What are the chances? There is a class with 20 students.
We pick 4 lucky students who can go to the cinema.
What are the chances? Andrew can go to the cinema, but his best friend John can't go with them.
 A: So the total number of ways to select the students is ${20}\choose{4}$, which serves as the denominator. For the numerator, we want the number of ways the students can be picked in which Andrew is chosen and John is not. Since we want Andrew chosen, there are only 3 open spots, and we want to pick from the remaining 18 students who are not John. Therefore, the answer is $\dfrac{{18}\choose{3}}{{20}\choose{4}}$
A: Total number of the possibilites are $\binom{20}{4}$.
Now we calculate the possibilities where Adam can go, and John can't. Thus, 18 students remain, and 3 of them gets a ticket. They are $\binom{18}{3}$.
The chance to get one of such combinations, are: $\frac{\binom{18}{3}}{\binom{20}{4}}$, which is $\frac{\frac{18!}{3! \cdot 15!}}{\frac{20!}{4! \cdot 16!}}$.
If you don't like calculators because you want to train your mind, you can simplify the result:
$\frac{\frac{18!}{3! \cdot 15!}}{\frac{20!}{4! \cdot 16!}}$ = $\frac{4! \cdot 16!}{20!} \cdot \frac{18!}{3! \cdot 15!} = \frac{4 \cdot 16}{20 \cdot 19} = \frac{16}{5 \cdot 19} = \underline{\underline{\frac{16}{95}}}$.
A: The probability that Andrew will be picked out is (as for every student) $\frac4{20}=\frac1{5}$. Under the condition that Andrew is picked out the probability that John will not be picked out is $\frac{16}{19}$.
So:$$\Pr(\text{Andrew lucky}\wedge\text{John unlucky})=$$$$\Pr(\text{Andrew lucky})\Pr(\text{John unlucky}\mid\text{Andrew lucky})=\frac15\frac{16}{19}=\frac{16}{95}$$
