Question

A snack-size bag of M&Ms candies contains 14 red candies, 10 blue, 5 green, 11 brown, 3 orange, and 12 yellow. If a candy is randomly picked from the bag, compute the following. (a) the odds of getting a green M&M (b) the probability of getting a green M&M

Answer 1

The odds of getting a green M&M are 5:55, and the probability of getting a green M&M is 5/55 or approximately 0.091 or 9.1%.

Step-by-step explanation:

(a) The odds of getting a green M&M can be calculated by dividing the number of green M&Ms by the total number of M&Ms in the bag. In this case, there are 5 green M&Ms out of a total of 55 M&Ms (14+10+5+11+3+12). So the odds of getting a green M&M are 5:55.

(b) The probability of getting a green M&M can be calculated by dividing the number of green M&Ms by the total number of M&Ms in the bag. In this case, there are 5 green M&Ms out of a total of 55 M&Ms. So the probability of getting a green M&M is 5/55 or approximately 0.091 or 9.1%.

Answer 2

The odds of getting a green M&M is
`(1)/(10)` and the probability of getting a green M&M is
`(1)/(11)`

Step-by-step explanation:

Red candies = 14

Blue candies = 10

Green candies = 5

Brown candies = 11

Orange candies = 3

Yellow candies = 12

Total No. of candies = 14+10+5+11+3+12 = 55

(a) the odds of getting a green M&M

Green candies = 5

Total number of candies excluding green or unfavorable = 50

Odds of getting a green M&M =
`\frac{\text{favorable outcome}}{\text{Unfavorable outcomes}}`

=
`(5)/(50)`

=
`(1)/(10)`

(b) the probability of getting a green M&M

Green candies = 5

Total No. of candies = 55

So, the probability of getting a green M&M =
`(5)/(55)`

=
`(1)/(11)`

Hence the odds of getting a green M&M is
`(1)/(10)` and the probability of getting a green M&M is
`(1)/(11)`