Exercise 2 Page P19

The event of selecting the first sock affects the probability of selecting the second one. This is because there is one fewer sock from which to choose. Thus, the events are dependent. Probability of Dependent Events If two events A and B are dependent, then the probability that A and B will occur is P(AandB)=P(A)* P(B|A) Let's start by calculating the probability of selecting a blue sock from the drawer. In drawer there are 4+6+8=18 socks, six of which are blue. P(A)&=6/18 l←blue socks ←total socks P(B|A) is the probability of selecting a black sock, given that the first sock selected is blue. Since we have already selected a blue sock, there are 17 socks, from which four of them are black. P(B|A)&= 4/17 l←black socks ←remainingsocks According to the formula, to calculate P(AandB) we have to multiply P(A) and P(B|A).

P(AandB)=P(A)* P(B|A)

SubstituteII

P(A)= 6/18, P(B|A)= 4/17

P(AandB)= 6/18* 4/17

ReduceFrac

a/b=.a /6./.b /6.

P(AandB)=1/3*4/17

MultFrac

Multiply fractions

P(AandB)=4/51

P(C|A andB) is the probability of selecting again a blue sock, given that the first sock selected is blue and the second is black. Since we have already selected two socks, and one of them was blue, there are 16 socks, from which five of them are blue. P(C | A andB)&= 5/16 l←blue socksleft ←socksleft Finally, according to the formula, to calculate P(C andA andB) we have to multiply P(A andB) and P(C|A andB).

P(C andA andB)=P(A andB)* P(C|A andB)

SubstituteII

P(A andB)= 4/51, P(C | A andB)= 5/16

P(C andA andB)= 4/51* 5/16

MultFrac

Multiply fractions

P(C andA andB)=20/816

ReduceFrac

a/b=.a /4./.b /4.

P(C andA andB)=5/204