One bag contains two coins. One is fair, the other is biased with Heads probability = $0.6$. One coin is randomly picked and it is tossed. It lands heads up. What is the probability that the same coin will land heads up if tossed again?
Now, the probability that for a coin randomly picked up the toss will result in a head is given by the formula for total probability. Writing $B_1, B_2$ the events that the fair coin and the biased coin is selected, respectively and $E_1$ the event that the coin will land heads up in the first toss, we obtain
$$
P(E_1) = P(E_1|B_1)P(B_1) + P(E_1|B_2)P(B_2) = 0.5\cdot0.5 + 0.6\cdot0.5 = 11/20.
$$
However, it is not clear to me how to calculate now the probability $P(E_2)$. I would like to use again the total probability formula for $E_2$, i.e.,
$$
P(E_2) = P(E_2|E_1)P(E_1) + P(E_2|E_1^c)P(E_1^c) = P(E_2|E_1)\cdot 11/20 + P(E_2|E_1^c)\cdot 9/20
$$
But I don't see how to calculate $P(E_2|E_1), \ P(E_2|E_1^c)$. So in the end I suspect that my method doesn't work and a better solution should be given.
Best Answer
Next you need to use bayes theorem to figure out the posterior probability $P(B_1|E_1)$ that the coin picked is fair $$P(B_1|E_1)=\frac {P(E_1|B_1)P(B_1)}{P(E_1)}=\frac {0.5 \cdot 0.5}{11/20} = 5/11$$
To determine the probability of $E_2$ given $E_1$ $$P(E_2|E_1) = P(E_2 B_1|E_1) + P(E_2 B_1^c|E_1) = P(E_2|B_1 E_1)P(B_1|E_1) + P(E_2|B_1^c E_1)P(B_1^c|E_1) $$ Using the fact that flips of the same coin are independent, $P(E_2|B_1 E_1)=P(E_2|B_1)=0.5$ and $P(E_2|B_1^c E_1)=P(E_2|B_1^c)=0.6$, and substituting these probabilities in gives $$P(E_2|E_1) = P(E_2|B_1)P(B_1|E_1) + P(E_2|B_1^c)P(B_1^c|E_1) = .5 \cdot 5/11 + .6 \cdot 6/11$$