Conditional Probability by counting

Question

I have stumbled upon this question and I was wondering about how to approach the entire question just by counting.

Two viewers are selected at random from a group of 6 local and 10 international viewers.

Events A = First viewer is international
Events B = Second viewer is international

We can use permutation to find the total arrangement. Therefore,

$$ \frac{16!}{(16-2)!} = 240 $$

To find event A,

(first international)(second international) and (first international)(second local)

$$ \frac{{10 \choose 1}{6 \choose 1} + {10 \choose 9}{9 \choose 1}}{240} = \frac {150}{240} = 0.625 $$

Similarly, to find event B

To find event B,

(first local)(second international) and (first international)(second international)

$$ \frac{{6 \choose 1}{10 \choose 1} + {10 \choose 1}{9 \choose 1}}{240} = \frac {150}{240} = 0.625 $$

The other part of the question asks me to find P(B|A)

to find $ P(A | B) $ we must first know $ P(A \cap B) $

In some source, it is mentioned that $ P(A \cap B) = P(A)P(B) = (0.625)(0.625) = 0.390625$

However, I am wondering if $ P(A \cap B) = \frac {{10 \choose 1}{9 \choose 1}}{240} = \frac {90}{240} = 0.375 $

Since $ P(A \cap B) $ can be interpreted as First is international and Second is international

From here, I am just doubting my approach to this question, how to obtain the value of $ P(A \cap B) $ and how to correctly solve for $ P(B | A)$.

Answer 1

No.  There are $\binom{16}1\binom {15}1$ ways to select a first and a second viewer from the sixteen.   That is not just selecting two viewers, but also an arrangement of those two.   Often this is referred to as "order matters". $$\binom{16}1\binom{15}1\quad=\quad\binom{16}2\, 2!$$

There are $\binom{10}1\left(\binom{6}1+\binom 91\right)$ ways to select a first viewer from the ten international candidates, as you had.   This equals $\binom{10}1\binom{15}1$. Thus: $$\mathsf P(A)\quad=\quad\dfrac{\dbinom{10}1\dbinom 61+\dbinom{10}1\dbinom 91}{\dbinom{16}2\,2!}\quad=\quad\dfrac{10}{16}$$

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The probability for event $B$ is found similarly. $\mathsf P(B)=10/16$.

And the probability for the intersection is $$\mathsf P(A\cap B)\quad=\quad\dfrac{\dbinom{10}2\,2!}{\dbinom{16}2\,2!}\quad=\quad\dfrac{10\cdot~~9}{16\cdot 15}\quad=\quad\dfrac 38$$

Answer 2

Your derivation by counting is fine, indeed we have

$$P(A)=P(B)=\frac{10\cdot 15}{16\cdot 15} =\frac58$$

$$P(A\cap B)=\frac{10\cdot 9}{16\cdot 15} =\frac{3}{8}$$

then by conditional probability we obtain

$$P(B | A)=\frac{P(A\cap B)}{P(A)}=\frac 3 5$$

Note that by definition

$$P(A \cap B) = P(A)P(B)$$

only if the two events are independent that is $P(B|A)=P(B)$ which is not true in this case.