I am having trouble trying to solve this problem. Anything helps!
Question: When doing a lab, the professor instructs students to get some rocks from a bucket and study them under a microscope. He says there are 90 sedimentary rocks (30 each of Chalk, Limestone, and Shale) and 180 Igneous rocks (60 each of Obsidian, Granite, and Diorite).
You draw 3 rocks at random from the bucket. What is the probability that you got exactly 2 sedimentary rocks, given that you know you got at least one sedimentary rock?
How many sedimentary rocks would you expect to receive in your 3-rock draw?
Your friends tell you that Diorite, Obsidian, and Limestone have interesting and unique features when placed under a microscope. You now draw 18 rocks at random to check if your friends are right. What is the expected number of the interesting rocks in your 18-rock draw?
Attempted answer:
- I went into this using Bayes' Theorem.
$P(2S, 1I | 1S)$ = $\frac{P(1S | 2S, 1I)*P(2S, 1I)}{P(1S)}$
I know that P(1S) = 33.3% and P(2S, 1I) = $\frac{90}{270}$ * $\frac{89}{269}$ * $\frac{180}{268}$ ≈ 7.4%.
The issue that I have is P(1S | 2S, 1I). Is the answer to this 100% because you are always going to get 1 sedimentary rock given that you get 2 sedimentary rocks and 1 Igneous rock? Am I doing this right?
- This is an expected value question
E(1st draw) = $\frac{90}{270}$ * 1 = $\frac{1}{3}$
E(2nd draw) = $\frac{89}{269}$ * 1 = $\frac{89}{269}$
E(3rd draw) = $\frac{88}{268}$ * 1 = $\frac{88}{268}$
ΣE(draws) = 0.993
I feel like my numerator for the probabilities are wrong but I don't know how I should change that. How would I determine the expected values if the probabilities change?
- This is also an expected value question.
This question continues on the same concept as part 2, as the question uses the same fundamentals of part 2 where the probabilities would change given your draw. Would I approach in the same method or would I use a different approach?
No. The condition is that you have at least one sedimentary rock.
Let $S$ be the count for sedimentary rocks among the three you pick. This will have a hypergeometric distribution.
The probability for drawing exactly $2$ from $90$ sedimentary rocks and $1$ from $180$ igneous rocks when drawing $3$ from $270$ rocks is $$\mathsf P(S=2)=\left.\binom{90}2\binom{180}1\middle/\binom{270}3\right.$$
The probability for drawing at least $1$ from $90$ sedimentary rocks is one minus the probablity for the complementary event. $$\mathsf P(S\geq 1)=1-\mathsf P(S=0)$$
Put it together to find$$\mathsf P(S=2\mid S\geq 1) =\dfrac{\mathsf P(S=2)}{\mathsf P(S\geq 1)}$$
The probabilities do not change. For example: take a standard deck of playing cards, select a card at random, now, what is the probability that it is an Ace?
Similarly, the probability that any particular rock among your three is sedimentary will be :
$$\mathsf E(S_1)=\mathsf E(S_2)=\mathsf E(S_3)=\dfrac{90}{270}$$
The third question is similar.