Solution:

Given, a fair dice is tossed repeatedly until six shows up 3 times.

Let X be the number of tosses needed to get 3 sixes.

We need to find the probability that exactly 5 tosses are needed.

To find this, we need to consider the following cases:

Case 1: The first four tosses do not result in 3 sixes

In this case, the first four tosses can result in any number other than six. The probability of getting any number other than six in one toss is 5/6. Hence, the probability of getting any number other than six in four tosses is (5/6)^4.

The fifth toss must be a six, so the probability of getting a six in one toss is 1/6.

Hence, the probability of getting 3 sixes in exactly 5 tosses given the first four tosses do not result in 3 sixes is (5/6)^4 × 1/6 = 25/7776.

Case 2: The first four tosses result in exactly 1 six

In this case, the first four tosses can result in any number other than six twice and a six once. The probability of getting any number other than six in one toss is 5/6, and the probability of getting a six in one toss is 1/6. Hence, the probability of getting exactly 1 six in four tosses is 4 × (5/6)^3 × 1/6.

The fifth toss must be a six, so the probability of getting a six in one toss is 1/6.

Hence, the probability of getting 3 sixes in exactly 5 tosses given the first four tosses result in exactly 1 six is 4 × (5/6)^3 × 1/6 × 1/6 = 5/486.

Case 3: The first four tosses result in exactly 2 sixes

In this case, the first four tosses can result in any number other than six once and a six twice. The probability of getting any number other than six in one toss is 5/6, and the probability of getting a six in one toss is 1/6. Hence, the probability of getting exactly 2 sixes in four tosses is 6 × (5/6)^2 × (1/6)^2.

The fifth toss must be a six, so the probability of getting a six in one toss is 1/6.

Hence, the probability of getting 3 sixes in exactly 5 tosses given the first four tosses result in exactly 2 sixes is 6 × (5/6)^2 × (1/6)^2 × 1/6 = 25/648.

Case 4: The first four tosses result in exactly 3 sixes

In this case, the first four tosses must result in exactly 3 sixes, so the probability of this happening is (1/6)^3 = 1/216.

The fifth toss must not be a six, so the probability of getting any number other than six in one toss is 5/6.

Hence, the probability of getting 3 sixes in exactly 5 tosses given the first four tosses result in exactly 3 sixes is (1/6

P(6 showing up) = 1/6= x

At 5th toss, we should get the 3rd '6' so that the process terminates.

In other words, we should get 2 sixes before the fifth toss. (I.e in four tosses)

Required probability= (1/6 *1/6)* (1-1/6)*(1-1/6).