+6)

Probability = 18/36 = 0.50

(c) Number of favorable outcomes = 16 (i.e. 1+4, 1+6, 2+3, 2+5, 3+2, 3+4, 3+6, 4+1, 4+3, 4+5, 5+2, 5+4, 5+6, 6+1, 6+3, 6+5)

Probability = 16/36 = 0.4444

(d) Number of favorable outcomes = 4 (i.e. 2+3, 2+5, 4+3, 4+5)

Probability = 4/36 = 0.1111

(e) Number of favorable outcomes = 12 (i.e. 3+1, 3+2, 3+3, 4+1, 4+2, 4+3, 5+1, 5+2, 5+3, 6+1, 6+2, 6+3)

Probability = 12/36 = 0.3333

(f) Number of favorable outcomes = 3 (i.e. 4+1, 4+3, 4+5)

Probability = 3/36 = 0.0833