Combinations and probabilities I'm tackling question below. I have a hard time understanding combinations and I do already have the numeric answers. If someone could please help me understand how to get there and reason in a pedagogic way?
There are seven ice cream flavors that can be combined in different ways. A flavor cannot be chosen twice.
Two of the seven flavors are vanilla and chocolate.
b) What is the probability to get an ice cream bowl with Van and Choc if five flavors total are selected randomly?
c) What is the probability to get an ice cream bowl without Van and Choc if five flavors total are selected randomly?
d) What is the probability to get an ice cream bowl with one of the flavors Van and Choc if five flavors total are selected randomly?
e) What is the probability to get an ice cream bowl with Van and Choc given that you get at least one of the ingredients of Van and Choc when five flavors total are selected randomly?
f) What is the probability to get an ice cream bowl with Van given that you get at least one of the ingredients of Van and Choc when five flavors total are selected randomly?
Thank you!
 A: For part b): 
We need to find
$$\tag{1}
\text{number of bowls that have both Chocolate and Vanilla}\over \text{total number of different bowls  } 
$$
For the denominator of $(1)$: the total number of ways to select five flavors from seven is  ${7\choose5}={7!\over 2!\cdot 5!}={7\cdot 6\over 2}=21$. (Generally, the number of ways to select $r$ objects from $n$ distinct objects where the order of selection does not matter is ${n\choose r}={n!\over (n-r)!r!}$).
For the numerator of (1), we need to find the number of ways to choose five flavors with both Vanilla and Chocolate amongst them. Hmm, two of the flavors are already selected: we need only select three flavors from the five flavors that aren't Vanilla or Chocolate. 
The number of ways to do this is ${5\choose3}={5!\over2!\cdot 3!}={5\cdot 4\over 2}=10$.
So, computing (1) now, the desired probability is $10\over 21$.



For part c): 
We need to compute:
$$\tag{2}
\text{number of bowls that have neither Chocolate nor Vanilla}\over \text{total number of different bowls  } 
$$
We already computed the denominator of (2) in part b), it is 21.

To find the numerator:
The number of ways of choosing five flavors from  the five flavors that aren't Vanilla or Chocolate  is ${5\choose5}={1}$.
So, computing (2) now, the desired probability is $1\over21$.



For part d): 
We need to compute 
$$\tag{3}
\text{number of bowls that have exactly one of Chocolate or Vanilla}\over \text{total number of different bowls  } 
$$
As before, the denominator in (3) is 21.
To find the number of ways to get a bowl with  exactly  one of the flavors Vanilla or Chocolate, it proves convenient to break this  up into cases:
Case 1): Vanilla only: we have one flavor already (Vanilla); we then need to choose four more flavors that are neither Vanilla nor Chocolate: This is ${5\choose4}={5!\over 1!\cdot 4!}= 5$.
Case 2): Chocolate only: we have one flavor already (Chocolate); we then need to choose four more flavors that are neither Chocolate nor Vanilla: This is ${5\choose4}= 5$.
So,
the number of ways to get a bowl with exactly one of the flavors Vanilla or Chocolate
is obtained by adding the preceeding two quantities: $ 5+ 5=10$.
So, computing (3) now, the desired probability is $10/21$.



Parts e) and f) are somewhat different as they are conditional probabilities.
For part e):
Let 
$\ \ \ \ \ \ \ A$ be the event that the bowl has both Chocolate and Vanilla 
and let 
$\ \ \ \ \ \ \ B$ be the event that at the bowl has at least one of the flavors Chocolate and Vanilla. 
We want to compute
$$
P(A|B)={P(A\cap B)\over P(B)}
$$
Now $A\cap B$ is just the event   that the bowl has both Vanilla and Chocolate. We found 
$P(A\cap B)$ in part b). It turned out to be $P(A\cap B)={10\over 21}$. 
To find $P(B)$, we need to find the number of bowls that have at least one of Vanilla or Chocolate.  To do this, we break it up into cases:
exactly one of Vanilla or Chocolate: 10 (part d)
both Vanilla and Chocolate: 10 (part b)
So the number of bowls with at least one of Vanilla or Chocolate is $10+10=20$ (note, we could have also computed this by taking the total number of bowls and subtracting the number that had neither Vanilla nor Chocolate: $21-1=20$).
So $P(B)= {10+10\over 21}={20\over21}$.
and
$P(A|B) =  {   10/21 \over 20/21}={1\over2}$.

I'll leave part f) for you; but leave a comment if you need help with it.
A: For simplification, let us call an icecream-cake any combination of $5$
different flavours (from the given $7$). There are $\dbinom{7}{5}=21$ ways of
getting an icecream-cake (there are $7$ flavours and we need to choose $5$
without repetition).
From these $21$ cases (which are equally probable to happen), there is/are:


*

*$\dbinom{5}{3}=10$  icecream-cakes with both $V$ (vanilla flavour) and
$C$ (chocolate flavour). This is because we only have to choose $3$ more
flavours ($2$ are already chosen: $V$ and $C$) of the $5$ flavours that are
not $V$ and $C$. Hence the first probability of getting an icecream-cake with
$V$ and $C$ is
$$p_{1}=\dfrac{10}{21}.$$

*one ($1$) icecream-cake without $V$ and $C$, since we need to choose all
the $5$ flavours that are not $V$ and $C$, hence
$$
p_{2}=\dfrac{1}{21}%
$$
and
$$
p_{3}=1-p_{2}=\frac{20}{21}
$$
(this is the probability of having an icecream-cake with at least one of $V$
and $C$).
Also, the probability of getting an icecream-cake:


*

*with only one of $V$ and $C$ is
$$p_{3^{\prime}}=\frac{10}{21}$$
since there are $20$ icecream-cakes with at least $V$ or $C$ (see $p_{3}$) and
in $10$ cases we have both $V$ and $C$ (see $p_{1}$) which we need to eliminate.

*with both $V$ and $C$ given that we already have $V$ or $C$ is
$$
p_{4}=\frac{10}{20}=\frac{1}{2}%
$$
since there are $10$ icecream-cakes with both $V$ and $C$ (see $p_{1}$) and
$20$ icecream-cakes with at least $V$ or $C$ (see $p_{3}$).

*with $V$ (and any option on $C$) given that we already have $V$ or $C$
is
$$
p_{5}=\frac{15}{20}=\frac{3}{4}
$$
since there are $\dbinom{6}{4}=15$ icecream-cakes with $V$ (after choosing
$V$, we have $4$ more flavours to choose from the remaing $6$) and $20$
icecream-cakes with at least $V$ or $C$ (see $p_{3}$).
