I have a bag with 3 coins in it. One of them is a fair coin, but the others are biased trick coins.

Question

When flipped, the three coins come up heads with probability 0.5, 0.3, 0.6 respectively. Suppose that I pick one of these three coins entirely at random and flip it three times. 1. What is P(HTT)? (i.e., it comes up heads on the first flip and tails on the last two flips.) 2.Assuming that the three flips, in order, are HTT, what is the probability that the coin that I picked was the fair coin? Don't need to reduce fractions

Work: 1. ((.5*.5)/(.5*.5))/3 + ((.3*.5)/(.7*.5))/3+ ((.6*.5)/(.4*.5))/3 - I think this is wrong 2. I dont know how to do

Answer 1

P(HHT) for each coin.

Coin 1 $P_1 = 0.5^3$

Coin 2 $P_2 = 0.3\cdot 0.7^2$

Coin 3 $P_2 = 0.6\cdot 0.4^2$

$P(HTT) = \frac 13 P_1 + \frac 13P_2 +\frac 13P_3$

$P(coin1|HHT) = \frac {P_1}{P_1+P_2+P_3}$

Answer 2

First: $$\frac13\cdot \frac18+\frac13\cdot 0.3\cdot 0.7^2+\frac13\cdot 0.6\cdot 0.4^2=0.122667.$$ Second: $$\frac{\frac13\cdot \frac18}{0.122667}=\frac{0.041667}{0.122667}=0.3397.$$

Answer 3

To set up Bayes' rule, first think the following way : FIX a coin. Now look at the probability of success.

Suppose a coin has probability $p$ of coming heads. From independence of successive flips, we conclude that the probability of the first three flips looking like $HTT$ is $p(1-p)^2$.

Now, you have three such coins, with $p=0.6,0.5,0.3$. So we can calculate the probability of , each one being FIXED, the first three flips looking like HTT, and these are (easy to see) respectively $0.096,0.125,0.147$.

Now, one of these coins are chosen uniformly at random. So, let us call the coins as coin $1,2,3$ in the order $p=0.6,0.5,0.3$.

Now, Bayes' rule would basically say this : the probability that you get HTT is the sum of the following:

1) The probability that you choose coin $1$, and it comes HTT.

2) The probability that you choose coin $2$, and it comes HTT.

3) The probability that you choose coin $3$, and it comes HTT.

What are these? Well, the choice is independent of the flipping, and is uniform, hence we have the probability of each one being chosen is $\frac 13$. Finally, the complete answer is $\frac{0.096+0.125+0.147}{3} = \frac{46}{375}$.

Now, the second question is asking this : given that HTT has occurred, what is the probability that we chose coin $2$. But then, coin $2$ is just case $2$ above, which is just the probability that with the second coin fixed, we get HTT.

So the answer is just the probability of getting HTT with coin $2$, divided by the probability of getting HTT with any of them(but here, there is no choice to be made, so we do not need to multiply by $\frac 13$). The numerator is then just $0.125 = \frac 18$ while the denominator is $\frac {46 \times 3}{375}$. Now, you can simplify and get $\frac{375}{1104}$.