To solve these questions, we can use the concept of probability and the properties of independent events.
Before we begin, it's important to note that the probability of getting a head on any given toss is p, and the probability of getting a tail is 1-p.
1. Let's find P(B|A), which is the probability of event B happening given that event A has already occurred (there are 6 Heads in the first 8 tosses).
Since event A has already occurred, we only need to consider the last two tosses to determine if event B happens or not. The probability of getting a Head on the 9th toss is p.
Therefore, P(B|A) = p.
2. To find the probability of getting 3 Heads in the first 4 tosses and 2 Heads in the last 3 tosses, we can consider each set of tosses separately.
The probability of getting 3 Heads in the first 4 tosses is given by the binomial probability formula:
P(3 Heads in 4 tosses) = C(4,3) * p^3 * (1-p)^(4-3),
where C(4,3) is the number of ways to choose 3 Heads out of 4 tosses. Using the formula for combinations, C(4,3) = 4.
The probability of getting 2 Heads in the last 3 tosses is:
P(2 Heads in 3 tosses) = C(3,2) * p^2 * (1-p)^(3-2),
where C(3,2) is the number of ways to choose 2 Heads out of 3 tosses. C(3,2) = 3.
Since these two sets of tosses are independent events, we can multiply their probabilities:
P(3 Heads in 4 tosses and 2 Heads in 3 tosses) = P(3 Heads in 4 tosses) * P(2 Heads in 3 tosses)
= 4 * p^3 * (1-p) * 3 * p^2 * (1-p)
= 12 * p^5 * (1-p)^2.
So, the probability is 12 * p^5 * (1-p)^2.
3. Given that there were 4 Heads in the first 7 tosses, we want to find the probability that the 2nd Heads occurred on the 4th toss.
The probability of the 2nd Heads occurring on the 4th toss is independent of the outcomes of the other tosses. We can treat it as a separate event.
Since there are 3 Heads out of the first 7 tosses, we know that the 2nd Heads can only occur on the 4th or 5th toss. Therefore, the event of the 2nd Heads occurring on the 4th toss has a probability of 1/2.
So, the numerical answer is 1/2.