Find the probability

Question

A bag contains 10 balls, of which 7 are green and 3 are white. A ball is picked at

random from the bag and its color is noted. The ball is not replaced and a second ball

is then picked out. Find the probability that;

a) The first ball is green

b) The second ball is white given that the first ball is green

c) The balls are of different colors.

Answer 1

Probability of P(C1) = number of color / total balls

for P(C2) you need to factor in "not replaced" so then C1-1, and total-1

So now we can say P(G1) = 7/10 and P(W1) = 3/10

(a) we have worked out, it's P(G1)

(b) is a "conditional" probability, given a certain condition.

To see the difference, let's consider what is the probability that the second ball is white?

Well, there are two combinations: WW, or GW. So P(W2) = 3/10 x 2/9 OR 7/10 x 3/9

= 1/15 + 7/30 = 9/30 = 3/10

However, given that the first ball is Green, the P(W2) must now be 3/9 or 1/3.

(c) balls are different colors gives 2 combinations Green first AND then White, OR White first AND then Green. AND = multiply probabilities, OR = add probabilities.

So we can say P(G1) = 7/10 as above, followed by P(W2) = 3/9 = 1/3

P(G1) AND P(W2) = 7/10 x 1/3 =

Other combination is P(W1) = 3/10 as above, followed by P(G2) = 7/9

P(W1) AND P(G2) = 3/10 x 7/9 =

Then add the two probabilities together.