Catchfire
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Hello,
I've been working on some questions from Rice's Mathematical Statistics and Data Analysis and I'm not sure about my solutions.
The question I'm working on is as follows:
1.8 #59c
A box has 3 coins: 1 with two heads, 1 with 2 tails and 1 fair coin. A coin chosen randomly is flipped and comes up heads. If the coin is flipped again and comes up heads, what's the probability it's the two headed coin?
My solution:
Let A be the event the coin has two heads and let B be the event that heads is flipped twice.
P(A|B) = P(A[itex]\cap[/itex]B) / P(B) = P(B|A)P(A) / P(B) = 1*(1/3) / P(B)
P(B) = 1/3*1 + 1/3*1/4 = 5/12
P(A|B) = 1/3 / 5/12 = 4/5
Is this correct?
I've been working on some questions from Rice's Mathematical Statistics and Data Analysis and I'm not sure about my solutions.
The question I'm working on is as follows:
1.8 #59c
A box has 3 coins: 1 with two heads, 1 with 2 tails and 1 fair coin. A coin chosen randomly is flipped and comes up heads. If the coin is flipped again and comes up heads, what's the probability it's the two headed coin?
My solution:
Let A be the event the coin has two heads and let B be the event that heads is flipped twice.
P(A|B) = P(A[itex]\cap[/itex]B) / P(B) = P(B|A)P(A) / P(B) = 1*(1/3) / P(B)
P(B) = 1/3*1 + 1/3*1/4 = 5/12
P(A|B) = 1/3 / 5/12 = 4/5
Is this correct?