Question

A snack-size bag of M&Ms candies is opened. Inside, there are 12 red candies, 12 blue, 7 green, 13 brown, 3 orange, and 10 yellow. Three candies are pulled from the bag in succession, without replacement. What is the probability that the first two candies drawn are brown and the third is brown?

Answer 1

We have to calculate the probability of pulling three candies in a row (without replacement).

We can think of each draw as a different event.

As there is no replacement of the candies, the probability changes with the result of the previous event.

We have 12+12+7+13+3+10 = 57 candies in the bag.

The probability of pulling a brown candy in the first draw is 13/57, as 13 of the candies are brown.

Now, for the second draw, the probability of picking a brown candy is 12/56, as we have one less candy and it is the brown one we picked in the first draw.

Lastly, we have a probability of 11/55 of picking a third brown candy from the bag.

Then, the probabiltiy of 3 brown candies in a row is the product of the probabilities of these three events:

`\begin{gathered} P=P_1*P_2*P_3 \\ P=(13)/(57)*(12)/(56)*(11)/(55) \\ P=(1716)/(175560)=\frac{}{}(143)/(14630)=(13)/(1330) \end{gathered}`

Answer: the probability of drawing 3 brown candies in a row without replacement is 13/1330.

NOTE:

If the last candy is green, then we have to change the third event.

The first two draw will remain the same, but the probabilty of drawing a green candy in the third draw will be 7/55, as 55 candies are left in the bag and 7 of them are green.

We then can calculate the probability of the two first candies being brown and the third being green as:

`\begin{gathered} P=P_(1b)*P_(2b)*P_(3g) \\ P=(13)/(57)*(12)/(56)*(7)/(55) \\ P=(13)/(57)*(3)/(14)*(7)/(55) \\ P=(13)/(19)*(1)/(2)*(1)/(55) \\ P=(13)/(2090) \end{gathered}`

Answer: the probability of the first two draws being a brown candy and the third being a green candy is 13/2090.