Pick up a coin C1 given C1+C2 with probability of trials p (h1) =.7, p (h2) =.6

Pick up a coin C1 given C1+C2 with probability of trials p (h1) =.7, p (h2) =.6 and doing 10 trials. And what is the probability that the given coin you picked is C1 given you have 7 heads and 3 tails?

Question Posted / kaptan singh

1 Answers
381 Views
Facebook, I also Faced
E-Mail Answers

Pick up a coin C1 given C1+C2 with probability of trials p (h1) =.7, p (h2) =.6 and doing 10 trials...

Answer / Mahesh Chauhan

To find the probability of picking coin C1 given that you have 7 heads and 3 tails in 10 trials, we need to use Bayes' theorem:nn1. Let H_1 be the event of picking coin C1, and H_2 be the event of picking coin C2.n2. Let H be the event of getting a head (since we are only considering heads and tails), and T be the event of getting a tail.n3. The probabilities of H_1 and H_2 are given as p(H_1) = 0.7, and p(H_2) = 0.6.n4. Let N be the number of trials (in this case, N = 10).n5. We have observed 7 heads and 3 tails, so let X = sum of H events occur (X = 7).n6. Using Bayes' theorem: P(H_1 | X) = [P(X|H_1) * p(H_1)] / [P(X|H_1) * p(H_1) + P(X|H_2) * p(H_2)]n7. Now, let's find the probabilities of getting X heads given H_1 and H_2:n - For H_1: P(X | H_1) = C(N, X) * (p(H_1))^X * (p(T|H_1))^(N-X)n where C(N, X) is the number of combinations of choosing X items from N, which can be calculated using the formula: C(N, X) = N! / [X!(N-X)!]n - For H_2: P(X | H_2) = C(N, X) * (p(H_2))^X * (p(T|H_2))^(N-X)n8. Plug the values in Bayes' theorem and calculate the probability of picking coin C1 given 7 heads and 3 tails.

Is This Answer Correct ?

0 Yes

0 No

Post New Answer