Question

X students took an exam on Math and Physics. The probability of passing the Mathematics exam is 70% while the probability of passing the Physics exam is 50%. If no student failed both subjects, and 8 students passed both subjects, how many students took the exam?

Answer 1

between 8 and infinity

Answer 2

......

Answer 3

you do realize it's impossible to have infinite students right?

Answer 4

on our mere mortal world. I dont think I understand.

Answer 5

if there is 70% chance of passing the physics exam? then there is some small chance that say 100,000 people pass the exam? I dont understand how you can work out how many people passed.

Answer 6

i believe a contingency table is used

Answer 7

Sorry, ive not heard of that, i will google

Answer 8

wonderful

Answer 9

Is this discreet math, or a probability course?

Answer 10

...well it's in the probability section....so i would say probability

Answer 11

i just noticed my question doesn't have a header "Probability"

Answer 12

its answer is 8

Answer 13

I mean, if they're independent events, then it's just \((.80)(.70)n = 8\) and you solve for \(n\).

Answer 14

it's not that simple..

Answer 15

my questions never are....

Answer 16

so the 70% and 50% is based on the students that took the exam this time were talking aobut

Answer 17

its not a constant 70% guaranteed

Answer 18

it's constant

Answer 19

so every time this exam happens, 70% of people pass. its always 70% not just 70% this time

Answer 20

You're right, it's unlikely we can assume they're independent. In fact since their complements seem mutually exclusive... they're most likely dependant.
Let \(A\) be the event you pass Math, \(B\) be the event you pass physics.
So no one failed both exams, meaning \[P(A^C\cap B^C)=0 \]We know what \[P(A^C\cup B^C) = P(A^C)+P(B^C)+P(A^C\cap B^C) = 0.30+0.50+0=0.80 \]

Answer 21

...drawing a contingency table is so much easier...

Answer 22

Then why didn't you do it?

Answer 23

he is encouraging debate

Answer 24

cool gys

Answer 25

because i don't know how

Answer 26

oh lol1

Answer 27

that's why i want to see how to do it

Answer 28

i have too much faith in you

Answer 29

there will be time for making OpenStudy users feud.

Answer 30

i still dont udnerstand why there cant have been infinate that passed one and failed the other, if the 70% is constant and doesnt just relate to this test

Answer 31

\[P(A^C\cup B^C) = P((A\cap B)^C) = 1-P(A\cap B)\]

Answer 32

becuse infinity is never accepted in probability and statistics

Answer 33

anybody please solve my qyestion its very intresting

Answer 34

hmm....it seems @wio is on to something....

Answer 35

okay, well what about 99999999 students, surely any amount above 8 is possible?

Answer 36

\[0.80=1-P(A\cap B) \rightarrow P(A \cap B) = 0.20\]

Answer 37

Thus \[0.2 n = 8\]

Answer 38

ahh yes. i remember now

Answer 39

40 people took the test.

Answer 40

40 people took the test, there was (7/10) x (5/10) = 35/100 chance of passing. so 40 x (35/100) should = 8, but it = 14?

Answer 41

@JamesWolf That only works if you assume that passing the tests are independent events. That isn't the case though. If you're good at math, you'll be decent at physics and vise versa.

Answer 42

right, so dont you need to know by how much being good at maths helps physics?

Answer 43

still thinking how to set up a contingency table though

Answer 44

like, what if its "slightly" independant, what if its completely not independat i.e. the same

Answer 45

You mean "slightly" dependent. Things are either independent or they are not. Anyway, look at algebra.

Answer 46

So those people who passed the math exam 7/10 will do better on the physics. but why dont you need to know HOW much better they will do?

Answer 47

Okay, want me to explain what I did...?

Answer 48

yes :)

Answer 49

\[P(A^C\cup B^C) = P(A^C)+P(B^C)-P(A^C\cap B^C) = 0.30+0.50-0=0.80\]
So remember that \(A\) is the event that you pass math, \(B\) is the event you pass physics. \(A^C\) is the event you don't pass physics.

\(P(A)\) would be the probability you pass math. \(P(A^C \cap B^C)\) is the probability you don't fail math and don't fail physics. Since they say no one failed both, we assume this to be zero.

\(P(A^C \cup B^C)\) is the probability you fail math or fail physics. We calculate this to be 80%.

Does this make sense so far?

Answer 50

hold on one sec, reading

Answer 51

P(AC∩BC) is the probability you don't fail math(YOU PASS) and don't fail physics (YOU PASS. Since they say no one failed both, we assume this to be zero. but 8 passed?

Answer 52

P(AC∩BC) = 0?

Answer 53

hahr ight i didnt see the n and u

Answer 54

sorry im not familiar with this notation

Answer 55

It's set notation. \(\cap\) means 'and' while \(\cup\) means 'or' when it comes to set representing events.

Answer 56

so this \[P(A^C \cap B^C)\] is probability you fail A AND fail B

Answer 57

\[-P(A^C \cap B^C)\] is the probability that you pass A AND pass B ?

Answer 58

how should i interpret the - sign?

Answer 59

remember that probability ranges from 0 to 1 (0% to 100%).

Also \(1-P(X) = P(X^C)\)

Answer 60

okay

Answer 61

Basically, the probability that something does happen OR doesn't happen is 100%

Answer 62

Which is why \[P(X)+P(X^C)=1 \rightarrow P(X^C)=1-P(X)\]

Answer 63

So \(-P(X)\) has no meaning... but \(1-P(X)\) does.

Answer 64

okay, im not sure about this bit

\[P(A^C\cup B^C) = P(A^C)+P(B^C)-P(A^C\cap B^C)\]

reading it...
the probability of not A or not b equals the probability of A, plus the probability of B minus the probability of A and B

Answer 65

am i reading that right?

Answer 66

the probability of not A or not b equals the probability of not A, plus the probability of not B minus the probability of not A and not B

Answer 67

sorry that ^

Answer 68

Yes, that is correct. The reason we subtract not A and not B is to get rid of overlap.

Answer 69

so since the equation balances could you divide through by "not" to get

the probability of A or B equals the probability of A plus the probability of B minus the probability of A and B

Answer 70

what do you mean by overlap?

Answer 71

Umm, yeah we could have done that. We didn't know the probability of A and B though.

Answer 72

Overlap as in...
suppose you have a group of twenty people. Have are guys, half are adults, and a quarter are both adult an guys. If you just add probability of guys and probability of adults together, you will not get the probability of being an adult or a guy, because you will count the people who are both guy and adult twice.

Answer 73

To only count them once, you have to subtract the overlap.