Question

A pharmacist receives a shipment of 21 bottles of a drug and has three of the bottles tested. If five of the 21 bottles are contaminated, what is the probability that less than two of the tested bottles are contaminated? Express your answer as a fraction or a decimal number rounded to four decimal places.

Answer 1

The probability is 0.8722

Explanation:

There are 21 bottles

5 of them are contaminated.

there are 21 - 5 = 16 non-contaminated bottles.

So, if we grab a bottle at random, the probability that this bottle is contaminated will be equal to the quotient between the number of contaminated bottles and the total number of bottles, this is:

p = 5/21

Now we want to find the probability that, for 3 tested bottles, that less than two (0 or 1 ) are contaminated.

Let's see each case on its own.

0 bottles:

The probability of getting a non-contaminated bottle in the first try is equal to the quotient between the number of non-contaminated bottles and the total number of bottles, this is:

p₁ = 16/21

For the second bottle is the same, but because one non-contaminated bottle was drawn before, now there are 15 non-contaminated bottles and 20 bottles in total, so now the probability is:

p₂ = 15/20

and similarly, for the third bottle the probability is:

p₃ = 14/19

The joint probability is the product of the individual probabilities, we get:

P = p₁*p₂*p₃ = (16/21)*(15/20)*(14/19) = 0.4211

Now the case that one bottle is contaminated.

Let's assume that the first one is contaminated.

The probability of getting a contaminated bottle in the first draw is equal to the quotient between the number of contaminated bottles and the total number of bottles, so:

p₁ = 5/21

For the second bottle, we want a non-contaminated one, there are 16 non-contaminated bottles and 20 bottles left, so here the probability is:

p₂ = 16/20

and for the third bottle we have the probability:

p₃ = 15/19

The joint probability is:

p = p₁*p₂*p₃ = (5/21)*(16/20)*(15/19)

Also notice that we only looked at the case where the first bottle is contaminated, we also have the case where the second one is contaminated and the case where the third one is contaminated, so there are 3 permutations, then the probability of having one contaminated bottle is:

Q = 3*p = 3*(5/21)*(16/20)*(15/19) = 0.4511

Then the probability of having less than two contaminated bottles is:

probability = P + Q = 0.4211 + 0.4511 = 0.8722