Question

A fair die is rolled 20 times. We define Ai as the event that number i appears i

times (i = 1, ..., 6).
(a) Calculate P(Ai) for all i.
(b) Calculate P(Ai ∩ Aj) for all i ̸= j.
(c) Calculate P(A1 ∩A2 ∩A3 ∩A4 ∩A5 ∩A6). (d) Calculate P(A2 ∩A3 ∩A4 ∩A5 ∩A6).

Answer 1

Part a)

The probability that the ith face appears on rolling a dice (success)= 1/6

Thus the probability of failure = 1 - 1/6 = 5/6

Assumption : the rolling of each die is independent of any other trial.

Given,
Ai is the event that number i appears i times (i = 1, ..., 6),
and the die is rolled 20 times
For i =1 (i.e 1 success and 19 failures)

For i =2(i.e 2 success and 18 failures)

For i =3(i.e 3 success and 17 failures)

For i =4(i.e 4 success and 16 failures)

For i =5(i.e 5 success and 15 failures)

For i =6 (i.e 6 success and 14 failures)

Thus,

Part b)

= i+j successes and 20 - (i+j) failures.

Such as, for i =2 and j =5, there must be 2 rolls with 'Face 2' and 5 rolls with 'Face 5' and rest out of 20 trials which has the probability as follows,

Part c)

The event A1 ∩A2 ∩A3 ∩A4 ∩A5 ∩A6 denotes that each of the i faces have appeared i times on rolling the dice 20 times.

This is an impossible event as the above event would imply that there are 1 roll with 'face 1', 2 roll with 'face 2', 3 roll with 'face 3', 4 roll with 'face 4', 5 roll with 'face 5', 6 roll with 'face 6', but sum of all number of events is 21; and given there are only 20 rolls of die.

THUS ;

Part d)