MHB Help with Math Homework: Conditional Probability - 19/30

[navi]

Hey! I need help with my Math homework :( The question is the following...

There are 5 history courses of interest to Howard, including 3 in the afternoon, and there are 6 psychology courses, including 4 in the afternoon. Howard picks a course by selecting a dept at random, then selecting a course at random. Find the pr that the course he selects is in the afternoon.

The answer is 19/30, but I get 13/30 by doing this:

For the history dept, the probability of an afternoon class is 1/2 x 3/5, or 3/10, and for the psych dept, the pr of an afternoon class is 1/2 x 4/6, or 1/3. I then add 1/3 and 3/10 and I get 13/30, but that is not the answer and I have no idea what I could be doing wrong...

Please help :(

 

[MarkFL]

Hello, and welcome to MHB! (Wave)

I would let event A be that Howard selected an afternoon course from the history department, and event B be that Howard selected an afternoon course from the psychology department.

$$P(A)=\frac{1}{2}\cdot\frac{3}{5}=\frac{3}{10}$$

$$P(B)=\frac{1}{2}\cdot\frac{4}{6}=\frac{1}{3}$$

And so:

$$P(A\text{ OR }B)=P(A)+P(B)=\frac{3}{10}+\frac{1}{3}=\frac{3\cdot3+10\cdot1}{30}=\frac{19}{30}$$

You just made an error in adding the fractions...your analysis of the problem was good. :)

 

[navi]

Hello, and welcome to MHB! (Wave)

I would let event A be that Howard selected an afternoon course from the history department, and event B be that Howard selected an afternoon course from the psychology department.

$$P(A)=\frac{1}{2}\cdot\frac{3}{5}=\frac{3}{10}$$

$$P(B)=\frac{1}{2}\cdot\frac{4}{6}=\frac{1}{3}$$

And so:

$$P(A\text{ OR }B)=P(A)+P(B)=\frac{3}{10}+\frac{1}{3}=\frac{3\cdot3+10\cdot1}{30}=\frac{19}{30}$$

You just made an error in adding the fractions...your analysis of the problem was good. :)

lol I see, thanks!