If you roll two fair dice repeatedly, what is the probability that you will get a sum of 4 before you get a sum of 5 ? (a) (b) (c (e) (d) 7

The probability that you will get a sum of 4 before you get a sum of 5 is

Step-by-step explanation:

Given : If you roll two fair dice repeatedly.

To find : What is the probability that you will get a sum of 4 before you get a sum of 5 ?

Solution :

When two dice are rolled the outcomes are

(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)

(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)

(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)

(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)

(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)

(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)

Total number of outcomes = 36

Favorable outcome get a sum of 4 before you get a sum of 5 is (1,3) ,(2,2) and (3,1) = 3

The probability that you will get a sum of 4 before you get a sum of 5 is

Therefore, The probability that you will get a sum of 4 before you get a sum of 5 is

idk lol xd

step-by-step explanation:

ez do it yourself lazy

1010: 3030

is it i think