Question

(2-1-14) ften in buying a product at a supermarket, there is a concern about the item being underweight. Suppose there are 20 "one-pound" packages of frozen ground turkey on display and 3 of them are underweight. A consumer group buys 5 of the 20 packages at random. What is the probability of at least one of the five being underweight?

Answer 1

Answer:

0.60

Step-by-step explanation:

Number of underweight packages = 3

Number of normal packages = 17

The probability of at least one of the five packages being underweight is 100% minus the probability that no packages are underweight:

$P(X\geq0)=1-P(X=0)\\P(X\geq0)=1-(\frac{17}{20} *\frac{16}{19}*\frac{15}{18}*\frac{14}{17}*\frac{13}{16}) \\P(X\geq0) = 0.60$

The probability of at least one of the five being underweight is 0.60

Answer 2

Answer:

60.09% probability of at least one of the five being underweight

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

The order in which the packages are chosen is not important. So the combinations formula is used to solve this question.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

Either

There are no underweight packers, or at least one is underweight. The sum of the probabilities of these events is 100%. So

Probability that there are no packages underweight.

Desired outcomes:

5 packets from a set of 17(that are not underweight). So

$D = C_{17,5} = \frac{17!}{5!(17-5)!} = 6188$

Total outcomes:

5 packets from a set of 20(total number of packages). So

$T = C_{20,5} = \frac{20!}{5!(20-5)!} = 15504$

Probability:

$p = \frac{D}{T} = \frac{6188}{15504} = 0.3991$

39.91% probability that none of the packages are underweight.

What is the probability of at least one of the five being underweight?

p + 39.91 = 100

p = 60.09

60.09% probability of at least one of the five being underweight