On a toss of two dice, a throws a total of 5. then the probability that he will throw another 5 before he throws 7 is
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Answered by
8
The total outcomes is 36. and the probability of 5 comes in 4 chances, and the probability of 7 comes in 6 chances.
The formula to find teh answer is 1 - p(5) plus p (7).
The answer is 26/36.
To find the probability of 5 is Σ (26 / 36)ⁿ⁻¹ (4 / 36)
= (4 / 36) Σ (26 / 36)ⁿ⁻¹
= 2 / 5.
Answered by
32
total probabilities for getting 5 = 4/36
total probabilities for getting 7 = 6/36
Total Probability = 10/36
We need only 5, hence prob of getting only 5 is (4/36)/(10/36)
=40%
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