Conditional Probability Difficult Problem Help... having my first encounter on conditional probability and then I meet this problem.. it includes several conditions and I am not sure if Im thinking the right procedure on this problem. Any idea will be of great help. Thanks in advance.
There are 3 coins. One is two headed coin, the other one comes heads 75 percent of the time and the third is a fair coin. A coin is selected at random and flipped.
1) What is the probability that the flipped coin will come up head?
2) Given that the coin that was flipped comes up head, what is the probability that it was fair coin?
 A: For the first, simply use the basic definition of probability
$$\dfrac{Favourable}{Total}=\frac{1+\frac{3}{4}+\frac{1}{2}}{3}$$
For the second, use Bayes Theorem. Let the events be :


*

*$U_1$: Coin with both heads flipped.

*$U_2$: Coin which shows head 3/4 times flipped.

*$F$: Fair coin flipped.

*$H$ : Head is the result.
Then we need to find $P(F|H)$
$$P(F|H) = \dfrac{P(H|F)\cdot P(F)}{P(H|U_1)\cdot P(U_1)+P(H|U_2)\cdot P(U_2)+P(H|F)\cdot P(F)}$$
$$=\dfrac{\frac{1}{2}\cdot \frac{1}{3}}{\frac{1}{1}\cdot \frac{1}{3}+\frac{3}{4}\cdot \frac{1}{3}+\frac{1}{2}\cdot \frac{1}{3}} = 2/9$$
A: For a)
Just create a tree diagram with all the different options:
then, follow the tree to find the probability of each event. It should equal 3/4
B is a little bit more complex
The formula for conditional probability is P(A|B) = P(A and B)/P(B)
In this case
The probability of a fair coin, if it is known that the outcome is a heads.
The probability of getting a heads is 3/4 as we solved for in a:
so we get P(A and B)/0.75
The probability that you choose the fair coin is 1/3, and the probability of heads on said coin is 1/2
So, probability of A and B is 1/6
if you do (1/6)/(3/4)
you get 2/9
A: Using tree diagram,
a) P[H] = P[C1H] + P[C2H]+P[C3H]= 1/3 + 1/4 + 1/6 = 3/4
b) P[C3|H] = P[C3H]/P[H] = (1/6)/ (3/4) = 2/9
