# I am looking for some advice on how to calculate the answer to this probability calculation. [closed]

I am looking for some advice on how to calculate the answer to this probability calculation.

A box contains $20$ components of which $15$ are good and $5$ are faulty. If $3$ components are chosen at random from the box, find the probability that at least $2$ are good.

## closed as off-topic by Did, Harish Chandra Rajpoot, Claude Leibovici, Michael Galuza, Milo BrandtAug 22 '15 at 15:15

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• Break it into two cases: A. All three are good. B. Exactly two are good. – lulu Aug 22 '15 at 11:36
• You could try drawing a tree diagram, perhaps? – David Quinn Aug 22 '15 at 11:44

Let $X$ be the number of successes (good components) drawn.

$P(X = 0) = \frac{\binom{15}{0} \binom{5}{3}}{\binom{20}{3}}$ is the probability that 0 are good.

$P(X = 1) = \frac{\binom{15}{1} \binom{5}{2}}{\binom{20}{3}}$ is the probability that 1 is good.

Thus $P(X \geq 2) =1 - P((X=0) \cup (X=1)) = 1 - [P(X = 0) + P(X = 1)]$.

This is an example of a hypergeometric distribution.

[EDIT] I am assuming that this experiment is without replacement. If it is with replacement then ignore my answer.

• Of course you can also compute $P(X\geq2)$ as $P(X=2)+P(X=3)$, but in questions with 'at least' or 'at most' in it focusing on the complement is often indeed very useful. – drhab Aug 22 '15 at 12:31

This is a route you can take.

Take the components one by one. Let $G_i$ denote the event that the $i$-th component is good and let $F_i$ denote the event that is faulty.

Then you are looking for: $$P(G_1G_2G_3\cup G_1G_2F_3\cup G_1F_2G_3\cup F_1G_2G_3)$$

(Here $AB$ stands for $A\cap B$.)

Remark that the $4$ events in this union exclude eachother. So?...

Also e.g. $P(G_1G_2F_3)$ can be calculated as: $$P(G_1)P(G_2\mid G_1)P(F_3\mid G_1G_2)$$

Practicizing this makes it more simple.

Edit:

Compare $P(G_1G_2F_3),P(G_1F_2G_3),P(F_1G_2G_3)$ with eachother and try to find a pattern. Wonder why. Thinking about it improves your intuition.

The total number of ways to choose $3$ out of $20$ items is:

$$\binom{20}{3}=1140$$

The number of ways to choose $2$ out of $15$ good items and $1$ out of $5$ faulty items is:

$$\binom{15}{2}\cdot\binom{5}{1}=525$$

The number of ways to choose $3$ out of $15$ good items and $0$ out of $5$ faulty items is:

$$\binom{15}{3}\cdot\binom{5}{0}=455$$

So the probability of choosing at least $2$ good items is:

$$\frac{525+455}{1140}\approx86\%$$