Conditional Probability

• Conditional probability: $P(B\vert A)$
  • Sometimes the occurrence of an event is influenced or related with some other event.
  • Hence we must take this restriction or the availability of certain limited information into consideration about the outcome of the experiment.
  • The probability of an event $B$ occurring when it is known that some event $A$ has occurred.
  • ``The probability that $B$ occurs given that $A$ occurs'' or ``The probability of $B$,given $A$''
  • The notion of conditional probability provides the capability of re-evaluating the idea of probability of an event in light of additional information.
• Example: $S = {1, 2, 3, 4, 5, 6},~ A={4,5,6},~ B={1,3,5}, \Longrightarrow P(B\vert A)?$
• Definition 2.9:
  

• Example: Our sample space $S$ is the population of adults in a small town. They can be categorized according to gender and employment status (see Table 2.1).
  

  Table 2.1: Categorized adult population in a small town.

  	Employed	Unemployed	Total
  Male	460	40	500
  Female	140	260	400
  Total	600	300	900

  
  
• One individual is to be selected at random for a publicity tour.
• The concerned events
  • $M$: a man is chosen
  • $E$: the one chosen is employed
  $$P(M\vert E)=\frac{460}{600}=\frac{23}{30}$$
  $$P(M\vert E)=\frac{n(E \cap M)/n(S)}{n(E)/n(S)}=\frac{P(E \cap M)}{P(E)} =\frac{\frac{460}{900}}{\frac{600}{900}}=\frac{23}{30}$$

• Example 2.33: The probability that a regularly scheduled flight departs on time is $P(D) = 0.83$;
• the probability that arrives on time is $P(A) = 0.82$;
• the probability that it departs and arrives on time is $P(D \cap A) = 0.78$.
• Find the probability that a plane
  • arrives on time given that it departed on time, and
    $$P(A\vert D) = \frac{P(D \cap A)}{P(D)} = \frac{0.78}{0.83} = 0.94$$

  • departed on time given that it has arrived on time.
    $$P(D\vert A) = \frac{P(D \cap A)}{ P(A)} = \frac{0.78}{0.82} = 0.95$$

• Definition 2.10:
• If knowing that event $B$ occurred doesn't change the probability that $A$ will occur, then $B$ must carry no information about $A$.
• The condition $P(B\vert A) = P(B)$ implies that $P(A\vert B) = P(A)$, and conversely.
• Example: Two cards are drawn in succession, with replacement
  • Event $A$: the first card is an ace
  • Event $B$: the second card is a spade
    $$P(B\vert A)=\frac{P(A \cap B)}{ P(A)} =\frac{1/52}{4/52}=\frac{13}{52}=\frac{1}{4} and P(B)=\frac{13}{52}=\frac{1}{4}$$

  • Since $P(B\vert A) = P(B)$, these two events are independent.