10.04 Conditional probability
Interactive practice questions
There are two positions available at a company and the applicants have been shortlisted down to $6$6: Patricia, Nadia, Amy, Aaron, Lachlan, Jimmy.
The two people to fill the position will be picked randomly.
Event $A$A: Patricia is chosen
Event $B$B: Aaron is chosen
$P$P$($($A$A|$B$B$)$) is the probability that:
Aaron is chosen second, after Patricia has been chosen first.
Patricia or Aaron are chosen, but not both.
Patricia is chosen first or second.
Patricia is chosen second, after Aaron has been chosen first.
$P$P$($($A$A$\mid$∣$B$B$)$) is:
less than $P$P$($($B$B$\cap$∩$A$A$)$).
equal to $P$P$($($B$B$\cap$∩$A$A$)$).
greater than $P$P$($($B$B$\cap$∩$A$A$)$).
The following are probabilities for an experiment in which $A$A and $B$B are two possible events.
$P\left(A\cap B\right)=0.48$P(A∩B)=0.48, and
$P\left(A\right)=0.6$P(A)=0.6.
Find $P\left(B|A\right)$P(B|A).
For events $A$A and $B$B we can find the probability of $A$A given $B$B using$P\left(A|B\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}$P(A|B)=P(A∩B)P(B).
An individual is chosen from the group. According to the Venn diagram,