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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?


Elaboration of @RyanVolpi's answer.

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,

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

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In your solution, you calculate the probability of getting tails twice as the square of the probability of getting tails on the first flip. This assumes that consecutive flips are independent. In fact, they are not, as the same coin is used throughout and so getting tails on the first throw means it is more likely to get tails on consecutive throws. This is because the coin landing tails on the first throw is more indicative of the case where you are flipping the fair coin. Your result is the correct solution for a case where the coin is reselected after the first flip. Instead, for the case where the same coin is used repeatedly, define:

$$H_2: \text{event that the coin comes up heads two times in a row.}$$

You can calculate the probability of this event simply as follows: $$ \begin{align} P(H_2) &= P(H_2 \cap C1) + P(H_2 \cap C2) \\ &= P(H_2 | C1)P(C1) + P(H_2 | C2)P(C2) \\ &= .5^2 * .5 + .1^2 * .5 \\ &= .13 \end{align} $$

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  • $\begingroup$ That's that made a lot of sense. I've elaborated your answer in my question which is a bit easier to follow IMO. $\endgroup$ – Quazi Irfan Jul 8 '20 at 3:31

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