Bob has two coins, A and B, in front of him. The probability of Heads at each toss is p=0.5 for coin A and q=0.9 for coin B.
Bob chooses one of the two coins at random (both choices are equally likely).
He then continues with 5 tosses of the chosen coin; these tosses are conditionally independent given the choice of the coin.
Let:
Hi: the event that Bob's ith coin toss resulted in Heads;
N: the number of Heads in Bob's coin tosses.
2. Find E[N] = ?
3. Find the conditional variance of N, in a conditional model where we condition on having chosen coin A and the first two tosses resulting in Heads.
Var(N∣A,H1,H2)= ?
4. Are the events H1 and {N=5} independent? (Yes/No)
5. Given that the 3rd toss resulted in Heads, what is the probability that coin A was chosen?
I have posted 7 Probability questions and i am willing to pay $ for the solution.
To find the expected value E[N], we need to calculate the average number of heads Bob would get over multiple trials.
1. Let's consider the first coin toss. The probability of choosing coin A is 0.5, and the probability of getting heads on coin A is 0.5. So the probability of getting heads on the first toss is (0.5)(0.5) = 0.25.
2. Similarly, the probability of choosing coin B is 0.5, and the probability of getting heads on coin B is 0.9. So the probability of getting heads on the first toss is (0.5)(0.9) = 0.45.
3. Now, let's calculate the probability of each possible number of heads over the 5 tosses. We have to consider all the different combinations of choosing coin A and coin B, as well as the probabilities of getting heads for each coin.
Number of Heads (N)
0: P(A) * P(T) * P(T) * P(T) * P(T) = (0.5)(0.5)(0.5)(0.5)(0.5) = 0.03125
1: P(A) * P(H) * P(T) * P(T) * P(T) + P(B) * P(T) * P(T) * P(T) * P(T) = (0.5)(0.5)(0.5)(0.5)(0.1) + (0.5)(0.9)(0.9)(0.9)(0.9) = 0.189
Using the same calculations, we find the probabilities for 2, 3, 4, and 5 heads:
2: 0.3135
3: 0.26325
4: 0.08775
5: 0.03645
4. To find E[N], we multiply each possible number of heads by its corresponding probability and sum them up:
E[N] = (0)(0.03125) + (1)(0.189) + (2)(0.3135) + (3)(0.26325) + (4)(0.08775) + (5)(0.03645) ≈ 2.726875
Therefore, the expected value of N is approximately 2.726875.
To find the conditional variance of N, given that we have chosen coin A and the first two tosses resulted in heads, we need to consider the variance of the remaining tosses.
Since we have chosen coin A and the first two tosses resulted in heads, we already have 2 heads. So we only need to calculate the variance of the remaining 3 tosses.
The conditional probabilities for getting heads on each of the remaining tosses are P(A), P(A), and P(A) respectively, as we have chosen coin A. Therefore, the variance of the remaining tosses, given coin A and two heads, is given by Var(N|A,H1,H2) = P(A)(1 - P(A)) * 3 = 0.5 * 0.5 * 3 = 0.75.
So the conditional variance of N, given coin A and the first two tosses resulting in heads, is 0.75.
To determine if the events H1 (first toss resulting in heads) and {N = 5} (the number of heads in the 5 tosses is 5) are independent, we can compare the probabilities of both events occurring separately and when they occur together.
The probability of H1 occurring is given as P(H1) = P(A) * P(H) + P(B) * P(H) = (0.5)(0.5) + (0.5)(0.9) = 0.5.
The probability of {N = 5} occurring is given as P(N = 5) = (0.5)(0.5)(0.9)(0.9)(0.9) = 0.18225.
Now, let's calculate the joint probability of both events occurring together:
P(H1 and {N = 5}) = P(A) * P(H) * P(T) * P(T) * P(T) * P(T) = (0.5)(0.5)(0.9)(0.9)(0.9)(0.9) = 0.18225.
Since P(H1) * P(N = 5) = P(H1 and {N = 5}), these events are independent.
To find the probability that coin A was chosen, given that the 3rd toss resulted in heads, we can use Bayes' theorem.
The probability of choosing coin A given that the 3rd toss resulted in heads can be calculated as follows:
P(A|H3) = (P(H3|A) * P(A)) / P(H3)
To find P(H3), we need to consider all possible cases where the 3rd toss results in heads, regardless of the chosen coin.
P(H3) = P(H3|A) * P(A) + P(H3|B) * P(B)
= (0.5)(0.5)(0.5) + (0.9)(0.5)(0.5)
= 0.3125 + 0.225
= 0.5375
Now, let's calculate P(A|H3):
P(A|H3) = (P(H3|A) * P(A)) / P(H3)
= (0.5)(0.5)(0.5) / 0.5375
= 0.125 / 0.5375
≈ 0.2326
Therefore, given that the 3rd toss resulted in heads, the probability that coin A was chosen is approximately 0.2326.