Consider 10 independent tosses of a biased coin with the probability of Heads at each toss equal to p, where 0<p<1. Given that there were 4 Heads in the first 7 tosses, find the probability that the 2nd Heads occurred at the 4th toss. Give a numerical answer.

To find the probability that the 2nd Heads occurred at the 4th toss, we need to use conditional probability.

Let's break down the problem step by step:

Step 1: Find the probability of getting 4 Heads in the first 7 tosses.

Since each toss is independent, the probability of getting a Head on any single toss is p. Therefore, the probability of getting a Tail on any single toss is 1 - p (since there are only two possible outcomes).

We are given that there were 4 Heads in the first 7 tosses. This means that there were 3 Tails in the first 7 tosses.

Using the binomial probability formula, we can calculate:

P(4 Heads in 7 tosses) = C(7, 4) * p^4 * (1 - p)^3

Step 2: Find the probability of the 2nd Heads occurring at the 4th toss, given that there were 4 Heads in the first 7 tosses.

Since there were 4 Heads in the first 7 tosses, that means the remaining 3 tosses must be Tails.

The possible positions for the 2nd Heads are at toss number 2, 3, 4, 5, 6, or 7. However, we are only interested in finding the probability of it occurring at the 4th toss.

Therefore, the probability of the 2nd Heads occurring at the 4th toss is equal to the probability of getting 4 Heads in the first 7 tosses multiplied by the probability of getting a Tails on the 3rd toss.

P(2nd Heads at 4th toss | 4 Heads in 7 tosses) = P(4 Heads in 7 tosses) * P(Tails on 3rd toss)

Step 3: Calculate the answer.

P(Tails on 3rd toss) = 1 - p (since it cannot be Heads)

P(2nd Heads at 4th toss | 4 Heads in 7 tosses) = P(4 Heads in 7 tosses) * P(Tails on 3rd toss)
= C(7, 4) * p^4 * (1 - p)^3 * (1 - p)

Now, you can plug in the value of p (the probability of Heads at each toss) to get the numerical answer.