Multiple Factors in finding the probability of a event

[Chuck Norris]

Well, hello again

This, I'm trying to work on, decided to look here first because not quite sure exactly what it's called but heres an example

The probability of drawing two hearts from a standard pack of cards is (3 over 51) [sorry still not sure how to type in math form]. What is the probability that the two cards will not be hearts.

Well, not quite sure what do here, i heard you need to multiply them by eachother but not quite sure.

I realise how to get the hearts, but I'm currently stuck on the 'how to get what isn't hearts'

Any help would be very helpful

If i figure it out before someone else does ill type it out to maybe help others

 

[galactus]

The probability of drawing 2 hearts would be (13/52)(12/51)=1/17

The probability of drawing 2 non-hearts would be (39/52)(38/51)=19/34

 

[soroban]

Hello, Chuck!

If you're familiar with "combinations", here's another approach.

The probability of drawing two hearts from a standard pack of cards is: \(\displaystyle \frac{3}{51} \,=\,\frac{1}{17}\)
What is the probability that the two cards will not be hearts?

There are: \(\displaystyle \:{52\choose2} \,=\,1326\) possible pairs of cards.

There are: \(\displaystyle \:{39\choose2} = 741\) ways to draw two non-hearts.

Therefore: \(\displaystyle \(\text{2 non-hearts}) \;=\;\frac{741}{1326} \;=\;\L\frac{19}{34}\)