Posted by **Preet** on Saturday, July 21, 2012 at 11:03pm.

John has three coins in his pocket. One is fair, one has two heads, and the third is unbalanced, showing head 60% of the time. John grabs a coin at random and tosses it. Find the probability that:

1) The coin shows a head

2) The coin is regular, give that the toss produced a head

3) The coin is regular, given that the toss produced a tail

- Finite Math -
**MathMate**, Sunday, July 22, 2012 at 7:16am
1. Probability of showing a head:

Coins are chosen at random, so each coin has a probability of (1/3) of being selected.

probability for a head:

fair coin: (1/3)(1/2)=1/6

two heads: (1/3)(1)=1/3

unbalanced: (1/3)(0.6)=1/5

Add up the three probabilities to get 7/10.

Conditional probabilty:

For events A and B, P(A|B) stands for probability of A happening given that B is true. In numerical terms,

P(A|B)=P(A∩B)/P(B)

2. Let

A=coin is regular

B=produced a head

P(A|B)=P(A∩B)/P(B)

P(A∩B)=[(1/3)(1/2)]=1/6 as calculated above

P(B)=7/10 as calculated above

P(A|B)=(1/6)/(7/10)

=5/21

3. similar to (2), left as exercise for you.

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