Q If 3 books are picked at random from a shelf containing 5 novels, 3 books of poems, and a dictionary, what is the probability that
(a) The dictionary is selected?
(b) 2 novels and 1 book of poems are selected?

Do you have any ideas how to approach these? I showed you the card problem, gratis. Please show me that you have a clue. These are worked the same way. For the first part, you are choosing 3 books from 9.
The dictionary must be among them, but the other two can be any of the other 5 novels or 3 poems. How would you set that up?

For part b, you are selectiong 2 from 5 novels and 1 from 3 poems. You are choosing 3 altogether from 9.

Do you think you could explain how to get this?

The probability of choosing 2 novels and 1 poem would be:

\frac{C(5,2)C(3,1)C(1,0)}{C(9,3)}=\frac{5}{14}

That is because we are choosing 2 out of 5 novels, 1 out of 3 poems, 0 out of 1 dictionary and 3 out of 9 in all.

Notice how the numbers in the numerator add up to the numbers in the denominator

You can also do it this way:

3(5/9)(4/8)(3/7)=5/14... see why?

I think I do. I read your other example of this, so is the top always going to be the break down of what you can get?

and also, did you do it for the probability of 2 novels and 1 poem? Because I meant for part a. I'm so sorry.

Choose the number from each you need divided by the total chosen.

For the dictionary, wouldn't it just be 1/3? There is only 1 dictionary out of 9 books.

that's what I thought the answer was at first, but it seemed to be too easy. I got confused and at first I was also thinking since you're picking 3 books, 1 would be a dictionary, so it is a 1/3 probability

Yep, that's it.

thank you. I'm sorry for all that lol. Probability is the one thing in math that I can never understand

Actually if I could bother you one more time, could you do the breakdown of the answer? Like you did previously for part b. it would help me a lot