MHB Probability Function of Z: 1/4 & 1/2 Explained

[oyth94]

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]

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]

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]

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]

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
$$P\left(Z=0\right) = P\left(X=0\right) = P\left(first \, coin \, is\, tails\right) = \frac12$$