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Probability function

oyth94

Member
Jun 2, 2013
33
Consider flipping two fair coins. Let X = 1 if the first coin is heads, and X = 0
if the first coin is tails. Let Y = 1 if the second coin is heads, and Y = 5 if the second
coin is tails. Let Z = XY. What is the probability function of Z?
how did you get 1/4 and 1/2 ?? and why? confused!
 

chisigma

Well-known member
Feb 13, 2012
1,704
Re: probability function

Consider flipping two fair coins. Let X = 1 if the first coin is heads, and X = 0
if the first coin is tails. Let Y = 1 if the second coin is heads, and Y = 5 if the second
coin is tails. Let Z = XY. What is the probability function of Z?
how did you get 1/4 and 1/2 ?? and why? confused!
Setting $P_{k}= P \{Z=k\}$ we have...


$$P_{0}= P \{X=0\} = \frac{1}{2}$$


$$P_{1} = P \{X=1\}\ P \{Y=1 \} = \frac{1}{4}$$


$$P_{5} = P \{ X=1\}\ P \{Y = 5\} = \frac{1}{4}$$


... and for any other k is $P_{k}=0$...

Kind regards

$\chi$ $\sigma$
 

oyth94

Member
Jun 2, 2013
33
Re: probability function

Setting $P_{k}= P \{Z=k\}$ we have...


$$P_{0}= P \{X=0\} = \frac{1}{2}$$


$$P_{1} = P \{X=1\}\ P \{Y=1 \} = \frac{1}{4}$$


$$P_{5} = P \{ X=1\}\ P \{Y = 5\} = \frac{1}{4}$$


... and for any other k is $P_{k}=0$...

Kind regards

$\chi$ $\sigma$
How come for the "P (X=0) = 1/2" you didn't multiply by P(Y=1) or P(Y=2)? I'm not sure how you ended up with 1/2 instead of 1/4 for this one...
 

chisigma

Well-known member
Feb 13, 2012
1,704
Re: probability function

How come for the "P (X=0) = 1/2" you didn't multiply by P(Y=1) or P(Y=2)? I'm not sure how you ended up with 1/2 instead of 1/4 for this one...
If X=0 then is Z = X Y = 0 no matter which is Y...

Kind regards

$\chi$ $\sigma$
 

TheBigBadBen

Active member
May 12, 2013
84
Re: probability function

How come for the "P (X=0) = 1/2" you didn't multiply by P(Y=1) or P(Y=2)? I'm not sure how you ended up with 1/2 instead of 1/4 for this one...
If it helps, you can also think of $P_0$ as follows:
Because y=1 and y=5 are mutually exclusive events, we can state:
$$P\left(Z=0\right) = P\left(X=0\right)
\\= P\left(X=0 \wedge Y=1\right)+P\left(X=0 \wedge Y=5\right)
\\= P\left(tails \wedge heads\right)+P\left(tails \wedge tails\right)$$
The probability of the first event, as you rightly stated, is
$$\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$$
The probability of the second even is the same.
Adding these two together, we have
$$\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$$
giving us our answer.

As $\chi \sigma$ rightly stated, we would have gotten the same answer if we had simply evaluated
$$P\left(Z=0\right) = P\left(X=0\right) = P\left(first \, coin \, is\, tails\right) = \frac12$$