My son Is having trouble working out the following:-

A box contains 2 Strawberry Yougurts, 4 Vanilla yogurts and 6 Cherry Yogurts.

Three yogurts are selected randomly from the box.

Calculate the probability that at least one of the selected yogurts is a cherry yogurt.

Any help greatly appreciated.

2. you need to calculate the probability that at least one yogurt is cherry, so that means you need to calculate the probability that one yogurt is cherry, two yogurts are cherry, and three yogurts are cherry and add them together.

3. Originally Posted by nightowl007
A box contains 2 Strawberry Yougurts, 4 Vanilla yogurts and 6 Cherry Yogurts.
Three yogurts are selected randomly from the box.
Calculate the probability that at least one of the selected yogurts is a cherry yogurt.
Find the the probability of no cherry yogurt: $\frac{\binom{6}{3}}{\binom{12}{3}}$.

Substract that number from 1. The oppsite of none is at least one.

4. ## Thanks Plato

Thanks Plato and o&apartyrock for such a quick answers........

Is there a formula to calculate answers for questions like this?

Thank you very much.

Nightowl007

5. There are six non-cherry yougurts, choose three.
$\binom{6}{3}=\frac{6!}{(3!)(3!)}=\frac{(6)(5)(4)}{ (3)(2)(1)}$.
[If your son does not understand this notation, he has no business being asked to do this problem.]
There are a total of $\binom{12}{3}$ ways to choose three from twelve.

6. Hello, nightowl007!

I must assume that your son is familiar with Combinations
. . which has a variety of notations.

There are 12 objects and we choose 3 of them.
The number of possible choices can be written: . $_{12}C_3\quad\text{ or }{12\choose3}\quad \text{or }C(12,3)$

They are all equal to: . $\frac{12!}{3!\,9!} \:=\:220$

A box contains 2 Strawberry yogurts, 4 Vanilla yogurts and 6 Cherry yogurts.

Three yogurts are selected randomly from the box.

Calculate the probability that at least one of the selected yogurts is a Cherry yogurt.

Plato's approach is usually appropriate with an "at least" or "at most" problem.

There are 6 Cherry yogurts and 6 Others.
Find the probability that we get no Cherry yogurts.
. . This means we get three Others.

There are: . $_6C_3 \:=\:20$ ways to get 3 Others.

And there are: . $_{12}C_3 \:=\:220$ possible outcomes.

Hence: . $P(\text{no Cherry}) \:=\:\frac{20}{220} \:=\:\frac{1}{11}$

Therefore: . $P(\text{at least 1 Cherry}) \;=\;1 - \frac{1}{11} \;=\;\boxed{\frac{10}{11}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

o&apartyrock's method usually takes longer, but it's correct.
. . And it is a great way to check our answer.

1 Cherry (and 2 Others): . $(_6C_1)(_6C_2) \:=\:6\cdot15 \:=\:90$

2 Cherries (and 1 Other): . $(_6C_2)(_6C_1) \:=\:16\cdot6 \:=\:90$

3 Cherries (and 0 Others): . $(_6C_3)(_6C_0) \:=\: 20\cdot1 \:=\:20$

Hence, there are: . $90 + 90 + 20 \:=\:200$ ways to get at least one Cherry.

Therefore: . $P(\text{at least 1 Cherry}) \;=\;\frac{200}{220} \;=\;\frac{10}{11}\quad\hdots\quad \text{ta-}DAA!$