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a) Students have two quizzes that they must take in their undergraduate studies unit. The probability of getting a perfect score on the first quiz is 68%, the probability of getting a perfect score on the second quiz is 74%, and the probability of falling short of a perfect score on both quizzes is 19%. What is the probability of scoring a perfect score on both quizzes?

My attempt: Pr(Scoring a perfect score on both quizes) $= 0.68 \cdot 0.74 = 0.5032$

b) Refer to the probabilities described in question a. If a student has achieved a perfect score on the first quiz, what is the conditional probability of getting a perfect score on the second quiz?

My attempt:

$$\frac{\text{Pr(Scoring a perfect score on both quizzes)}}{\text{Pr(Perfect score on second Quiz)}} = 0.68$$

but I feel like there is more to both questions

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  • $\begingroup$ what has been tried ? what is the context ( where did you find the question) ? $\endgroup$ – user451844 Aug 19 '17 at 14:22
  • $\begingroup$ a) Pr(Scoring a perfect score on both quizes) = 0.68 * 0.74 = 0.5032 b) Pr(Scoring a perfect score on both quizes)/Pr(Perfect score on second Quiz) = 0.68 but I feel like there is more to both questions $\endgroup$ – user473207 Aug 19 '17 at 14:29
  • $\begingroup$ if you fall short 19% of the cases, what happens the other 81% of the cases ? $\endgroup$ – user451844 Aug 19 '17 at 14:44
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    $\begingroup$ @user473207 Your calculations assume that the two events (obtainoing a perfect score on quiz 1 and obtaining a perfect score on quiz 2) are independent. BUt just conceptually, that is unlikely: SOmeone who gets a perfect score on quiz 1 is probably more likely to get a perfect score on quiz 2 than someone who does not get a perfect score on quiz 1 ... so you are right to suspect there is more to this! $\endgroup$ – Bram28 Aug 19 '17 at 14:47
  • $\begingroup$ @user473207 Also, the expression 'falling short of a perfect score on both quizzes'' is a bit ambiguous (is it that you don't get a perfect score on quiz 1 and also don't get a perfect score on quiz 2, or is it that it is not true that you do perfect on both quizzes (so you can get a perfect score on one of the two quizzes, just not on both quizzes)). What they undoubtedly mean is the first interpretation. SO, you know that there is a 19% chance of not getting a perfect score on quiz 1 and also not a perfect score on quiz 2. NOw, as Roddy asked: what does the other 81% then represent? $\endgroup$ – Bram28 Aug 19 '17 at 14:49
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Let $A$ be the event that a perfect score is obtained on the first quiz; let $B$ be the event that a perfect score is obtained on the second quiz. We are given \begin{align*} P(A) & = 0.68\\ P(B) & = 0.74\\ P(A^C \cap B^C) & = 0.19 \end{align*} Hence, \begin{align*} P(A^C) & = 1 - P(A) = 0.32\\ P(B^C) & = 1 - P(B) = 0.74 \end{align*} The probability of not obtaining a perfect score on at least one of the quizzes is $$P(A^C \cup B^C) = P(A^C) + P(B^C) - P(A^C \cap B^C) = 0.32 + 0.26 - 0.19 = 0.39$$ Thus, the probability of obtaining a perfect score on both quizzes is $$P(A \cap B) = 1 - P(A^C \cup B^C) = 1 - 0.39 = 0.61$$ so your assumption that the probabilities of obtaining a perfect score on each quiz are independent was false.

The probability that a student who received a perfect mark on the first quiz obtains a perfect mark on the second quiz is $$P(B \mid A) = \frac{P(A \cap B)}{P(A)} = \frac{0.61}{0.68} = \frac{61}{68}$$

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  • $\begingroup$ For the 2nd part of the question isn't a perfect score on the first quiz the given condition. Which means P(B | A) = 61/68 ? $\endgroup$ – user473207 Aug 21 '17 at 3:22
  • $\begingroup$ You are correct. I had reversed the probabilities of getting a perfect score on the first and second quizzes. $\endgroup$ – N. F. Taussig Aug 21 '17 at 7:53

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