Question

A bag of M and Ms contains 5 yellow, 11 red, 4 green, 12 blue and 7 brown candies.What is the probability, to the nearest percent, that Julie pulls a green M and M fromthe bag, eats it, then pulls a blue candy from the bag?A. 48%B. 16%C. 3%D. 1%

Answer 1

In this question, we need to find the probability of an event in which the final result depends on one of the events to happen. It is a conditional probability.

We have a bag with the following amount of different candies:

• 5 yellows candies

,

• 11 red candies

,

• 4 green candies

,

• 12 blue candies

,

• 7 brown candies.

If we add all of these candies, we have a total of 39 candies.

Now, we need to find the probability of pulling a green candy from the bag is:

`P(green)=(4)/(39)_{}`

Since Julie ate one of the candies, there will be 38 candies in the bag after that event. Now the probability of pulling a blue candy is:

`P(\text{blue)}=(12)/(39)`

Hence, the probability, in this case, is the product of both previous probabilities as follows:

`\begin{gathered} P=(4)/(39)\cdot(12)/(39)_{}\approx0.0323886639676\approx3.24\% \\ \end{gathered}`

If we round the probability to the nearest percent, we have that the probability will be, approximately, 3%.

In summary, we have that the probability, to the nearest percent, that Julie pulls a green M and M from the bag, eats it, then pulls a blue candy from the bag is, approximately, 3% (option C.)