Question

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).

Answer 1

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:

• N = 1, P(N = 1, S = 5) = 0.5*(1/6) = 1/12
• N = 2, P(N = 2, S = 5)
  This can happen as: (14, 23, 32, 41) that is: Prob. = 0.5²*(4/36) = 1/36
• N = 3, P(N = 3, S = 5)
  Thie can happen as: (113, 131, 311, 122, 212, 221) that is Prob. = 0.5³*(6/216) = 1/288
• N = 4, P(N = 4, S = 5)
  This can happen as: (4111, 1411, 1141, 1114), that is Prob. = 0.5⁴*(4/6⁴) = 1/5184
• N = 5, P(N = 5, S = 5) can happen only as (11111) that is,
  Prob. = 0.5⁵ / 6⁵ = 1/248832

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.