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

AI Thread Summary
The discussion focuses on calculating the probability function of Z, defined as the product of two variables X and Y based on the outcomes of flipping two fair coins. The probabilities are derived as follows: P(Z=0) = P(X=0) = 1/2, P(Z=1) = P(X=1) * P(Y=1) = 1/4, and P(Z=5) = P(X=1) * P(Y=5) = 1/4, with all other probabilities being zero. Confusion arises regarding why P(X=0) is not multiplied by P(Y=1) or P(Y=5), but the explanation clarifies that if X=0, Z is always zero regardless of Y's value. The final conclusion is that the calculations correctly reflect the probabilities based on the defined conditions.
oyth94
Messages
32
Reaction score
0
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!
 
Mathematics news on Phys.org
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
$$P\left(Z=0\right) = P\left(X=0\right) = P\left(first \, coin \, is\, tails\right) = \frac12$$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top