# very basic probability question

• Mar 3rd 2010, 11:01 AM
pandakrap
very basic probability question
I was just confused if I am thinking this one right.

so the question is:

"In a set of numbers from 1 to 10. If two numbers are to be selected from these 10 numbers with replacement, then what is the probability that at least one of them is even?"

So I thought. there are four possible outcome,

Even*Odd
Odd*Even
Odd*Odd
Even*Even

and since 3 out of them contains even number So I thought the probability
was 3/4. Is this the right way of thinking this problem??

• Mar 3rd 2010, 12:14 PM
mr fantastic
Quote:

Originally Posted by pandakrap
I was just confused if I am thinking this one right.

so the question is:

"In a set of numbers from 1 to 10. If two numbers are to be selected from these 10 numbers with replacement, then what is the probability that at least one of them is even?"

So I thought. there are four possible outcome,

Even*Odd
Odd*Even
Odd*Odd
Even*Even

and since 3 out of them contains even number So I thought the probability
was 3/4. Is this the right way of thinking this problem??

Calculate 1 - Pr(none are even) = $1 - \frac{{5 \choose 2} \cdot {5 \choose 0}}{{10 \choose 2}}$.
• Mar 3rd 2010, 12:24 PM
pandakrap

and I'm sorry, but could you explain the formula you have put up? I don't think I clearly understand the logics of that formula.

thank you
• Mar 3rd 2010, 12:27 PM
mr fantastic
Quote:

Originally Posted by pandakrap

and I'm sorry, but could you explain the formula you have put up? I don't think I clearly understand the logics of that formula.

thank you

Have you learned about combinatorics eg. the number of ways of choosing r objects from n objects etc.?
• Mar 3rd 2010, 12:40 PM
pandakrap
yes I have I know why you did 5 choose 2, because there are 5 even and 5 odd, and i know why you did 10 choose 2, because there are total of 10 numbers and you need to choose 2,

but What I don't get is why you did 1 minus, that whole rationalized combinatorics.

So I guess, I kind of can understand why you did each part of the equation. but I can't seem to understand why you have decided to put those numbers together.
• Mar 3rd 2010, 12:42 PM
mr fantastic
Quote:

Originally Posted by pandakrap
yes I have I know why you did 5 choose 2, because there are 5 even and 5 odd, and i know why you did 10 choose 2, because there are total of 10 numbers and you need to choose 2,

but What I don't get is why you did 1 minus, that whole rationalized combinatorics.

So I guess, I kind of can understand why you did each part of the equation. but I can't seem to understand why you have decided to put those numbers together.

Read my first reply again .... The part where I said "Calculate 1 - Pr(none are even)" .....
• Mar 3rd 2010, 12:51 PM
pandakrap
oh......... okokok hahaha, now I feel stupid.. thanks, I just got that.. (Headbang)

thank you very much
• Mar 3rd 2010, 05:39 PM
matheagle
actually the choose is incorrect, that's the hypergeometric situation which is correct when you are choosing WITHOUT replacement

This is binominal, since it is with replacement.

The 3/4 is correct, which can be achieved via 1-(1/2)(1/2) as well.
• Mar 3rd 2010, 06:49 PM
With replacement,

$P(both\ odd)=\frac{\binom{5}{1}}{\binom{10}{1}}\frac{\bino m{5}{1}}{\binom{10}{1}}=\frac{5}{10}\ \frac{5}{10}=\frac{1}{2}\ \frac{1}{2}=\frac{1}{4}$

Therefore,

$P(at\ least\ 1\ even)=1-P(both\ odd)=1-\frac{1}{4}$
• Mar 4th 2010, 12:18 AM
mr fantastic
Quote:

Originally Posted by matheagle
actually the choose is incorrect, that's the hypergeometric situation which is correct when you are choosing WITHOUT replacement

This is binominal, since it is with replacement.

The 3/4 is correct, which can be achieved via 1-(1/2)(1/2) as well.

I could have sworn it said without .... So I'm going blind as well ....
• Mar 4th 2010, 12:21 AM
matheagle
and who calls me oldeagle?
am eagled eyed as well