Question

A fair 6-faced die is tossed twice. Letting E and F represent the outcomes of the each toss (which are independent), compute the following probabilities. a. The sum of E and F is 11 b. The sum of E and F is even c. The sum of E and F is odd and the sum is 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

Answer 1

(a)
`(1)/(18)` (d)
`(1)/(9)`

(b)
`(1)/(2)` (e)
`(1)/(3)`

(c)
`(4)/(9)` (f)
`(1)/(12)`

Explanation:

On tossing a 6-face die twice the outcomes of E and F are:

{1, 2, 3, 4, 5 and 6}

And there are total 36 outcomes of the form (E, F).

(a)

The sample space of getting a sum of 11 is: {(5, 6) and (6, 5)}

The probability of getting a sum of 11 is:

`P(Sum 11) =(2)/(36) \\=(1)/(18)`

(b)

The sample space of getting an even sum is:

{(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)}

The probability of getting an even sum is:

`P(Even Sum)=(18)/(36)\\ =(1)/(2)`

(c)

The sample space of getting an odd sum more than 3 is:

{(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) and (6, 5)}

The probability of getting an odd sum more than 3 is:

P (Odd Sum more than 3) =

`=(16)/(36)\\=(4)/(9)`

(d)

The sample space of (E, F) such that E is even and less than 6, and F is odd and greater than 1 is:

{(2, 3), (2, 5), (4, 3) and (4, 5)}

The probability such that E is even and less than 6, and F is odd and greater than 1 is:

`P(Even E<6, Odd F>1) =(4)/(36) \\=(1)/(9)`

(e)

The sample space of E and F such that E is more than 2, and F is less than 4 is:

E = {3, 4, 5 and 6} F = {1, 2 and 3}

Then the total outcomes of (E, F) will be 12.

The probability such that E is more than 2, and F is less than 4 is:

`P(E>2, F<4)=(12)/(36) \\=(1)/(3)`

(f)

The sample space of (E, F) such that E is 4 and sum of E and F is odd is:

{(4, 1), (4, 3) and (4, 5)}

The probability such that E is 4 and sum of E and F is odd is:

`P(E=4, E+F = Odd)=(3)/(36) \\=(1)/(12)`