Question

On the HPC Ch.7B Test, 24 out of 46 students earned a sticker on their test. If a WASC member randomly pulls 5 tests out of the folder of tests, what is the probability that more than 1 of them have a sticker?

Answer 1

Answer:

85.27% probability that more than 1 of them have a sticker.

Step-by-step explanation:

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

The order of in which the tests are chosen is not important. So we use the combinations formula 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.

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

We have that

24 students earned a sticker

22 students did not earn a sticker.

What is the probability that more than 1 of them have a sticker?

Either one or less student earned a sticker, or more than 1 did. The sum of these probabilities is 100%. So i will find the probability that one or less did.

Probability that one or no student earned a sticker.

Desired outcomes

1 sticker

1 with sticker from a set of 24, 4 without a sticker from a set of 22. So

$C_{24,1}*C_{22,4} = \frac{24!}{1!23!}*\frac{22!}{4!18!} = 24*7315 = 175560$

0 sticker

5 without a sticker from a set of 22. So

$C_{22,5} = \frac{22!}{5!17!} = 26334$

Then

$D = 175560 + 26334 = 201894$

Total outcomes

5 from a set of 46

$T = C_{46,5} = \frac{46!}{5!41!} = 1370754$

Probaiblity:

$P = \frac{D}{T} = \frac{201894}{1370754} = 0.1473$

14.73% probability that 1 or none have a sticker

More than 1:

100 - 14.73 = 85.27

85.27% probability that more than 1 of them have a sticker.