Fundamentals of Probability and Statistics for Engineers

22 Fundamentals of Probability and Statistics for Engineersand the third axiom holds.The definition of conditional probability given by Equation (2.24) can beused not only to compute conditional probabilities but also to compute jointprobabilities, as the following examples show.Example 2.9. Problem: let us reconsider Example 2.8 and ask the followingquestion: what is the conditional probability that the first two components aregood given that (a) the first component is good and (b) at least one of the twois good?Answer: the event S 1 S 2 means both are good components, and S 1 [ S 2 is theevent that at least one of the two is good. Thus, for question (a) and in view ofEquation (2.24),P…S 1 S 2 jS 1 †ˆP…S 1S 2 S 1 †P…S 1 †ˆ P…S 1S 2 †P…S 1 †ˆ p1p 2p 1ˆ p 2 :This result is expected since S 1 and S 2 are independent. Intuitively, we see thatthis question is equivalent to one of computing P(S 2 ).For question (b), we have[Now, S 1 S 2 S 1 [ S 2 ) ˆ S 1 S 2 . Hence,P…S 1 S 2 jS 1 [ S †ˆP‰S 1S 2 …S 1 [ S 2 †Š2 :P…S 1 [ S 2 †P…S 1 S 2 jS 1 [ S 2 †ˆ P…S 1S 2 †P…S 1 [ S 2 † ˆˆp 1 p 2p 1 ‡ p 2 p 1 p 2:P…S 1 S 2 †P…S 1 †‡P…S 2 † P…S 1 S 2 †Example 2.10. Problem: in a game of cards, determine the probability ofdrawing, without replacement, two aces in succession.Answer: let A 1 be the event that the first card drawn is an ace, and similarlyfor A 2 . We wish to compute P(A 1 A 2 ). From Equation (2.24) we writeP…A 1 A 2 †ˆP…A 2 jA 1 †P…A 1 †:…2:25†Now, PA 1 ) ˆ 4/52 and PA 2 jA 1 ) ˆ 3/51 (there are 51 cards left and three ofthem are aces). Therefore,P…A 1 A 2 †ˆ 351 452ˆ 1221 : TLFeBOOK