posted by Preet on .
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
Add up the three probabilities to get 7/10.
For events A and B, P(A|B) stands for probability of A happening given that B is true. In numerical terms,
A=coin is regular
B=produced a head
P(A∩B)=[(1/3)(1/2)]=1/6 as calculated above
P(B)=7/10 as calculated above
3. similar to (2), left as exercise for you.