# Quick help in easy-ish probability

1. Jan 22, 2008

### rock.freak667

1. The problem statement, all variables and given/known data

Here is the basic info. There are 24 prizes(4 cars,8 bicycles and 12 watches) to be given to 24 prize winners. A,B,C are 3 of the prize winners.
Find
(i)The probability that both A and B get cars given that C gets a car.
(ii)The probability that either A or C or both gets a car
(iii)The probability that A gets a car and B gets either a car of a bicycle
(iv)The probability that A gets a car given then B gets either a car or a bicycle.

2. Relevant equations

3. The attempt at a solution

(i)P($A_c&B_c|C_c$)=$$\frac{\frac{3}{23}*\frac{2}{22}}{\frac{4}{24}}$$

(ii)simply I got this one by finding when they both did not get cars and then 1 minus that value.

(iii)$\frac{4}{24}*\frac{3}{23}+\frac{4}{24}*\frac{8}{23}$

(iv)This part confused me as I got a way wrong answer.(ANS=$\frac{11}{69}$, which I got something way smaller than that)

2. Jan 23, 2008

### Shooting Star

(iv) Does the sentence read "The probability that A gets a car given that B gets either a car or a bicycle."

EDIT: correct method follows.

Last edited: Jan 23, 2008
3. Jan 23, 2008

### rock.freak667

Aren't you supposed to use
$$P(A|B)=\frac{P(A\cap B)}{P(B)}$$

4. Jan 23, 2008

### Shooting Star

All right. What is P(B) in that case? More importantly, what would be P(A_int_B)?

5. Jan 23, 2008

### rock.freak667

B should be the probability of getting a car or a bicycle which is 4/24+8/24.

and so P(A_int_B)=4/24*(4/24+8/24)

6. Jan 23, 2008

### Shooting Star

Shouldn't it be P(A_int_B)=4/24*(4/23+8/23)? Take a sec...

7. Jan 23, 2008