# Calculating pmf of different coins

I have 6 coins, with 2 coins being double-headed and the rest normal. A coin is chosen at random and tossed twice. If I decided that the number of heads obtained is a random variable $X$, How can I find the pmf of $X$?

So fair it is obvious that $X = 0, 1, 2$, but I am unsure of how to calculate $P(X=0), P(X=1)$ and $P(X=2)$. How can I do this? I am learning the basics of probability and would appreciate any help for solving this problem.

• One way to look at it is as a two-step process. First pick a coin, then flip. Generally we multiply probabilities for two step processes such as this. Can you think of what you will get? HInt: there are two types of coins you can get, and the flips of each have different probabilities. You can also think of it as three-steps as there are two flips.
– Arby
Apr 5, 2017 at 3:01

The probability of selecting a two-headed coin is 2/6 or 1/3. The probability of a selecting a normal coin is 4/6 or 2/3. If you get a two-headed coin, you're guaranteed to get 2 heads and a normal coin gives two heads 1/4th of the time for two flips, so the probability of two heads is $\frac 1 3\cdot \frac 1 1 + \frac 2 3\cdot \frac 1 4=\frac 12$.

One or zero heads must come from the normal coins, so we won't consider the two-headed coins any more. The probability of one head on two flips is 1/2 so the probability of first selecting a normal coin and then getting one head is $\frac 2 3\cdot \frac1 2=\frac 1 3$. The probability of zero heads on two flips is 1/4 so the probability of first selecting a normal coin and then getting zero heads is $\frac 2 3\cdot \frac 1 4=\frac 1 6$

• Thanks for your answer. Their is one thing I don't understand here, why does $P(X=1)$ not consider double head coins? Is it not possible to flip a double headed coin, the flip a normal coin and receive tails? This would mean that only one head was tossed. Apr 5, 2017 at 4:04
• In your problem, a single coin is selected at random and tossed twice, so it's not possible to flip two different coins. In other words, every time you select a two-headed coin, you'll get two heads from the two flips.
– Arby
Apr 5, 2017 at 4:18
• My bad, for some reason I was in the mindset of picking 2 at random and tossing both once.I should have read my own question better. Thanks :). Apr 5, 2017 at 4:20
• That would be a different problem, worth doing as well to flex your mathematical muscle. :D
– Arby
Apr 5, 2017 at 4:21

$$P(X) = P(Bias) \cdot P(X|Bias)^2+P(Fair) \cdot P(X|Fair)^2$$

We either pick a bias or fair coin.

$$P(Bias) = 1/3$$

$$P(Fair) = 1-P(Bias) = 2/3$$

And we know the odds of each outcome of X given which type of coin we have.

$$P(X=2|Bias) = 1$$

$$P(X=0|Fair) = 1/4$$

$$P(X=1|Fair) = 1/2$$

$$P(X=2|Fair) = 1/4$$

Thus,

$$P(X=0) = P(Fair) \cdot P(X=0|Fair) = 2/3 \cdot 1/4 = 1/6$$

$$P(X=1) = P(Fair) \cdot P(X=1|Fair) = 2/3 \cdot 0.5 = 2/6$$

$$P(X=2) = P(Bias) \cdot P(X=2|Bias) + P(Fair) \cdot P(X=2|Fair) = 1/3 + 2/3 \cdot 0.25 = 3/6$$