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Zielflug [23.3K]

3 years ago

5

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

1 answer:

Nitella [24]3 years ago

4 0

Answer:

The odds of getting a green M&M is $\frac{1}{10}$ and the probability of getting a green M&M is $\frac{1}{11}$

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 = $\frac{\text{favorable outcome}}{\text{Unfavorable outcomes}}$

= $\frac{5}{50}$

= $\frac{1}{10}$

(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 = $\frac{5}{55}$

= $\frac{1}{11}$

Hence the odds of getting a green M&M is $\frac{1}{10}$ and the probability of getting a green M&M is $\frac{1}{11}$

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