# A probability question literally about homework

#### megatkhairoll

##### New member
3. The homework diaries and completed homework of two Students A and B, are examined.

There is a probability of 0.4 that Student A does not make a note in her diary homework set. She always does the homework if it is written in her diary, but never does the homework if it is not written in.

There is a probability of 0.8 that Student B writes the homework given

in her diary. When she does this, she will do the homework 90% of the time. If she has nothing written in her diary then she checks with a friend,

who knows what the homework is 50% of the time; Student B always does the homework if she is told what it is by her friend.

C) If one piece of homework was given to a student but wasn’t done, find the probability that it was given to Student A.

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#### Subhotosh Khan

##### Super Moderator
Staff member
3. The homework diaries and completed homework of two Students A and B, are examined.

There is a probability of 0.4 that Student A does not make a note in her diary homework set. She always does the homework if it is written in her diary, but never does the homework if it is not written in.

There is a probability of 0.8 that Student B writes the homework given

in her diary. When she does this, she will do the homework 90% of the time. If she has nothing written in her diary then she checks with a friend,

who knows what the homework is 50% of the time; Student B always does the homework if she is told what it is by her friend.

C) If one piece of homework was given to a student but wasn’t done, find the probability that it was given to Student A.

Please share your work with us ...even if you know it is wrong.

If you are stuck at the beginning tell us and we'll start with the definitions.

http://www.freemathhelp.com/forum/announcement.php?f=33

#### megatkhairoll

##### New member

Please share your work with us ...even if you know it is wrong.

If you are stuck at the beginning tell us and we'll start with the definitions.

http://www.freemathhelp.com/forum/announcement.php?f=33
Pardon my mistake there,

Well, what I've did was actually making a ratio of ( probability of student A not doing the work) to ( the probability of students A and B not doing the work) because i believe i could find the chance of student A not doing the work.

Values being

0.4 / ( 0.4 + 0.18)

Giving out 0.69.
However I missed and the correct answer is 0.752.

I couldn't upload my working due to a network errror

Last edited by a moderator:

#### stapel

##### Super Moderator
Staff member
3. The homework diaries and completed homework of two Students A and B, are examined.

There is a probability of 0.4 that Student A does not make a note in her diary homework set. She always does the homework if it is written in her diary, but never does the homework if it is not written in.

There is a probability of 0.8 that Student B writes the homework given in her diary. When she does this, she will do the homework 90% of the time. If she has nothing written in her diary then she checks with a friend, who knows what the homework is 50% of the time; Student B always does the homework if she is told what it is by her friend.

C) If one piece of homework was given to a student but wasn’t done, find the probability that it was given to Student A.
Well, what I've [done] was actually making a ratio of (probability of student A not doing the work) to (the probability of students A and B not doing the work) because i believe i could find the chance of student A not doing the work.

Values being 0.4 / ( 0.4 + 0.18)

Giving out 0.69.

However I missed and the correct answer is 0.752.

I couldn't upload my working due to a network errror
That's okay; you can type it out, just like you typed out the results, as displayed above.

So please reply with your work and reasoning for the results you've posted, so that we can "see" what's going on. Thank you!

#### j-astron

##### Junior Member
Pardon my mistake there,

Well, what I've did was actually making a ratio of ( probability of student A not doing the work) to ( the probability of students A and B not doing the work) because i believe i could find the chance of student A not doing the work.

Values being

0.4 / ( 0.4 + 0.18)

Giving out 0.69.
However I missed and the correct answer is 0.752.

I couldn't upload my working due to a network errror
It sounds like, from your original post, that you only want help with part C? Because it also seems like that's what you are trying to figure out above.

Did you get the correct answers for parts A and B already?

For part C, I think where you are going wrong is that the probability of student A not doing the work is something different from what is being asked. What's being asked is, given that the homework being examined/checked was not completed, what's the probability that it was the one assigned to student A? In other words, the question has been flipped around. Instead of, given you assign the homework to student A, what's the probability it will get done, you're being asked the reverse: given the assigned homework wasn't done, what's the probability that you assigned it to student A?

The keyword above is given. You're being asked to evaluate the probability under a certain given condition. This is a conditional probability. Do you know what these are? The conditional probability you want to compute is

Probability it is student A given that the work was not done i.e. P(student A | not done)

The probability you already know how to compute is

P(not done | student A)

So you have to figure out how to flip around the conditional probability. Can you think of any equations or theorems you've been given in class that would tell you how to flip around a conditional probability? ;-)