Consider 10 independent tosses of a biased coin with the probability of Heads at each toss equal to p, where 0<p<1.
1. Let A be the event that there are 6 Heads in the first 8 tosses. Let B be the event that the 9th toss results in Heads.
Find P(B∣A) and express it in terms of p using standard notation. (You can click on the “STANDARD NOTATION" button below.)
2. 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. Remember not to use ! or combinations in your answer.
3. 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.
4. We are interested in calculating the probability that there are 5 Heads in the first 6 tosses and 3 Heads in the last 5 tosses. Give the exact numerical values of a, b, c, d that would match the answer ap7(1−p)3+bpc(1−p)d.
a=
b=
c=
d=
4.a =30
4.b =4
4.c =8
4.d =2
1. = p
2. = 12*p^5*(1-p)^2
3. = 9/35
4.a =
4.b =
4.c =
4.d =
Can someone find the 4.a, 4.b, 4.c, 4.d? :)
4.c = 8
4.b = 4
4.c = 8
1. P(B∣A):
To find P(B∣A), we need to find the probability of event B occurring given that event A has occurred. In this case, event A is the occurrence of 6 Heads in the first 8 tosses.
Since A has already occurred, we now have 2 tosses remaining (9th and 10th tosses). The probability of getting a Head on the 9th toss is p, which is given. Therefore, P(B∣A) = p.
2. Probability of 3 Heads in the first 4 tosses and 2 Heads in the last 3 tosses:
Let's break this down into two events, C and D.
Event C is the occurrence of 3 Heads in the first 4 tosses. The probability of getting 3 Heads in 4 tosses is given by the binomial distribution formula:
P(C) = (4 choose 3) * p^3 * (1-p)^(4-3) = 4p^3(1-p)
Event D is the occurrence of 2 Heads in the last 3 tosses. The probability of getting 2 Heads in 3 tosses is given by the binomial distribution formula:
P(D) = (3 choose 2) * p^2 * (1-p)^(3-2) = 3p^2(1-p)
To find the probability of both events C and D occurring, we multiply their probabilities:
P(C and D) = P(C) * P(D) = 4p^3(1-p) * 3p^2(1-p) = 12p^5(1-p)^2
3. Probability that the 2nd Heads occurred at the 4th toss:
Given that there were 4 Heads in the first 7 tosses, we already know that the 2nd Heads occurred within these 7 tosses. Therefore, the 2nd Heads cannot occur at the 4th toss.
The probability is 0.
4. Probability of 5 Heads in the first 6 tosses and 3 Heads in the last 5 tosses:
The probability can be found using the binomial distribution formula.
a = (6 choose 5) = 6
b = p^5 = p^5
c = (5 choose 3) = 10
d = (1-p)^3 = (1-p)^3
Therefore, the answer is ap7(1−p)3+bpc(1−p)d = 6p^7(1-p)^3+10p(1-p)^3.
1. To find P(B|A), we need to find the probability of event B occurring given that event A has already occurred.
Event A is defined as getting 6 Heads in the first 8 tosses, so we can use the binomial probability formula to find the probability of getting exactly 6 Heads in 8 tosses:
P(A) = C(8, 6) * p^6 * (1-p)^2
Now, given that event A has occurred and we have 6 Heads in the first 8 tosses, we have 2 tosses left. We want to find the probability of getting a Head in the 9th toss (event B). Since events A and B are independent, we can simply multiply the probability of event B given that A has occurred by the probability of A:
P(B|A) = p * P(A)
Therefore, P(B|A) = p * C(8, 6) * p^6 * (1-p)^2 = C(8, 6) * p^7 * (1-p)^2
2. To find the probability of having 3 Heads in the first 4 tosses and 2 Heads in the last 3 tosses, we can again use the binomial probability formula:
P(3 Heads in 4 tosses) = C(4, 3) * p^3 * (1-p)^1
P(2 Heads in 3 tosses) = C(3, 2) * p^2 * (1-p)^1
Since these events are independent, the probability of both events happening is simply the product of their individual 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)
Therefore, the probability is:
P(3 Heads in 4 tosses and 2 Heads in 3 tosses) = C(4, 3) * p^3 * (1-p)^1 * C(3, 2) * p^2 * (1-p)^1
3. Given that there were 4 Heads in the first 7 tosses, we want to find the probability that the 2nd Heads occurred at the 4th toss.
In this case, we are assuming that the 4th toss is the 2nd Heads. So, we need to calculate the probability of getting a Head on the 4th toss, given that there are already 3 Heads in the first 3 tosses and 1 Head in the next 3 tosses.
Since these events are independent, the probability of the 2nd Heads occurring at the 4th toss is simply:
P(2nd Heads at 4th toss | 4 Heads in first 7 tosses) = p
So, the probability is simply the value of p.
4. To find the exact numerical values of a, b, c, and d that match the given expression ap7(1−p)3+bpc(1−p)d, we need to calculate the probabilities of the individual events.
Specifically, we want to calculate the probability of getting 5 Heads in the first 6 tosses, which can be calculated using the binomial probability formula:
P(5 Heads in 6 tosses) = C(6, 5) * p^5 * (1-p)^1
Also, we want to calculate the probability of getting 3 Heads in the last 5 tosses:
P(3 Heads in 5 tosses) = C(5, 3) * p^3 * (1-p)^2
Now, the values of a, b, c, and d from the given expression correspond to the coefficients in these probabilities.
Therefore:
a = C(6, 5) = 6
b = 1
c = C(5, 3) = 10
d = 2
So, the exact numerical values of a, b, c, and d that match the expression ap7(1−p)3+bpc(1−p)d are:
a = 6
b = 1
c = 10
d = 2