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So heres the problem: In the exam there are 25 questions, the student knows the answer to 20. Each exam ticket contains 2 questions. The student randomly takes 1 ticket. What is the probability:

1)That the student knows both answers to questions

2)Knows only 1 answer

3)doesnt know any answers

4)Knows no less than one answer

Heres what I have done:

P(A)= 20/25 (The student knows the answer to the first question)

P(B)=19/24(The student knows the answer to the second question assuming that he also knew the first one)

P(C)=20/24(The student knows the answer to the second question assuming that he Didnt know the first one)
I'm not entirly sure about the B and C, so I calculated for both.

A and B
1)P(A∩B) = P(A) × P(B)=(20/25) * (19/24) = 19/30 (0.633)
2)P(AΔB) = P(A) + P(B) - 2P(A∩B)= (20/25) + (19/24) - 2 * (19/30) = 13/40 (0.325)
3)P(A∪B) =P(A) + P(B) - P(A∩B) = 20/25 + 19/24 - 19/30 = 23/24 (0.958)
P((A∪B)') = 1 - P(A∪B) = 1- 23/24 = 1/24 (0.042)
4) I also dont know how to calculate this.

And the same would be if it were A and C but I'm not sure which one is right

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    $\begingroup$ You seem to have confused your notation about whether you are talking about $Pr(\text{knows second question})$ versus $Pr(\text{knows second question}\mid\text{knows first question})$. I recommend using $B$ as the event that he knows second question. You would have here $Pr(B)=\frac{20}{25}$ and $Pr(B\mid A)=\frac{19}{24}$, don't confuse these with one another. You'd have here $Pr(A\cap B)=Pr(A)\times Pr(B\mid A) = \frac{20}{25}\times\frac{19}{24}$ as you wrote before, but now with correct notation and correctly identifying that this is a conditional probability. $\endgroup$ – JMoravitz Oct 3 '19 at 13:32
  • $\begingroup$ Welcome to MathSE. This tutorial explains how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Oct 3 '19 at 16:53
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What is the probability that the student knows answers to both questions given that the student knows the answers to $20$ of the $25$ questions that may be posed?

We solve the problem in two ways.

Method 1: Let $\Pr(F)$ be the probability that the student knows the answer to the first question; let $\Pr(S)$ be the probability that the student knows th answer to the second question; let $\Pr(S \mid F)$ be the probability that the students know the answers to the second question given that the student knows the answer to the first question. Then the probability that the student knows the answer to both questions is $$\Pr(F \cap S) = \Pr(F)\Pr(S \mid F) = \frac{20}{25} \cdot \frac{19}{24} = \frac{19}{30}$$

Method 2: The student will be able to answer both questions if they are drawn from the $20$ questions the student knows how to answer. Since there are a total of $25$ questions from which the two questions on the ticket are drawn, this has probability $$\Pr(F \cap S) = \frac{\dbinom{20}{2}}{\dbinom{25}{2}} = \frac{190}{300} = \frac{19}{30}$$ of occurring.

What is the probability that the student knows only one answer?

We again solve the problem in two ways.

Method 1: The student either knows the answer to the first question but does not know the answer to the second question, or the student knows the answer to the second question but does not know the answer to the first question. Hence, the probability that the student can answer exactly one of the questions is $$\Pr(F \triangle S) = \Pr(F)\Pr(S' \mid F) + \Pr(F')\Pr(S \mid F') = \frac{20}{25} \cdot \frac{5}{24} + \frac{5}{25} \cdot \frac{20}{24} = \frac{1}{6} + \frac{1}{6} = \frac{1}{3}$$

Method 2: If the student can answer exactly one of the questions, the ticket must contain one of the $20$ questions the student knows and one of the five the student does not know. Hence, the probability that the student can answer exactly one of the questions is $$\Pr(F \triangle S) = \frac{\dbinom{20}{1}\dbinom{5}{1}}{\dbinom{25}{2}} = \frac{100}{300} = \frac{1}{3}$$

What is the probability that the student does not know either answer?

Method 1: We multiply the probability that the student does not know the answer to the first question by the probability that the student does not know the answer to the second question given that the student did not know the answer to the first question. $$\Pr(F' \cap S') = \Pr(F')\Pr(S' \mid F') = \frac{5}{25} \cdot \frac{4}{24} = \frac{1}{30}$$

Method 2: If the student cannot answer either question, both questions on the ticket must have been drawn from the five questions the student did not know. Hence, $$\Pr(F' \cap S') = \frac{\dbinom{5}{2}}{\dbinom{25}{2}} = \frac{10}{300} = \frac{1}{30}$$

What is the probability that the student knows the answer to at least one of the questions?

We can do this in two ways. We can add the probabilities that the student can answer both questions to the probability that the student can answer exactly one of the questions.

$$\Pr(F \cup S) = Pr(F \cap S) + \Pr(F \triangle S) = \frac{19}{30} + \frac{1}{3} = \frac{29}{30}$$

Alternatively, we can subtract the probability that the student can answer neither question from $1$.

$$\Pr(F \cup S) = 1 - \Pr(F' \cap S') = 1 - \frac{1}{30} = \frac{29}{30}$$

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