What Is the Probability a Student Doesn't Get an A on Either Exam?

[Potatochip911]

Homework Statement

Probability a student gets an A on the midterm is 13.96%
Probability a student gets an A on the final is 13.12%
Probability a student gets an A on both the final and the midterm is 5.7%
What is the probability that a student doesn't get an A on the final or the midterm?

Homework Equations

3. The Attempt at a Solution [/B]
I would think that the probability of not getting an A on the final or the midterm would be $$P(a_{M}^c)+P(a_{F}^c)-P(a_{F}^c \space \space \mbox{&}\space \space a_{M}^c)$$ Unfortunately I'm not sure what the probability of not getting an A on both the midterm and the final is. Also I'm confused as to why the probability that a student gets an A on the final and the midterm is 5.7%, shouldn't that be equal to Probability of an A on the midterm*probability of an A on the final?

 

[paisiello2]

I think they are trying to say that they are NOT two independent events (maybe there are other tests to study for or something).

I think you are on the right track. Maybe if you plug in some numbers you can figure out what P(acF & acM) is?

 

[ehild]

You need the probability of the event (NOT getting A on midterm) OR (NOT getting A on final ), that is a_M^c OR a_F^c. I is equivalent to the event... (Apply de Morgan's law.)

Getting a mark is not accidental. If the student studies hard it is high probability that he/she gets good marks on both exams.

 

[Potatochip911]

I think they are trying to say that they are NOT two independent events (maybe there are other tests to study for or something).

I think you are on the right track. Maybe if you plug in some numbers you can figure out what P(acF & acM) is?

You need the probability of the event (NOT getting A on midterm) OR (NOT getting A on final ), that is a_M^c OR a_F^c. I is equivalent to the event... (Apply de Morgan's law.)

Getting a mark is not accidental. If the student studies hard it is high probability that he/she gets good marks on both exams.

I tried taking $$P(A \space \space \mbox{or} \space \space B)=P(A)+P(B)-P(A \space \space \mbox{&} \space \space B)$$ and changing it to $$P(A^{c} \space \space \mbox{or} \space \space B^{c})=P(A^c)+P(B^c)-P(A^c \space \space \mbox{&} \space \space B^c)$$ but after I plugged in the values I still got the wrong answer.

 

[ehild]

Think the event first. What is the complementer of (A and B)?

 

[Potatochip911]

Think the event first. What is the complementer of (A and B)?

$A^{c}\mbox{&}B, A\mbox{&}B^c A^c\mbox{&}B^c$?

 

[ehild]

What do you mean?
Read https://en.wikipedia.org/wiki/De_Morgan's_laws

 

[Potatochip911]

What do you mean?
Read https://en.wikipedia.org/wiki/De_Morgan's_laws

Okay I managed to get the answer by doing 1-P(A & B)=0.943 since P(A & B) is the only time both scores will be A's so the complements of that event will be when the student doesn't get an A on the midterm and the final.

 

[ehild]

Okay I managed to get the answer by doing 1-P(A & B)=0.943 since P(A & B) is the only time both scores will be A's so the complements of that event will be when the student doesn't get an A on the midterm and the final.

It is OR instead of AND.

 

[Potatochip911]

It is OR instead of AND.

Whoops, I see why now.

 

[Ray Vickson]

I tried taking $$P(A \space \space \mbox{or} \space \space B)=P(A)+P(B)-P(A \space \space \mbox{&} \space \space B)$$ and changing it to $$P(A^{c} \space \space \mbox{or} \space \space B^{c})=P(A^c)+P(B^c)-P(A^c \space \space \mbox{&} \space \space B^c)$$ but after I plugged in the values I still got the wrong answer.

One could argue that the question is ambiguous. To me, it asks for the probability of not getting an A at all, which is $(A\, \text{or} \,B)^c$, with probability $1 - P(A \, \text{or} \,B)$. However, others have suggested it is asking for the probability of $(A^c\, \text{or} \,B^c)$. Which interpretation are you assuming?

 

[ehild]

The question was 'What is the probability that a student doesn't get an A on the final or the midterm?'. It does not mean not getting an A at all.
.

 

[Ray Vickson]

The question was 'What is the probability that a student doesn't get an A on the final or the midterm?'. It does not mean not getting an A at all.
.

If M = get A on mid-term and F = get A on final, then "doesn't get an A on the final or midterm" looks like $(M \cup F)^c$.

It all depends on whether you distribute things like this: "doesn't get an A on (the final or midterm)", or whether you expand out the 'doesn't' to "doesn't get an A on the final or doesn't get an A on the midterm". It may be 100% clear to you , but it is not to me.

 

[insightful]

The question was 'What is the probability that a student doesn't get an A on the final or the midterm?'. It does not mean not getting an A at all.
.

If you ask a student, "Did you get an A on the midterm or the final?" and he says, "No," how could he get an A on either?

 

[haruspex]

You need the probability of the event (NOT getting A on midterm) OR (NOT getting A on final ), that is a_M^c OR a_F^c.

English can be poor at expressing logical relationships.
As a native English speaker, I read the question

What is the probability that a student doesn't get an A on the final or the midterm?

as meaning the student is to get an A on neither the final nor the mid-term. I agree it could be worded better, but I believe that is the intent.
If the intent had been the event that the student doesn't get a double A, I would expect the wording to be more like "What is the probability that a student doesn't get an A on the final or doesn't get an A on the midterm?"

P(a^c_M)+P(a^c_F)−P(a^c_F & a^c_M)​

With my reading, P(a^c_F & a^c_M) is what you are to determine.

 

[ehild]

As a native English speaker, I read the question

as meaning the student is to get an A on neither the final nor the mid-term.

With my reading, P(a^c_F & a^c_M) is what you are to determine.

You are certainly right. I am not an English speaker. I tried to interpret "or" in problem on probability as it was logical "OR" instead of "AND".
We do not know what was the intent of the problem writer. We do not know if the original text was in English. But the OP was happy with his result in Post#8 P = 1-P(A & B).