# Probability that a sample of sweets is rejected

I am answering the following three simple questions, and I just want to make sure I'm not involved in confusions at this early stage:

A sweet shop owner receives a delivery of a decorative sweet. If the batch is not properly prepared then a significant proportion of the batch is fragile and can break when compressed with a small amount of pressure. The owner selects 16 sweets randomly from the delivery to test. If 2 or more of these sweets are fragile then she will return the batch to the manufacturer.

Imagine that the owner receives 400 sweets of which 87 are fragile (will break under a small amount of pressure). Give your answers to the following questions to 3 significant figures:

1. What is the probability that the owner finds no fragile sweets?

2. What is the probability that the owner finds exactly one fragile sweet?

3. What is the probability that the owner returns the batch?

My thinking is that both of these are combinations problems. For the first one, surely want to know how many ways there are of arranging 16 non-fragile sweets given 313 non-fragile sweets, and then we want to divide that by the total number of ways to arrange 16 sweets regardless of fragility. I.e. we want:

$$P(f=0) = \frac{313 \choose 16}{400 \choose 16}$$

$$= 0.0181$$

(I'm assuming 3 significant figures starts from the first non-zero digit... is that right?)

For the second one, we essentially want to know how many different ways there are to choose 16 sweets out of a batch of 400 and get just 1 fragile sweet in there. To work that out we want:

$$313 \choose 15$$

Since we want to know how many distinct arrangments of the remaining 15 non-fragile sweets there are from the 313 non-fragile ones once we've already picked a fragile one. Assuming that each sweet is equally likely to be picked, our sample space is just the set of different possible combinations of 16 sweets out of 400. Thus,So, if 'f' stands for 'fragile sweet', we should have:

$$P(f = 1) = \frac{313 \choose 15 }{400 \choose 16}$$

$$= 0.0009735015578820893 = 0.000973$$

For the third question, we are told that the owner returns the batch if two or more sweets are fragile, so we basically want to the union of the events wherein 2 fragile sweets are picked, 3 fragile sweets are picked... 16 fragile sweets are picked. To avoid calculating factorials for all of those, we could just calculate the probability that the owner gets 1 or less fragile sweet, and then the probability that we get 2 or more is just one minus that. So:

$$P(f\ge2) = 1 - P(f<2)$$

$$= 1 - \frac{313 \choose 15}{400 \choose 16} + \frac{313 \choose 16}{400 \choose16}$$

which gives:

$$1 - 0.019104968073436004 = 0.980895031926564 = 0.981$$

$$P(f = 0) = 0.0181$$

$$P(f = 1) = 0.000973$$

$$P(f \ge 2) = 0.981$$

Is my thinking right on these questions, or have I made a blunder (or several)? I am perhaps a little bit concerned that it is a lot less likely that the owner gets 1 fragile sweets 0. Let me know! I used the math.comb() function in python to get the relevant combinations.

For the second question, if the owner selects exactly one fragile sweet, then she must pick $$15$$ of the $$313$$ good sweets and one of the $$87$$ fragile sweets when she selects $$16$$ of the $$400$$ sweets in the batch. Hence, $$\Pr(f = 1) = \frac{\dbinom{313}{15}\dbinom{87}{1}}{\dbinom{400}{16}}$$
Hence, the probability that she returns the batch is $$Pr(f \geq 2) = 1 - \Pr(f < 2) = 1 - \Pr(f = 0) - \Pr(f = 1) = 1 - \frac{\dbinom{313}{16}}{\dbinom{400}{16}} - \frac{\dbinom{313}{15}\dbinom{87}{1}}{\dbinom{400}{16}}$$