when coin 1 is flipped, it lands on heads with a probability of .6; when coin 2

Question

when coin 1 is flipped, it lands on heads with a probability of .6; when coin 2 is flipped it lands on heads with a probability of .4. one of these coins is randomly chosen and flipped 6 times.

a) what is the conditional probability that the chosen coin is coin 1 given that the first flip is heads

b) what is the conditional probability that the chosen coin is coin 2 given that the first flip is heads

c) what is the conditional probablity that the coin lands on heads exactly 4 times given that the first of the 6 flips was on heads

Explanation / Answer

a) Let A be the event of coin 1 being flipped and B be the event that heads lands on first flip.

P(B|A) = 0.6 P(B|Ac) = 0.4

P(A) = 0.5 P(Ac) = 0.5

a) P(A B) = P(B|A) P(A) = 0.6 * 0.5 = 0.3

P(Ac B) = P(B|Ac) P(Ac) = 0.4 * 0.5 = 0.2

P(B) = P(A B) + P(Ac B) = 0.3 + 0.2 = 0.5

P(A|B) = P(A B) / P(B) = 0.3 / 0.5 = 0.6

b) P(Ac|B) = 1 - P(A|B) = 1 - 0.6 = 0.4

c) Since the first flip was on heads, we need to calculate the probability that out of the next 5 flips, there are exactly 3 heads.

The 3 heads can come in 5C3 = 10 ways.

Probability of exactly 3 heads = 10 * (0.5)3 * (0.5)2

= 10 * 0.55

= 0.3125.

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