5 coins are put in a bag. 3 of the coins are weighted with the probability of flipping heads being three times as great than the probability of flipping tails; the remaining coins are fair. One of these coins is selected at random and then flipped once. What is the probability that a weighted coin was selected given that heads was flipped?

9/13 = 0.6923

Step-by-step explanation:

We start by defining

A as event that head was flipped

B1 = event that coin is biased

B2 = event that it is unbiased

P(B1) = 3/5

P(B2) = 2/5

P(A|B1) = 3/4

P(A|B2) = 2/4 = 1/2

When we solve this using bayes theorem we have to find

p(B1|A) = [P(B1) x P(A|B1)]/[P(B1) x P(A|B1) + P(B2) x P(A|B2)

= 0.6 x 0.75 / 0.6 x 0.75 + 0.4x0.5

= 0.45/0.45+0.2

= 0.45/0.65

= 0.6923

$27.5

step-by-step explanation:

$110.00 divided by 4

the answer is 45 i just did it on oddesy ware