# Part 1. Probability P(A or B), P(A and B) There are M&M candies in the box, red and blue and yellow. Assign your numbers of candies for each color. Use your numbers to calculate probability...

Part 1. Probability P(A or B), P(A and B)

There are M&M candies in the box, red and blue and yellow.

Assign your numbers of candies for each color.

Use your numbers to calculate probability that

a) randomly taken one candy is blue or red: P (Blue or Red)

b) first taken candy is blue and second is red: P (Blue and Red).

In case b) keep in mind that first taken candy is not returned back.

So, when you take the second candy there are one less candies in the box.

Part 2. Counting Outcomes (Permutations and Combinations)

Assign two numbers: N and X (X is portion of N).

Calculate in how many ways can you select X people out of N candidates.

Consider two cases:

a) for the same position (order of selection doesn't matter - use Combinations)

b) for different positions (order of selection makes a difference - use Permutations)

*print*Print*list*Cite

### 2 Answers

Part 1: M&M Candies: 3 yellow, 4 red, 5 blue = 12 candies

a: P(Blue or Red) = `5/12 + 4/12 = 9/12 = 3/4`

b: P(Blue and Red) = `5/12 + 4/11 = 55/132+ 48/132 = 103/132`

Part 2: Two numbers, let 4 be x and 8 be N.

a: select 4 people out of 8 (order does not matter-use combinations)

`(8!) / ((8-4)!x4!) = (8x7x6x5x4x3x2x1)/((4x3x2x1)(4x3x2x1)) = 70`

`` b: select 4 people out of 8 (order does matter-use permutations)

`(8!)/((8-4)!) = 1680`

Let’s say that there are 5 red, 15 blue and 10 yellow

a) randomly taken one candy is blue or red: P (Blue or Red)

**n(S)=30**

**n(A)=blue, 15**

**n(B)=red, 5**

P(A)=n(A)/n(S)

P(B)=n(B)/n(S)

**P(A or B)= P(A)+P(B)**

= 15/30 + 5/30

= 20/30

= **0.67**

b) first taken candy is blue and second is red: P (Blue and Red).

In case b) keep in mind that first taken candy is not returned back.

So, when you take the second candy there is one less candy in the box.

**P(A and B)= P(A) * P(B)**

=15/30 * 5/29

= **0.086**

Part 2. Counting Outcomes (Permutations and Combinations)

Assign two numbers: N and X (X is portion of N).

Calculate in how many ways can you select X people out of N candidates.

Consider two cases:

a) for the same position (order of selection doesn't matter - use Combinations)

b) for different positions (order of selection makes a difference - use Permutations)

**Let n be 4 and x be 3**

**Formula for combinations= n!/[x!(n-x)!]**

= 4!/[3!(4-3)!]

= 4!/3! 1!

= 4*3*2*1/3*2*1

** = 4**

b) for different positions (order of selection makes a difference - use Permutations)

**Formula for Permutations= n!/(n-x)!**

= 4!/(4-3)!

= 4!/1!

= 4*3*2*1

=** 24**