- Thread starter
- #1

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!

- Thread starter oyth94
- Start date

- Thread starter
- #1

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!

- Feb 13, 2012

- 1,704

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

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!

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

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

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

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

Kind regards

$\chi$ $\sigma$

- Thread starter
- #3

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...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

$\chi$ $\sigma$

- Feb 13, 2012

- 1,704

If X=0 then is Z = X Y = 0 no matter which is Y...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...

Kind regards

$\chi$ $\sigma$

- May 12, 2013

- 84

If it helps, you can also think of $P_0$ as follows: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...

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

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