Conditional Probability Question Help (1 Viewer)

[KevDa5'9"]

Found this question but didn't have answers. In the final exam, 55% of students failed chemistry, 25% failed physics and 16% failed both. What is the probability that a student passed physics given that they passed chemistry?
Thank you.

 

[Trebla]

First question I would ask is what is meant by “16% failed both”? Is the population of students exactly the same in both Chemistry and Physics? If there are different students (i.e. people only doing one and not the other) it is not clear what that statement means.

 

[CM_Tutor]

I was contemplating the same question as Trebla. If the question means "amongst a group of students all of whom take both physics and chemistry", then the question is straight-forward:

• the population consists of four non-overlapping groups that can be represented on a Venn Diagram
  • students who passed both chemistry and physics, call them the "P" group
  • students who passed chemistry but failed physics, call them the "failP" group
  • students who failed chemistry but passed physics, call them the "failC" group
  • student who failed both chemistry and physics, call them the "F" group
• You could take the question to mean that
  • failC = 55%
  • failP = 25%
  • F = 16%
  • Everyone must belong to one of these four non-overlapping groups, and so P = (100 - 55 - 25 - 16) % = 4%
  • but I don't think that is what it is meant to mean, as it means interpreting "failed chemistry" as meaning "failed chemistry but passed physics."
• Much more likely, in my view, is that:
  • 55% failed chemistry, coming from the F and failC groups, so F + failC = 55 . . . . . (eqn1)
  • 25% failed physics, coming from the F and failP groups, so F + failP = 25 . . . . . (eqn2)
  • 16% failed both, this being the F group, so F = 16 . . . . . (eqn3)
  • Put (eqn3) into (eqn1) to get that failC = 55 - 16 = 39
  • Put (eqn3) into (eqn2) to get that failP = 25 - 16 = 9
  • Everyone must belong to one of these four non-overlapping groups, and so P = 100 - 16 - 39 - 9 = 36

The conditional probability formula is that



which here means that we seek P(passes physics given passed chemistry), which depends on

• P(passes both chemistry and physics) = P(P) = 36%, and
• P(passes chemistry) = P(P or failP) = 45%

If, however, the group is not entirely made of students doing both subjects, then it is unclear how to approach the question, as there would be students who passed chemistry, students who failed it, and students who do not take it and so neither passed nor failed the exam.

I think the interpretation above is what is meant, but the question is poorly worded.