For my stats summer work, we were givin data to evaluate and perform problems with. They were the grades of a stats test between two class. A first period and last period class. I've gotten up to the conditional probability section. I've looked it up but I am not sure if I am doing it right. One example is the conditional of the first period class that got an A on the test. The data given is that one person in the first class got an A and 3 people got an A in the last class. There are 21 students in the first and 30 in the last, making the total 51. Now I tried, and did them all. For the A's, I got 2/51 for the first class and 2/17 for the last class. I did this for all the grades and added up the fractions to 2. Is this correct?
2026-04-13 05:53:43.1776059623
Conditional distribution
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1
Your question is very unclear - what conditional probability are you after?
Also, if all the probabilities ever add up to anything other than 1 then you are, beyond a shadow of a doubt, wrong. If any of the probabilities are negative or greater than one you are wrong. If any subset of the non-zero probabilities is greater than 1 then you are wrong.
To demonstrate, define the events:
$$F\text{ - student is in the first class}$$ $$L\text{ - student is in the last class}$$ $$A\text{ - student got an A}$$
From the data stated:
$$P(F) = \frac{21}{51}$$ $$P(L) = \frac{30}{51}$$ $$P(A) = \frac{4}{51}$$ $$P(A|F) = \frac{1}{21}$$ $$P(A|L) = \frac{3}{30}$$
A typical question would be "Given that this student got an A, what is the probability that they are in the first class"
$$\begin{align} P(F|A)&=\frac{P(A|F)P(F)}{P(A)}\\ &=\frac{\frac{1}{21}\frac{21}{51}}{\frac{4}{51}}\\ &=\frac{1}{4}\\ \end{align}$$