Question

Suppose you have a bag of M&M’s with 20 blue M&M’s, 15 green M&M’s, 35 red M&M’s, and 17 yellow M&M’s. You are very hungry and eat all M&M’s in this bag, but just one at a time (despite being very hungry).

1. a) (4 pts) How many possible orders are there to eat all M&M’s in the bag?

2. b) (4 pts) If you pick M&M’s from the bag in a random order to eat them, what is

   the probability that you first eat all the red M&M’s?

3. c) (2 pts) If you pick M&M’s from the bag in a random order to eat them, what is the probability that the red M&M’s are the ones you eat last?

Answer 1

20 BLUE
15 GREEN
35 RED
17 YELLOW

Total = 87

a)
Possible orders to eat them all (at random)
= ways
= 1.8025827*10⁴⁷ ways

b)
Number of ways to select red M&Ms first
= ⁸⁷C₃₅
= 2.52894603 * 10²⁴

P(selecting red M&M's first) = n(selecting red M&M's first)/n(ways to select all M&M's)
P(selecting red M&M's first) = 2.52894603 * 10²⁴/1.8025827*10⁴⁷

P(selecting red M&M's first) = 1.402957007 * 10^-23

c)In this part we are eating red M&M's last so first 52 M&M's are of different color

Number of ways to select red M&Ms last
= ⁸⁷C₅₂
= 2.52894603 * 10²⁴

P(selecting red M&M's last) = n(selecting red M&M's last)/n(ways to select all M&M's)
P(selecting red M&M's last) = 2.52894603 * 10²⁴/1.8025827*10⁴⁷

P(selecting red M&M's last) = 1.402957007 * 10^-23