Q:

# An urn is filled with marbles of various colors (but each marble only has one color). If the probability of pulling a red marble at random is 1/4 and the probability of pulling a blue marble at random is 1/3, what is the probability that a marble randomly pulled from the urn will be one of those two colors (red or blue)?

Accepted Solution

A:
Answer:$$\frac{7}{12}$$Step-by-step explanation:We are given that The probability of pulling a red marble=$$\frac{1}{4}$$The probability of pulling a blue marble=$$\frac{1}{3}$$We have to find the probability that a marble at randomly pulled from the urn will be one of those two colors ( red or blue)When a marble is of red color then it can not be of blue color .When a marble is of blue color then it can not be of red color.$$P(Red\cap blue)=0$$We know that $$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$A=Red marble B=Blue marble $$P(A)=\frac{1}{4}$$$$P(B)=\frac{1}{3}$$Substitute the values then, we get $$P(Red\cup blue)=\frac{1}{4}+\frac{1}{3}=\frac{3+4}{12}=\frac{7}{12}$$Hence, the probability of pulling red or blue marble from the urn=$$\frac{7}{12}$$