help please please

Question

Two balls are drawn, without replacement, from a bag containing 11 red balls numbered 1−11 and 3 white balls numbered 12−14. (Enter your probabilities as fractions.)

(a) What is the probability that the second ball is red, given that the first ball is white? 

(b) What is the probability that both balls are even-numbered? 

(c) What is the probability that the first ball is red and even-numbered and the second ball is even-numbered? 

Answer 1

(a) (3 / 14)(11 / 13) = ?

(b) (7 / 14)(6 / 13) = ?

(c) (5 / 14)(1 / 13) = ?

Answer 2

(a)

Given that the first ball is white, the second ball is drawn from a bag containing

11 red balls and 2 white balls.

The probability of drawing a red ball out of such a bag is

11/(11+2) = 11/13

(b)

The answer is the ratio between the number of ways of drawing two even numbered balls,

divided by the number of ways of drawing any two balls.

First count how many ways we can be draw two even balls.

The bag starts with 7 even balls and 7 odd balls.

There are 7 ways the first ball can be even.

For each, there are 6 ways the second ball can be even. (Note that only 6 even balls are left.)

Therefore there are 7×6 ways of drawing two even balls.

Now count the number of ways of drawing any two balls.

The first ball can be chosen in 14 ways.

And for each there 13 ways of drawing the second ball.

So there are 14×13 ways of drawing any pair of balls.

Consequently the probability of drawing a pair of even balls is

7×6/14×13 = 3/13

(c)

There are 5 even-numbered red balls.

Therefore there are 5 ways the first ball can be chosen.

And for each, there are 6 ways of getting an even ball on the second draw.

(Note that only 6 even balls are left.)

Therefore the desired probability is

5×6/14×13 = 15/91