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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 _______________
probability
asked
Jan 11, 2018
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Probability
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Saswat Senapati
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is it approx 0.16666
0
I am getting 0.44 :/ @Saswat Senapati what is the answer?
0
Can you please tell how to solve such problems.
0
@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.
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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. 11/4=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.
answered
Jan 11, 2018
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MiNiPanda
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Jan 26, 2018
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