Probability questions - picking two random numbers

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MaryStew

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So I have some probability questions below, and I did the first 3 questions and was wondering if someone could confirm my answers for them. I'm not exactly sure how I would go about doing 4 and 5 and was hoping for some clarification. My friends are all getting different answers for the last 2 questions and I'm not sure how to do them. Any help would be appreciated!

We independently pick two random numbers R1 and R2 from the set {1,…,1000}. (Note: Independence means we may pick R1=R2.)

1. What is the probability that R1 and R2 are both even?
P(A and B) = P(A) * P(B) = (1/2) * (1/2) = 1/4

2. Suppose I tell you that at least one of R1 or R2 is even. What is the (conditional) probability that R1 and R2 are both even?
P(A|B) = (1/4)/(3/4) = 1/3

3. What is the probability that at least one of R1 or R2 is equal to 1000?
P(A or B) = 1 - P(A' and B') = 1 - 0.998001 = 0.001999

4. Suppose I tell you that at least one of R1 or R2 is even. What is the probability that at least one of R1 or R2 is equal to 1000?

5. Suppose I tell you that at least one of R1 or R2 is even. What is the probability that at least one of R1 or R2 is equal to 999?
 
1st I need to say that you can use any variables you like, but you must define YOUR variables, specially if you use them multiple times with different definitions!!

#3 is so painfully hard to follow since I do not know what you mean by A or B since you chose not to tell us what A and B are! Since you want us to check YOUR work it would have been nice if you told us that .998001 = .999^2. I thought that was what it is but I had to verify it when you could have just said 1-.999^2 = 1-.998001 = .001999

4 and 5. You can't do these problems until you have well defined variables. Try again with well defined variables and I bet that you can get somewhere.

On this forum we really have no problem helping students but please understand that you can also help us help you. If you do multiplication without showing what you multiplied it takes us some time to do the computation to see if you are correct. Also, please define your variables. You may think they are obviously defined but if you used the wrong formula or something like that then in fact it may be hard to figure out what they mean.

Thanks, and please post back.
 
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Hint for 4:
Suppose R1 is the result of the 1st pick and R2 is the result of the 2nd pick.
Can you define a conditional probability using R1 and R2 that corresponds to p(at least one of R1 or R2 is equal to 1000 | at least one of R1 or R2 is even) or do the calculation using the reduced sample space?

You need to really understand what you are given and what you need to find.
 
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MaryStew said:
We independently pick two random numbers R1 and R2 from the set {1,…,1000}. (Note: Independence means we may pick R1=R2.)
1. What is the probability that R1 and R2 are both even?
P(A and B) = P(A) * P(B) = (1/2) * (1/2) = 1/4 CORRECT
2. Suppose I tell you that at least one of R1 or R2 is even. What is the (conditional) probability that R1 and R2 are both even?
P(A|B) = (1/4)/(3/4) = 1/3CORRECT
3. What is the probability that at least one of R1 or R2 is equal to 1000?
P(A or B) = 1 - P(A' and B') = 1 - 0.998001 = 0.001999CORRECT

4. Suppose I tell you that at least one of R1 or R2 is even. What is the probability that at least one of R1 or R2 is equal to 1000?

5. Suppose I tell you that at least one of R1 or R2 is even. What is the probability that at least one of R1 or R2 is equal to 999?
Click to expand...
For # Three you could have done \(\mathscr{P}(A)+\mathscr{P}(B)-\mathscr{P}(A)\cdot\mathscr{P}(B)\).
For \(\#4~\&~\#5\) it is easier to realize that the sample space has \(10^6\) pairs. You are picking one pair.
How many of those \(10^6\) pairs have at least one even number?
Of those only three contain at least one \(\small{1000}\).
 
Sorry for the late reply. For 1) I let P(A) equal the probability that R1 is even, and P(B) equal the probability that R2 is even. 2) I let P(A) equal the probability that R1 is even, and P(B) equal the probability that R2 is even. And for 3) I let P(A) represent the probability that R1 is equal to 1000, and P(B) be the probability that R2 is equal to 1000.

For 2), P(A and B) in the numerator part of the formula P(A|B) was reduced to P(A)

An expansion of 3) is P(A'and B')=P(A')× P(B')= 0.999 × 0.999 = 0.998001 and then I used this value in P(A or B) = 1 - P(A' and B') = 1 - 0.998001 = 0.001999
 
MaryStew said:
Sorry for the late reply. For 1) I let P(A) equal the probability that R1 is even, and P(B) equal the probability that R2 is even. 2) I let P(A) equal the probability that R1 is even, and P(B) equal the probability that R2 is even. And for 3) I let P(A) represent the probability that R1 is equal to 1000, and P(B) be the probability that R2 is equal to 1000. For 2), P(A and B) in the numerator part of the formula P(A|B) was reduced to P(A)
An expansion of 3) is P(A'and B')=P(A')× P(B')= 0.999 × 0.999 = 0.998001 and then I used this value in P(A or B) = 1 - P(A' and B') = 1 - 0.998001 = 0.001999
Click to expand...
I told that you have done the first three correctly. I gave you hints for #'s 4 & 5.
What have you done on those two?
 
What about 4 and 5. Just use the reduced sample space that pka and I spoke about. If you need help with that then please reply back.
 
Hi! So for 4 I was told that answer was around 0.0026653 but in my attempt I seem to be off by a step.

I did P(A) = the probability that one of R1 or R2 is even and P(B) = the probability that one or R1 or R2 is 1000.

I calculated P(A) = 1/2 + 1/2 - 1/4 = 3/4 = 0.75

I calculated P(B) = 1/1000 + 1/1000 - 1/10,000,000 = 0.002 - 0.0019999 = 0.0000001

Using the formula P(B|A) = P(A and B)/ P(A) = P(B)/P(A) = 0.0000001/0.75 = an extremely large number that is not 0.0026653.

I figured out that I can get 0.0026653 if I skipped the '1/1000 + 1/1000' portion of P(B) but that doesn't make sense to me. I'm not sure why P(B) should be 0.0019999. I added 1/1000 + 1/1000 to find the probability of R1 being 1000 and R2 being 1000, then subtracted (1/1000)*(1/1000) because I know I'm supposed to subtract that overlapping value in the formula:

P(A or B) = P(A) + P(B) - P(A and B)

but it's not adding up. I'm not sure what I'm doing wrong.
 
Here is how I would do #4

You are given that at least R1 or R2 is even.

If R1 is even, then R1 can be any of 500 values while R2 can be any of 1000 values so at least R1 or R2 will be even----this gives us 500000 outcomes where at least one of R1 and R2 is even.

If R2 is even then by symmetry we have another 500000 outcomes where at least one of R1 and R2 is even.

Although we do allow for both R1 and R2 to be even we should not count then twice as we did.
So we need to remove (one time) the cases where R1 and R2 are even which is 500*500=250,000 cases

So our sample size is 500,000 + 500,000 - 250,000 = 750,000.

Now we want to know how many of these 750,000 cases have at least one of R1 and R2 equal to 1000.
R1 can be any of 500 values while R2 =1000. This gives us 500 cases.
R2 can be any of 500 values while R1= 1000. This gives us 500 cases.
We counted R1=R2=1000 twice so we need to remove one of them.
Total is 500+500-1 =999.

So p(that at least one of R1 or R2 is equal to 1000 | at least one of R1 or R2 is even) = 999/750000 ~0.001332

Now can you do number 5.
 
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