pmf not summing to 1: bag contains 6 coins, 2 of which have a head on both sides

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joshjoshjosh

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Hi guys currently I'm working on this question. I've found the pmf for the random variable X and am not sure where I got wrong, since the pmf does not sum to 1.

The question is:

A bag contains 6 coins, 2 of which have a head on both sides while the other 4 coins are normal. A coin is chosen at random from the bag and tossed 2 times. The number of heads obtained is a random variable, say X. What are the possible values of X? Find the pmf for each value of X.

My answer is:
The possible values of X are 0, 1 and 2.
If X = 0, then the coin has to be the 1-headed one. So P(X=0) = P(choose a 1 headed coin) x P(2 heads obtained) = 4/6 x (1/2)^2 = 1/6
If X = 1, then the coin has to be the 1-headed one. So P(X=1) = P(choose a 1 headed coin) x P(1 head, 1 tail obtained) = 4/6 x 1/2 x 1/2 = 1/6
If X = 2, there are 2 scenarios:
- If the coin chosen is 2-headed, then the probability of getting 2 heads is: P(choose a 2 headed coin) x P(2 heads obtained) = 2/6 x 1^2 = 2/6
- If the coin chosen is 1-headed, then the probability of getting 2 heads is: P(choose a 1 headed coin) x P(2 heads obtained) = 4/6 x (1/2)^2 = 1/6
So P(X=2) = 2/6 + 1/6 = 3/6

However, if I sum up the three pmf's above, I get 1/6 + 1/6 + 3/6 = 5/6, not 1. Where was the problem?
 
Your case X=1 neglects the fact that the order can be either HT or TH. You need to double the probability, making it 1/3.
 
joshjoshjosh said:
The question is:

A bag contains 6 coins, 2 of which have a head on both sides while the other 4 coins are normal. A coin is chosen at random from the bag and tossed 2 times. The number of heads obtained is a random variable, say X. What are the possible values of X? Find the pmf for each value of X.
My answer is:
The possible values of X are 0, 1 and 2.
If X = 0, then the coin has to be the 1-headed one. So P(X=0) = P(choose a 1 headed coin) x P(2 heads obtained) = 4/6 x (1/2)^2 = 1/6
If X = 1, then the coin has to be the 1-headed one. So P(X=1) = P(choose a 1 headed coin) x P(1 head, 1 tail obtained) = 4/6 x 1/2 x 1/2 = 1/6
If X = 2, there are 2 scenarios:
- If the coin chosen is 2-headed, then the probability of getting 2 heads is: P(choose a 2 headed coin) x P(2 heads obtained) = 2/6 x 1^2 = 2/6
- If the coin chosen is 1-headed, then the probability of getting 2 heads is: P(choose a 1 headed coin) x P(2 heads obtained) = 4/6 x (1/2)^2 = 1/6
So P(X=2) = 2/6 + 1/6 = 3/6
However, if I sum up the three pmf's above, I get 1/6 + 1/6 + 3/6 = 5/6, not 1. Where was the problem?
Click to expand...
\[\begin{array}{*{20}{c}}
T&T&{}&{{{\left( {\frac{1}{2}} \right)}^2}\left( {\frac{2}{3}} \right)}&{ = \frac{1}{6}}\\
T&H&{}&{{{\left( {\frac{1}{2}} \right)}^2}\left( {\frac{2}{3}} \right)}&{ = \frac{1}{6}}\\
H&T&{}&{{{\left( {\frac{1}{2}} \right)}^2}\left( {\frac{2}{3}} \right)}&{ = \frac{1}{6}}\\
H&H&{}&{{{\left( {\frac{1}{2}} \right)}^2}\left( {\frac{2}{3}} \right) + (1)\left( {\frac{1}{3}} \right)}&{ = \frac{3}{6}}\end{array}\]
As you can see my total is one. I leave it to you to see why.
 
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