0
$\begingroup$

Two integers are to be selected from random and without replacement from the set of the first $25$ positive integers. What's the probability that product of the two integers will be even?

I'm stuck on this question for calculation, I know there are total $25\choose{2}$ ways to randomly select two numbers. Within $25$ integers we have $13$ odd integers and $12$ even integers. So maybe we have combination of one odd and one even, or two even.

But not sure how to calculate the probability when including selection.

Any response is helpful, thanks.

$\endgroup$
1
$\begingroup$

Alternative approach that yields identical math as TKA's answer.

$$1 - \frac{N\text{(umerator)}}{D\text{(enominator)}}$$

where $N =$ total # ways of selecting 2 odd numbers and
$D =$ total # ways of selecting any 2 numbers.

Answer is

$$1 ~-~ \frac{\binom{13}{2}}{\binom{25}{2}}.$$

$\endgroup$
0
$\begingroup$
  • There are 13 odd and 12 even numbers.
  • Probability of drawing a odd first number is $\frac{13}{25}$ and odd second number is $\frac{12}{24}$ giving probability that product is odd = $\frac{13}{25} * \frac{12}{24}$
  • In any other case the product will always be even. So probability that product is even = $1 - \frac{13}{25} * \frac{12}{24} = \frac{156}{600}$
  • Another ways is to calculate the probabilities of all selections resulting into an even product and sum them up. which will be $\frac{12}{25} * \frac{24}{24} + \frac{13}{25} * \frac{12}{24} = \frac{156}{600}$.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.