1. ## Conditional Probability

A class consists of seven boys and 9 girls. Two different members of the class are chosen at random. A is the event {the first person is a girl}, and B is the event {the second person is a girl}. Find the probability of:

a) B|A'

2. Originally Posted by creatively12
A class consists of seven boys and 9 girls. Two different members of the class are chosen at random. A is the event {the first person is a girl}, and B is the event {the second person is a girl}. Find the probability of:

a) B|A'

$\displaystyle P(B|A)$ is asking us what if the prob a randomly getting a girl is you have already pick one.

This just reduces your sample space. After you have picked the first girl there are now 7 boys and 8 girls to choose from.

To the probability is $\displaystyle P(B|A)=\frac{8}{15}$

after thought: does $\displaystyle B|A'$ mean A compliment i.e $\displaystyle B|A^c$ if that is the case the same reasoning as above would still work exept you would remove a boy instead of a girl from your sample space.

3. thnx for the reply, but why u have to take out not from a girl, but from a boy, cause A represents girls, and A' would be 1-A right, so if we say lik, 1-(9/16), then why is it wrong, can you explain this part, i'd really appreciate that

4. Originally Posted by creatively12
thnx for the reply, but why u have to take out not from a girl, but from a boy, cause A represents girls, and A' would be 1-A right, so if we say lik, 1-(9/16), then why is it wrong, can you explain this part, i'd really appreciate that
TES showed you an example of what to do. To paraphrase:

After you have picked the first boy there are now 6 boys and 9 girls to choose from
Therefore ....

5. Hello, creatively12!

A class consists of 7 boys and 9 girls.
Two members of the class are chosen at random.
$\displaystyle A$ = (1st person is a girl}, and $\displaystyle B$ = (2nd person is a girl}.

Find: .$\displaystyle (a)\;P(B\,|\,A')$

We want: .$\displaystyle P(B\,|\,A') \;=\;P(\text{2nd is girl}\,|\,\text{1st is a boy})$

Since the first chosen is a boy,
. . there are 6 boys and 9 girls to choose from.
Then the probability of choosing a girl is: .$\displaystyle \tfrac{9}{15} \:=\:\tfrac{3}{5}$

Therefore: .$\displaystyle P(\text{2nd is girl}\,|\,\text{1st is boy}) \;=\;\frac{3}{5}$