Question

Back to your Grandma's house for Christmas, with the bowl of holiday-themed red and green M&Ms (100 of each color).

What's the probability that the first 3 you eat will all be the same color?

Group of answer choices

(100/200 x 99/199 x 98/198) + (100/200 x 99/199 x 98/198)

1 - (100/200 x 99/199 x 98/198)

100/200 x 99/199 x 98/198

100/200 + 99/199 + 98/198

(100/200 x 100/200 x 100/200) + (100/200 x 100/200 x 100/200)

Answer 1

There are a total of 200 M&M's (100 red and 100 green)

There are two cases -

Case I: The first 3 M&M's that one eats is of red color

Probability that one eats 3 M&M's of red color = (100C1/200C1) x (99C1/199C1) x (98C1/198C1)

= (100/200) x (99/199) x (98/198)

(Since, for the first pick 100 choices are available out of 200, for the second pick 99 choices are available out of 199 because one was eaten earlier, and similarly it follows for the third pick)

Case II: The first 3 M&M's that one eats is of green color

Probability that one eats 3 M&M's of green color = (100C1/200C1) x (99C1/199C1) x (98C1/198C1)

= (100/200) x (99/199) x (98/198)

(Since, for the first pick 100 choices are available out of 200, for the second pick 99 choices are available out of 199 because one was eaten earlier, and similarly it follows for the third pick)

Total probability that the first 3 one eats

will be of the same color = (100/200) x (99/199) x (98/198) + (100/200) x (99/199) x (98/198)

So, the first option is the appropriate option

and the probability equals 0.2462 (rounded up to 4 decimal places)