Any help on this problem is greatly appreciated! I'm completely stuck

School board officials are debating whether to require all high school seniors to take a proficiency exam before graduating. A student passing all three parts (mathematics, language skills, and general knowledge) would be awarded a diploma; otherwise, he or she would receive only a certificate of attendance. A practice test given to this year’s ninety-five hundred seniors resulted in the following numbers of failures: $$\begin{array}{|l|l|} \hline \\ \textbf{Subject Area} & \textbf{Number Failing} \\\hline \\ \text{Mathematics} & 3325 \\\hline \\ \text{Language Skills} & 1900 \\\hline \\ \text{General Knowledge} & 1425\\ \hline \end{array}$$

If “Student fails mathematics,” “Student fails language skills,” and “Student fails general knowledge” are independent events, what proportion of next year’s seniors can be expected to fail to qualify for a diploma?

  • $\begingroup$ See the answers to this question and apply it to the event that a student fails at least one test and so fails to qualify for a diploma. $\endgroup$ – Dilip Sarwate May 26 '15 at 2:50
  • $\begingroup$ Inclusion/Exclusion Principle: P(A or B or C) = P(A) + P(B) + P(C) - P(A and B) - P(A and C) - P(B and C) + P(A and B and C) $\endgroup$ – Frank Newman May 26 '15 at 2:51

We can use the given data to estimate the probability that a student passes in each of the three categories.

For instance, $p_1:=\frac{9500-3325}{9500}$ is the proportion of students who passed the mathematics exam. Similarly you can compute $p_2,p_3$ for the language and general exams.

Since the events of passing in each exam are assumed to be independent, the probability of a student passing can be estimated by the product $p_1p_2p_3$, so the estimate for failing is estimated by $1-p_1p_2p_3$.


Not 100% sure if this is right but this gave me gave me 0.568 and the answer was 0.56

let A = people failed math, B = people language skills, C = people general knowledge since we know they are independent we can use...

p(A ∩ B ∩ C)=p(A)p(B)p(C)

p(A ∩ B)=p(A)p(B)

p(A ∩ C)=p(A)p(C)

p(B ∩ C)=p(C)p(B)

theses are all % so to find out how p(A ∩ B)=p(A)p(B) it would be..

(failed math/total fail)(failed language skill/ total fail)(total fail)

[(3325/6550)*(1900/6550)] =

the % of people who failed both math and language skills.

after figuring out all of the formulas above, you can use that information to find all the people who failed at least one of the 3 subjects p(A u B u C).

with p(A) = (fail math)/(total fail) and same idea for p(B) and p(C)

P(A u B u C) = p(A) + p(B) + p(C) + p(A ∩ B ∩ C) - p(A ∩ B) - p(A ∩ C) - p(C ∩ B)

  • $\begingroup$ Why would you post an answer you aren't sure is right? $\endgroup$ – The Count Jan 29 '17 at 3:35

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