SOLUTION: A fair die is tossed twice.find the probability of getting 4,5, or 6 on the first toss and a 1,2,3 or 4 on the second toss.

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Question 1177052: A fair die is tossed twice.find the probability of getting 4,5, or 6 on the first toss and a 1,2,3 or 4 on the second toss.
Found 2 solutions by ewatrrr, ikleyn:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!

Hi
the probability of getting 4,5, or 6 on the first toss 
and a 1,2,3 or 4 on the second toss.

 P = 12/36 = 1/3

(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)

Wish You the Best in your Studies.


Answer by ikleyn(52852) About Me  (Show Source):
You can put this solution on YOUR website!
.

The probability of getting 4, 5, or 6 on the first toss  is  3%2F6 = 1%2F2.


The probability of getting 1, 2, 3, or 4 on the second toss is  4%2F6 = 2%2F3.


The outcomes of the first and the second toss are INDEPENDENT, so the final answer is the product


    P = %281%2F2%29%2A%282%2F3%29 = 1%2F3.    ANSWER

Solved.

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To solve the problem, there is NO any need to present the table of the 36 outcomes, as @ewatrrr does.

There is no any need to spend your time to construct and to write this table,
as well as there is no any need to spend your time counting the number of favorable outcomes from this table.