Let N be the number of times we will toss a fair die, where N ∈ Z+ and P(N=k)=0.5^k for any k∈Z+ .Let S be the sum of all the throws of the die.
For example, say N turned out to be 5, then we toss the die 5 times. Say the outcomes are 6, 1, 1, 3, 4, 5, then S = 6 + 1 + 1 + 3 + 4 = 15.
Calculate P(N = 2 | S = 5).
The conditional probability P(N = 2 | S = 5) is computed using
Bayes theorem here as:
P(N = 2 | S = 5) = P(N = 2, S = 5) / P(S = 5)
S = 5 here could be obtained in the following cases:
Therefore the probability here is computed as:
P(S = 5) = (1/12) + (1/36) + (1/288) + (1/5184) + (1/248832)
P(S = 5) = 0.1148
Therfore the probability here is computed as:
= P(N = 2, S = 5) / P(S = 5)
= (1/36) / 0.1148
= 0.2420
Therefore 0.2420 is the required probability here.
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