Finite Math
posted by Preet .
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

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(AB) stands for probability of A happening given that B is true. In numerical terms,
P(AB)=P(A∩B)/P(B)
2. Let
A=coin is regular
B=produced a head
P(AB)=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(AB)=(1/6)/(7/10)
=5/21
3. similar to (2), left as exercise for you.
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