Question

In a bag of skittles 12 are yellow, 10 purple, 8 red, 9green and 22 orange. If 5 were selected from the bag. Calculate using counting technique what is the probability that:
A). exactly 2 are red
B). At most 2 are red

Answer 1

Answer:

a) 11.03% probability that exactly two are red.

b) 98.64% probability that at most 2 are red.

Step-by-step explanation:

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

In this question, the order in which the skittles 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.

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

In this question:

12 + 10 + 8 + 9 + 22 = 61 skittles.

A). exactly 2 are red

Desired outcomes:

2 red, from a set of 8.

3 non-red, from a set of 61 - 8 = 53.

So

$D = C_{8,2}*C_{53,3} = \frac{8!}{2!(8-2)!}*\frac{53!}{3!(53-3)!} = 655928$

Total outcomes:

Five skittles from a set of 61. So

$T = C_{61,5} = \frac{61!}{5!(61-5)!} = 5949147$

Probability:

$p = \frac{D}{T} = \frac{655928}{5949147} = 0.1103$

11.03% probability that exactly two are red.

B). At most 2 are red

Desired outcomes:

None red(5 from a set of 53)...

One red(from a set of 8), and four non-read(4 from a set of 53).

Two red(655928), as found in a.

So

$D = C_{53,5} + C_{8,1}*C_{53,4} + 655928 = \frac{53!}{5!48!} + \frac{8!}{1!7!}*\frac{53!}{4!49!} + 655928 = 5868213$

Total outcomes:

Five skittles from a set of 61. So

$T = C_{61,5} = \frac{61!}{5!(61-5)!} = 5949147$

Probability:

$p = \frac{D}{T} = \frac{5868213}{5949147} = 0.9864$

98.64% probability that at most 2 are red.