Hello averera

Welcome to Math Help Forum! Originally Posted by

**averera** 1. I have a card deck with 52 cards and I have to pick 3 cards and what are the probability of these events and also a opposite events

p(A)- 1 black and 2 red

p(B) -3 reds

p(C) - 1 club, 1 diamond and 1 heart

2. There are 10 white and 22 black balls. How big is the probability if I pick 2 white and 2 black balls?

I out of my mind and I doesnt understand how to solve them. And excuse me for my english.

**Here's how you do 1A:**

The total number of ways of choosing 3 cards from 52 is$\displaystyle \binom{52}{3}=\frac{52!}{3!49!}=\frac{52.51.50}{3. 2}=26.17.50$

The number of ways of choosing 1 black from 26 is 26; and the number of ways of choosing 2 red from 26 is $\displaystyle \binom{26}{2}$. So the number of ways of choosing 1 black and 2 red is$\displaystyle 26\times\binom{26}{2}=\frac{26\times26!}{2!24!}=\f rac{26.26.25}{2}=26.13.25$

Therefore the probability that this occurs when 3 cards are chosen from 52 is$\displaystyle \frac{26.13.25}{26.17.50}=\frac{13}{34}\approx 0.382$

I'm not quite sure what you mean by the 'opposite events'. Do you mean the probability that this doesn't happen? If so, then it's $\displaystyle 1 - 0.382 = 0.618$.

** Do 1B and 1C in a similar way.**

Start B by saying "The number of ways of choosing 3 reds from 26 is $\displaystyle \binom{26}{3}$ ..."

And start C by saying "The number of ways of choosing 1 club from 13 is 13...

Can you complete these now?

** For question 2:**

The number of ways of choosing 4 balls from 32 is $\displaystyle \binom{32}{4} =$ ... ? (a)

The number of ways of choosing 2 white from 10 is ... ?

The number of ways of choosing 2 black from 12 is ... ?

So the number of ways of doing each of these things, one after the other is ... ? (b) (*Hint*: Multiply these two answers together.)

Therefore the probability that this occurs is ... ? (*Hint*: divide the second answer (b) by the first, (a).)

Can you do this one now?

Grandad

PS Here are the answers.