Question

Suppose you have a bag of M&Ms with 19 M&Ms. Suppose 4 of them are red, 3 are green, and 12 are yellow.

(a) If one M&M is chosen at random from the bag, find the probability that it is yellow.

(b) If one M&M is chosen at random from the bag, and eaten, and then a second M&M is chosen at random from the bag, find the probability that they are both red.

Answer 1

Solution:

Given: a bag of M&Ms have 9 M&Ms.

4 of them are red,

3 are green, and

12 are yellow.

Part a) Find:

P( it is yellow) =.............?

P( it is yellow) = Number of Yellow M&Ms / Total M&Ms

P( it is yellow) = 12 / 19

P( it is yellow) = 0.6316

Part b) If one M&M is chosen at random from the bag, and eaten, and then a second M&M is chosen at random from the bag. That means selection is without replacement. That is first selected M&M will not be returned in bag.

Find:

P(  they are both red) =...........?

P(  they are both red) = P(First is red ) X P( Second is red)

For first M&M , we have 4 red M&M and total 19 M&M

For second M&M , we have 3 red M&M and total 18 M&M.

Thus

P(they are both red) = ( 4/19) X ( 3/18)

P(they are both red) = 0.2105263 X 0.1666667

P(they are both red) = 0.0350877

P(they are both red) = 0.0351

Note: ( Round final answer to specified number of decimal places)