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.
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.
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.
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 the event that there are 6 heads in the first 8 tosses. So, we have already fixed the outcome of the first 8 tosses as 6 heads.
Now, to find P(B|A), we need to find the probability that the 9th toss results in heads. Since the coin is biased, the probability of heads at each toss is p.
Therefore, P(B|A) = P(9th toss is heads given 6 heads in the first 8 tosses) = p
2. To find the probability of getting 3 heads in the first 4 tosses and 2 heads in the last 3 tosses, we need to find the probability of the specific arrangement HHHHTTH.
The probability of getting heads in any toss is p, and the probability of getting tails is (1 - p).
Therefore, the probability of this specific arrangement is:
P = p * p * p * (1 - p) * (1 - p) = p^3 * (1 - p)^2
3. Given that there were 4 heads in the first 7 tosses, we need to find the probability that the 2nd heads occurred at the 4th toss.
We can calculate this by considering the number of ways this can happen. The first heads can occur in any of the first 3 tosses (since the 2nd heads occurs at the 4th toss). The remaining 3 heads can occur in any of the remaining 6 tosses.
So, the number of favorable outcomes (number of ways 2nd heads occurs at the 4th toss) is 3 * 6 = 18.
The total number of possible outcomes is 2^7 (since each toss can result in 2 possibilities: heads or tails).
Therefore, the probability is given by:
P = favorable outcomes / total outcomes = 18 / 2^7 = 18 / 128 = 9 / 64
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.
1. We have 8 coins that, of which 6 will turn heads. Out of 8 Coins, we come up with a numerical representation: 8C6, where C=combination.
Answer: 28⋅p6⋅(1−p)^2*p/(28⋅p6⋅(1-p)^2)
2. Answer= 12p^5*(1-p)^2
3. For the following equation: fill in letters a-f to show problem 3.
ap7(1−p)3+bpc(1−p)d+epf(1−p)f.
a=15
b=60
c=6
d=4
e=10
f=5