# probability!

• Mar 26th 2012, 01:53 AM
lawochekel
probability!
three-digit numbers are to be formed from 1,2,3,4 and 5. if repetition is not allowed how many can be formed?

what is the probability of selecting an odd number from the numbers formed in the above?

pls guys help on this, thanks.
• Mar 26th 2012, 02:16 AM
princeps
Re: probability!
Quote:

Originally Posted by lawochekel
three-digit numbers are to be formed from 1,2,3,4 and 5. if repetition is not allowed how many can be formed?

what is the probability of selecting an odd number from the numbers formed in the above?

pls guys help on this, thanks.

Let's denote number of three-digit numbers as N . Then :

$N=\frac{5!}{(5-3)!}=\frac{5 \cdot 4 \cdot 3 \cdot 2!}{2!}=60$
• Mar 26th 2012, 02:58 AM
biffboy
Re: probability!
First digit can be any of 5 numbers, then second any of 4, then third any of 3. Hence answer is 5x4x3=60
Probability number ends in 1 is 1/5. Same for 3 and 5. Hence probability odd is 3/5
• Mar 26th 2012, 05:21 AM
bling
Re: probability!
60
• Mar 26th 2012, 08:48 AM
Wilmer
Re: probability!
2 evens and 3 odds...so simply 3/5 will be odd.