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

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Discussion Overview

The discussion revolves around the probability function of a random variable Z defined in the context of flipping two fair coins. Participants explore the calculations leading to the probabilities of different outcomes for Z, specifically addressing the values of 1/4 and 1/2. The scope includes mathematical reasoning and clarification of probability concepts.

Discussion Character

  • Mathematical reasoning, Conceptual clarification, Debate/contested

Main Points Raised

  • Some participants define Z as the product of two variables, X and Y, based on the outcomes of two coin flips.
  • One participant states that the probability P(Z=0) is calculated as P(X=0) = 1/2, questioning why it is not multiplied by P(Y=1) or P(Y=2).
  • Another participant explains that if X=0, Z will always equal 0 regardless of Y's value, suggesting that the calculation for P(Z=0) does not require multiplication by Y's probabilities.
  • A later reply provides a detailed breakdown of calculating P(Z=0) by considering mutually exclusive events for Y, arriving at the conclusion that P(Z=0) = 1/2 through a summation of probabilities.
  • Participants express confusion about the calculations and the reasoning behind the probabilities assigned to Z.

Areas of Agreement / Disagreement

Participants express differing views on the calculation of P(Z=0) and whether it should involve probabilities of Y. There is no consensus on the correct approach to determining the probability function of Z.

Contextual Notes

Some assumptions about the independence of events and the definitions of X and Y may not be fully articulated. The discussion includes unresolved questions about the application of probability rules in this context.

oyth94
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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!
 
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Re: probability function

oyth94 said:
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

$\chi$ $\sigma$
 
Re: probability function

chisigma said:
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...
 
Re: probability function

oyth94 said:
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

$\chi$ $\sigma$
 
Re: probability function

oyth94 said:
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
$$P\left(Z=0\right) = P\left(X=0\right) = P\left(first \, coin \, is\, tails\right) = \frac12$$
 

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