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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!

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Do you require that the five selected flavors are different? This is the difference between sampling with and without replacement. It makes a big difference. –  Ross Millikan Jan 13 '12 at 22:13
    
If this is homework, it should be marked as such. –  Scaramouche Jan 13 '12 at 22:16
    
Hi! They need to be different. Vanilla can't be chosen twice for example. –  samuelf Jan 13 '12 at 22:17
    
It's not really homework as I'm just trying to understand. Not for school that is.. –  samuelf Jan 13 '12 at 22:17
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2 Answers

up vote 1 down vote accepted

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.

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Thank you so much for this detailed answer, I cannot express how grateful I am that you take of your time to do this. I will study it for a little while and get back to you shortly. –  samuelf Jan 13 '12 at 22:50
    
Regarding (b), what I don't understand is if I do 5C3, that makes me think that Van or Choc could turn up again. Please look at wolframalpha.com/input/… where I simulated this. In the combinations they turn up again, why is that so? We only wanted to select them once? I don't get the part why 5C3 would only pick the ones we haven't already selected. Regarding part (c), couldn't it be the complement instead? if 10 was the answer for part B, wouldn't the complement of 11 be the answer for part (c)? –  samuelf Jan 13 '12 at 23:23
    
@samuelf It's $5\choose3$ where the "5" is the five flavors that are not V or C. We select three of those: from $\{ strawberry, mint, peach, rasberry, peanut butter\}$, we select three. For part c), the complement would include, e.g., selections with Vanilla but not Chocolate. I interpreted part c) as having neither Vanilla nor Chocolate (note I rephrased part c) as such. And, your very welcome; I'm glad if this helps. –  David Mitra Jan 13 '12 at 23:48
    
This helps a lot! How can I know that the "5" doesn't contain V or C? What is the logic behind that? Is it really someting like ${7-2 \choose 3}$ ? I'm thinking that the 5 could contain V or C, we don't know because it was randomly selected. –  samuelf Jan 14 '12 at 0:17
    
@samuelf Not random at all: you decide what the 5 things are, but you decide so that you can solve the problem. I'm not sure how to explain it... Simply put, if you know the selection includes only one V and only one C, then to get the other three, you choose from the other five flavors. I guess you might call it "real world" reasoning. –  David Mitra Jan 14 '12 at 0:23
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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 $p_{3}$).

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