Repeated die toss experiment. A fair die is tossed repeatedly, and the sequence of numbers that turn up is recorded, until a 1 is

obtained. When a 1 is obtained, the experiment is over, but the 1 is included in the sequence. Thus every outcome is a sequence of integers ending
in 1. Let X be the sum of the sequence of numbers obtained in the experiment. So the smallest possible value of X is 1, corresponding to the outcome
in which a 1 is obtained on the �first toss.

Compute P(X=5).

Probability of NOT getting a 1 = 5/6, getting a 1 = 1/6.

If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.

For five tosses:

(5/6)^4 * 1/6 = ?

To compute the probability that the sum X is equal to 5, we need to determine all the possible sequences of numbers that result in a sum of 5.

First, let's consider the possible number combinations for sums of 5:

1 + 1 + 1 + 1 + 1 = 5
1 + 1 + 1 + 2 = 5
1 + 1 + 2 + 1 = 5
1 + 2 + 1 + 1 = 5
2 + 1 + 1 + 1 = 5
1 + 1 + 3 = 5
1 + 3 + 1 = 5
3 + 1 + 1 = 5
2 + 2 + 1 = 5
2 + 1 + 2 = 5
1 + 2 + 2 = 5

Note that all the sequences end with a 1.

Now, let's calculate the probability for each of these sequences:

The probability of getting a 1 on the first toss is 1/6, and the remaining tosses can be any number from 1 to 6 with a probability of 1/6 for each. So, the probability for the first sequence (1 + 1 + 1 + 1 + 1) is (1/6)^5 = 1/7776.

For the other sequences, we can calculate the probabilities using combinatorics. Since there can be different arrangements of the numbers, we need to consider the number of ways each sequence can occur.

For example, for the sequence 1 + 3 + 1, there are 3 positions where we can place the 1s, and the remaining position is a 3. So, the probability for this sequence is (3/6) * (1/6) * (1/6) * (1/6) = 3/7776.

We can do the same calculations for all the sequences and sum up their probabilities:

P(X=5) = (1/7776) + (3/7776) + (3/7776) + (3/7776) + (3/7776) + (6/7776) + (6/7776) + (6/7776) + (3/7776) + (3/7776) + (3/7776) = 42/7776

Simplifying this fraction, we get:

P(X=5) = 7/1296

Therefore, the probability that the sum X is equal to 5 is 7/1296.