# Probability and conditional?

A bag contains four red balls and four white balls. A ball is selected at random, removed and replaced by a ball of the opposite colour. A second ball is then selected at random. Calculate the probability that

(a) the first ball selected was white;

(b) the second ball was white, given that the first ball was white;

(c) the second ball was white;

(d) the first ball was white, given that the second ball was white.

a) 1/2

b) 3/8

c) 15/64

d) 1/2

I'm not sure if I've done this correctly, can someone please check my answers?

• The answer of c is $\frac 12$, because with both change $\frac 12$, the bag contains three red and five white balls or five red and three white balls, and $\frac 12\frac 58+\frac 12\frac38=\frac 12$. For the last one, you will need Bayes' rule: $P(A|B)=\frac{P(B|A)P(A)}{P(B)}$, and the quantities in the right side of the equation are sub-problem a,b,c. – Ragnar Dec 18 '13 at 0:42
• Thank you, I will go through Bayes' rule again, so are the rest of answers correct – user116487 Dec 18 '13 at 0:47
• a and b are correct, but c and d are not. – Ragnar Dec 18 '13 at 0:47
• Is d same as b? – user116487 Dec 18 '13 at 0:52
• that seems right to me. Do you understand why that makes sense (when just using your intuition)? – Ragnar Dec 18 '13 at 0:53

For part (a), because only the first ball has to be white, this means that the first ball chosen has to be red, and the second ball chosen can be either white or red.

This is represented by $\frac{4}{8} * \frac{7}{7} = \frac{1}{2}$ So you are correct.

For part (b), the P(second ball was white | first ball was white) = P(first and second ball are white)/P(first ball is white) = $$\frac{\frac{1}{2}*\frac{3}{8}}{\frac{1}{2}} = \frac{3}{8}$$ So you are correct for part (b) as well.

For part (c), we must have the second ball as white. This falls into 2 cases:

1. The first ball chosen is red
2. The first ball chosen is white

For 1. the probability is $\frac{4}{8}*\frac{5}{8} = \frac{5}{16}$. For 2. the probability is $\frac{4}{8}*\frac{3}{16} = \frac{3}{16}$ Adding the two probabilities together gets us $\frac{1}{2}$

For part (d), we will approach it with similar steps we've used in part (b). It's asking for the P(First ball was white | Second ball was white) = P(First and second ball are white)/P(Second ball is white) = $$\frac{\frac{1}{2}*\frac{3}{8}}{\frac{1}{2}} = \frac{3}{8}$$

Summing it all up: Parts (a) and (c) are $\frac{1}{2}$ Parts (b) and (d) are $\frac{3}{8}$

(a). This one's easy. $4$ white balls out of $8$ = $4/8 = 1/2$.

(b). If the first one was white, then it's replaced by a red ball. Now there are 3 white balls and 5 red balls. $3$ white balls out of $8$ = $3/8$.

(c). You must take cases and multiply the "and" conditions. The second ball's probability depends on the first ball's value.

$P(b2=white) = P(b1=white) * P(b2=white,given b1=white) + P(b1=red) * P(b2=white,given b1=red) =$

$4/8 * 3/8 + 4/8 * 5/8 = 3/16 + 5/16 = 1/2$

(d). There are 5 ways out of 8 to choose the white ball second if the first one is red. There are 3 ways out of 8 to choose the white ball second if the first one is red. The probability that the first was white, given that the second is white, is:

$\frac {3} {(5 + 3)} = 3/8$