$(1)$ What is the probability that that a family with $3$ children
has at least one girl, given that at least one is a boy?
$(2)$ And what's the probability that they have at least one boy AND one girl?
$(3)$ And what's the probability that they have at least one boy OR one girl?
All possible combinations:
BBB
BBG
BGB
BGG
GBB
GBG
GGB
GGG
Question $(1)$:
Exclude the 'GGG' option.
$\implies$ $P$(at least $1$ boy) $= 7/8$
The new sample space is $7$ of which $6$ have at least $1$ girl
$\implies$ $P$(at least $1$ girl $\mid$ at least $1$ boy) $= 6/7$
Question $(2)$:
By looking at the combinations I can quickly say that there are $6$ out of $8$ that satisfy (at least $1$ girl AND at least $1$ boy).
$\implies 6/8 = 3/4$.
However, considering larger data sets, this becomes impossible.
According to:
$P(A\ \text{and}\ B) = P(A)*P(B|A) = \frac{7}{8} \times \frac{6}{7} = \frac{6}{8} = \frac{3}{4} $
I get the same result.
So, why does multiplying $7/8 \times 6/7$ provide the correct result?
I know that the denominators $7\times 8$ produce a new sample space (permutations), and the $7\times 6$ new desirable outcomes (permutations). But these are permutations of $(x,y)$ ($2$ children) not $(x,y,z)$ ($3$ children).
But I cannot explain why it works, since the original sample space does not consider permutations. Or am I wrong?
Many thanks for assistance in explaining why this works.
$(3)$ What is the difference between $(2)$ and $(3)?$ Are they not the same?
Best Answer
For (2), it is a standard formula that $$ P(AB)=P(B|A)P(A) $$ Often conditional probability is defined as $$ P(B|A)=P(AB)/P(A) $$ For (3), let $A$ be event of having at least one boy and $B$ be event of having at least one girl. Then $P(A)=P(B)=7/8$. And the probability of having at least one boy and at least one girl is $P(A\cap B)=6/8=3/4$.
So we have $$ P(A\cup B)=P(A)+P(B)-P(A\cap B)=1 $$