Is this a trick question??

[kay43]

Question: Two tests were given to a class of 50 students. As an outcome, 25% of the students cleared both the tests and 42% of the students cleared only the first test. Compute the percent of students who cleared the first test who also cleared the second test.

I began the problem like this:
1st test: 42% of 50 students would mean 21/50 students passed the first test
but 25% of students passed both tests, so 25% of 50 students would be 12.5, making the total students who passed both 33.5 students, giving me 67% passed both tests.

But then in the problem it says that 25% of students cleared both tests, so would it just be 25%?

 

[Romsek]

They are asking (I think) for
\(\displaystyle P[\text{student passes 2nd test}|\text{student passed 1st test}]\)

 

[Dr.Peterson]

Compute the percent of students who cleared the first test who also cleared the second test.

It is easy to misread, but reading carefully is part of what is being tested.

Restate it a little, and it is easier to see that it is a conditional probability:

Of students who cleared the first test, compute the percent who also cleared the second test.​

That is,

If a student cleared the first test, what is the probability that they cleared the second test.​

 

[kay43]

It is easy to misread, but reading carefully is part of what is being tested.

Restate it a little, and it is easier to see that it is a conditional probability:

Of students who cleared the first test, compute the percent who also cleared the second test.​

That is,

If a student cleared the first test, what is the probability that they cleared the second test.​

So, in essence, a conditional formula is p(ab) = p(a and b)/ p(b) I would take the 42% who passed the first test, then take 25% of that number? so 42/25 = 17% (rounded)

 

[Dr.Peterson]

Check your work carefully. First, 42/25 is not 17% (in fact, it's greater than 1); second, which number is P(A and B), and which is P(B)?