# Probability of two tails with biased coins

Here is the problem I am trying to solve.

Coin 1 is fair. When flipped, it has a probability of 0.5 for heads and 0.5 for tails. Coin 2 is biased. When flipped, it has a probability of 0.9 for heads and 0.1 for tails. You grab a coin at random and flip it twice. What's the probability that it comes up tails both times?

Here is my attempted solution:

We have to calculate $$P(T \cap T)$$.

Using the law of total probability, we can calculate the probabliy of getting a head, \begin{align} P(H) &= P(H \cap C1) + P(H \cap C2) \\ &= P(H | C1)P(C1) + P(H | C2)P(C2) \\ &= .5 * .5 + .9 * .5 \\ &= .25 + .45 \\ &= .7 \end{align}

Similarlity we can calculate the probability of gettting a tail,

\begin{align} P(T) &= P(T \cap C1) + P(T \cap C2) \\ &= P(T | C1)P(C1) + P(T | C2)P(C2) \\ &= .5 * .5 + .1 * .5 \\ &= .25 + .05 \\ &= .3 \end{align}

Now we can make a truth table of 2 coin tosses, $$\begin{array} {|r|r|}\hline Toss1 & Toss2 & Probability \\ \hline H & H & .7 * .7 = .49 \\ \hline H & T & .7 * .3 = .21 \\ \hline T & H & .3 * .7 = .21 \\ \hline T & T & .3 * .3 = .09 \\ \hline \end{array}$$

So the answer to the question is $$P(T \cap T) = .09$$.

But it is the wrong answer. The correct answer is $$0.13$$.

What did I do wrong?

The original solution would be correct if we were to fetch a new coin in every toss. But in this problem, we do not replace the coin. We choose one coin and make both tosses with it. Therefore we need to construct a truth table for each coin. And also calculating the probabilities of both tosses resulting in tail from each coin.

Truth table for Coin 1 is, $$\begin{array} {|r|r|}\hline Toss1 & Toss2 & Probability \\ \hline H & H & .5 * .5 = .25 \\ \hline H & T & .5 * .5 = .25 \\ \hline T & H & .5 * .5 = .25 \\ \hline T & T & .5 * .5 = .25 \\ \hline \end{array}$$

Truth table for Coin 2 is, $$\begin{array} {|r|r|}\hline Toss1 & Toss2 & Probability \\ \hline H & H & .9 * .9 = .81 \\ \hline H & T & .9 * .1 = .09 \\ \hline T & H & .1 * .9 = .09 \\ \hline T & T & .1 * .1 = .01 \\ \hline \end{array}$$

From these tables we can see that $$P(T1, T2 | C1 ) = .25$$ and $$P(T1, T2 | C2 ) = .01$$

Using law of total probability,

\begin{align} P(T1, T2) &= P(T1, T2 \cap C1) + P(T1, T2 \cap C2) \\ &= P(T1, T2 | C1) P(C1) + P(T1, T2 | C2)P(C2) \\ &= .25 * .5 + .01 * .5 \\ &= .5 * .5 + .1 * .5 \\ &= .125 + .005 \\ &= .13 \end{align}

$$H_2: \text{event that the coin comes up heads two times in a row.}$$
You can calculate the probability of this event simply as follows: \begin{align} P(H_2) &= P(H_2 \cap C1) + P(H_2 \cap C2) \\ &= P(H_2 | C1)P(C1) + P(H_2 | C2)P(C2) \\ &= .5^2 * .5 + .1^2 * .5 \\ &= .13 \end{align}