Question

Consider the experiment of rolling a six-sided fair die. Let X

denote the number of rolls it takes to obtain the first 5,

Y denote the number of rolls until the first 2, and Z denote

the number of rolls until the first 4. Numerical answers are needed only for parts (a) and

(b). Expressions are sufficient for parts (c), (d), and (e).

a) E[X|Y = 1 or Z = 1]

b) E[X|Y = 1 and Z = 2]

c) E[X|Y = 1 and Z = 3]

d) E[X|Y = 3 and Z = 4]

e) E[X^2 |Y = 3 and Z = 4]

Answer 1

a) E[ X | Y = 1 or Z = 1 ]

Here, we have either Y = 1 or Z = 1, therefore the first toss cannot be a 5, it has to be one of 2 or 4. Therefore after the first toss, the expected number of tosses required to get a 5 is computed as: 1/p where p = 1/6 is the probability of getting a 5 in any toss.

Therefore E[ X | Y = 1 or Z = 1 ] = 1 + 6 = 7

Therefore 7 is the expected value here.

b) E [ X | Y = 1 and Z =2 ]

Y = 1 and Z = 2 means that the first 2 tosses were 2 and 4 respectively.

Therefore after the first two tosses as the expected number of tosses required to get a 5 is six. Therefore, we get here:

E [ X | Y = 1 and Z =2 ] = 2 + 6 = 8

Therefore 8 is the expected value here.

c) E[X|Y = 1 and Z = 3]

This means that the first toss was a 2 and the third toss was a 4, also there cannot be a 4 on the second toss.

Therefore there is a 1/5 probability that the second toss was a 5

Also there is a 4/5 probability that the second toss was not a 5.

Therefore the expected number of tosses required here is computed as:

E[X|Y = 1 and Z = 3] = (1/5)*2 + (4/5)*(3 + 6) = 2/5 + 36/5 = 38/5 = 7.6

d) E[X|Y = 3 and Z = 4]

Here the third toss was the first 2 and the fourth toss was the first 4

Therefore there is a (1/4) probability that the first 5 occurs on first toss, (3/4)*(1/4) probability that the first 5 occurs on the second toss and also a (3/4)² probability that there was no 5 in the first 4 tosses. Therefore the expected number of tosses required here is computed as:

E[X|Y = 3 and Z = 4] = (1/4)*1 + (3/4)*(1/4)*2 + (3/4)² *(4 + 6)

This is the required expression here.

e) Using the same cases as in the previous part, and just squaring X, we get here:

E[X²|Y = 3 and Z = 4] = (1/4)*1 + (3/4)*(1/4)*2² + (3/4)² *(4 + 6)² = (1/4) + (3/4) + 100*(3/4)²

This is the required expression here.