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A pair of dice is rolled and the sum is determined. The probability that a sum of 5 is rolled before the sum of 8 is rolled in a sequence of rolls of dice is _______________
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is it approx 0.16666
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I am getting 0.44 :/ @Saswat Senapati what is the answer?
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Can you please tell how to solve such problems.
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@MiNiPanda yes given solution is 0.44 but . will the probability be just, p(geting 5)  given that 5 or 8 have already occured. ???? i have a doubt over there

i need the method to approach the problem, given solution may be wrong.
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@anu007 can you please elaborate the method ???
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i did calculation mistake same method using like him.

Step 1: Find probability of getting sum as 5. It is 4/36 =1/9. Name it P(F).

Step 2: Find probability of getting sum as 5 or 8. It is 9/36=1/4.

Step 3: Find probability of not getting sum as 5 or 8. 1-1/4=3/4. Name if P(N).

Step 4:

You want to get sum=5 before sum=8.

In the 1st trial if you get sum as 5 then it's success. P(F)

If you don't then you should also ensure that sum=8 also doesn't appear in the 1st trial. So P(N)*P(F).

If you don't get 5 in the 2nd trial also then 8 should also not come in the 1st two trials. So P(N)*P(N)*P(F)

Keep doing this and add them up. You will get a gp series.

1/9+ 3/4*1/9+ 3/4*3/4*1/9...

Solve it to get 0.44.
by Boss (23.5k points)
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