# Probability on numbers

• Jan 5th 2011, 01:44 PM
Ka9
Probability on numbers
We randomly select 4 prime numbers without replacement from the first 17 prime numbers. What is the probability that the sum of the four selected numbers is odd? Express answer in common fraction.

I did this problem as
2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,61

(1/17 * 1/16 * 1/15 * 1/14) * 9

I multiplied by I could form sum as odd.

Please correct me. I am not myself sure with the approach.

============

We select 5 numbers at random, with replacement, from the set of integers from 1 to 400 inclusive. What is the probability that the product of 5 numbers is even? Answer should be in common fraction.

(1/400)^5

Please guide me in the above problems.

Thanks
• Jan 5th 2011, 01:56 PM
Quote:

Originally Posted by Ka9
We randomly select 4 prime numbers without replacement from the first 17 prime numbers. What is the probability that the sum of the four selected numbers is odd? Express answer in common fraction.

I did this problem as
2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,61

(1/17 * 1/16 * 1/15 * 1/14) * 9

I multiplied by I could form sum as odd.

Please correct me. I am not myself sure with the approach.

Thanks

Even+Even=Even
Odd+Odd=Even
Odd+Even=Odd

(Odd+Odd)+(Odd+Odd)=Even+Even=Even
(Odd+Even)+(Odd+Odd)=Odd+Even=Odd
(Even+Even)+(Odd+Even)=Even+Odd=Odd

Hence, what type of selection of 4 numbers do we need such that the sum is Odd ?

How many of the primes you've listed are Odd?
How many are Even?
• Jan 5th 2011, 02:38 PM
Plato
Note that there is only one even prime.
The sum of four odd numbers is even.
The sum of three odds and an even is odd.
• Jan 6th 2011, 12:22 PM
tom@ballooncalculus
... And then also consider... 2 will complete a foursome with $\binom{16}{3}$ threesomes, and this is out of a total number $\binom{17}{4}$ of possible foursomes.

===========================

The other one... Precisely how can the product of all 5 numbers fail to be even? (Employ similar reasoning as with addition, above.) And work out the probability of this failure occurring.

Probability of the first number being odd = ... ? ... AND... probability of the second number being odd as well = 199/399... AND... probability of the third being odd = ... ? ...etc.

Edit: whoops! WITH replacement - sorry. So all five probabilities are the full 200/400.
• Jan 6th 2011, 04:38 PM
Ka9
I understand 17c4 part. But not 16c3, as choose 3 numbers out of 16 and how to include "2" in the 16c3. Not sure. Please reply me.
Thanks,
• Jan 6th 2011, 04:48 PM
Quote:

Originally Posted by Ka9
I understand 17c4 part. But not 16c3, as choose 3 numbers out of 16 and how to include "2" in the 16c3. Not sure. Please reply me.
Thanks,

The $\binom{17}{4}$ part is the denominator of the probability fraction.

Remember that the sum of the 4 primes needs to be odd ?
There is only 1 way this can happen,
which is the even prime "2" summed to 3 other primes (all of which are odd).

Hence, you may consider the number "2" as already chosen, since it must be one of the 4 numbers.

You need 3 more and you have 16 to choose from.
• Jan 6th 2011, 05:04 PM
Plato
Quote:

Originally Posted by Ka9
I understand 17c4 part. But not 16c3, as choose 3 numbers out of 16 and how to include "2" in the 16c3.

@Ka9, I have some questions for you.
Are you doing a self-directed study of topics in probability?
Are you comfortable with the dominate language of this website which is English?
If the answer to the first question is no, then I must tell you that I think whoever is teaching you probability is doing you a enormous disservice.
If the answer to the first is yes and the answer to the second is no then that may explain a lot of what your confusion results from.

Now to this current problem.
In order four prime numbers to sum to an odd number, three of them must be odd and one must be even. There is only one even prime, 2.
So the number 2 must be among the four primes which sum to an odd number; the other three must be odd. From the first seventeen primes one is even and sixteen are odd numbers.
There are $\binom{16}{3}$ ways to choose three of those odd primes. So there are that many ways to have three odd primes and one even prime from the first seventeen primes.
• Jan 6th 2011, 05:31 PM
Ka9
Hi Plato,
Answer to the first question is yes. Here and there confusion with the dominate language. But, I am able to follow.
Sorry for the inconvience.
• Jan 7th 2011, 01:03 AM
chisigma
Selecting randomly 2 primes among the first 17, the probability that their sum is odd is...

$\displaystyle p_{o}= \frac{1}{17} + \frac{1}{16}\ \frac{16}{17} = \frac{2}{17}$ (1)

Selecting randomly 3 primes among the first 17, the probability that their sum is even is...

$\displaystyle p_{e}= \frac{1}{17} + \frac{1}{16}\ \frac{16}{17} + \frac {1}{15}\ \frac{15}{16}\ \frac{16}{17}= \frac{3}{17}$ (2)

Selecting randomly 4 primes among the first 17, the probability that their sum is odd is...

$\displaystyle p_{o}= \frac{1}{17} + \frac{1}{16}\ \frac{16}{17} + \frac {1}{15}\ \frac{15}{16}\ \frac{16}{17} + \frac {1}{14}\ \frac {14}{15}\ \frac{15}{16}\ \frac{16}{17}= \frac{4}{17}$ (3)

... and so one...

Kind regards

$\chi$ $\sigma$
• Jan 7th 2011, 01:57 AM
CaptainBlack
Quote:

Originally Posted by Ka9
We select 5 numbers at random, with replacement, from the set of integers from 1 to 400 inclusive. What is the probability that the product of 5 numbers is even? Answer should be in common fraction.

For the product to be even at least one of the five numbers must be even. Sampling with replacement means that evry one of the numbers selected has the same chance of being even so the number of evens in our sample of 5 numbers has a binomial distribution B(5,0.5).

So the probability of one or more evens is 1-Pr(zero evens)=1-b(0,5,0.5)

CB
• Jan 7th 2011, 04:13 AM
Ka9
Thanks, everyone.
I got it.