Question

A bag of Mand Ms contains 5 yellow, 11 red, 4 green, 12 blue and 7 brown candies.What is the probability that a red or brown candy is pulled from the bag?A. 18/39B. 18/78C. 9/20D. 77/1521

Answer 1

In this question, we need to find the probability of pulling a red or brown candy from a bag.

We know that the bag contains the next amount of candies:

• 5 yellow candies

,

• 11 red candies

,

• 4 green candies

,

• 12 blue candies

,

• 7 brown candies

Therefore, in total, we have 39 candies.

Then the probability of pulling a red candy is:

`P(\text{red)}=(11)/(39)`

The probability of pulling a brown candy is:

`P(\text{brown)}=(7)/(39)`

Now, we know that the general formula for the probability of two events is given by:

`P(A\cup B)=P(A)+P(B)-P(A\cap B)`

However, in this case, we do not have any probability that both events happen at the same time - in other words, they are mutually exclusive events. Therefore, we have:

`\begin{gathered} P(R\cup B)=P(R)+P(B)-P(R\cap B) \\ P(R\cup B)=(11)/(39)+(7)/(39)=(18)/(39) \\ P(R\cup B)=(18)/(39) \end{gathered}`

Therefore, in summary, the probability that a red or brown candy is pulled from the bag is 18/39 (option A.)