Find the probability that there are 3 Heads in the first 4 tosses and 2 Heads in the last 3 tosses. Express your answer in terms of p using standard notation.
To find the probability of getting a certain number of heads in a series of coin tosses, we will use the binomial probability formula.
The binomial probability formula is given by P(x) = (nCx) * (p^x) * ((1-p)^(n-x)), where
P(x) is the probability of getting x successes,
n is the total number of trials,
x is the number of successful outcomes,
p is the probability of a single successful outcome,
(1-p) is the probability of a single unsuccessful outcome,
and (nCx) is the number of combinations of choosing x successes from n trials.
In this case, we want to find the probability of getting 3 Heads in the first 4 tosses and 2 Heads in the last 3 tosses.
Let's break it down:
First, we find the probability of getting 3 Heads in the first 4 tosses, which means n = 4 and x = 3.
P(3) = (4C3) * (p^3) * ((1-p)^(4-3))
Next, we find the probability of getting 2 Heads in the last 3 tosses, which means n = 3 and x = 2.
P(2) = (3C2) * (p^2) * ((1-p)^(3-2))
Finally, we multiply the two probabilities together to get the overall probability:
P = P(3) * P(2)
So, the probability of getting 3 Heads in the first 4 tosses and 2 Heads in the last 3 tosses is given by P = (4C3) * (p^3) * ((1-p)^(4-3)) * (3C2) * (p^2) * ((1-p)^(3-2)).