Actually, I was suggesting a sort of probability/proportion table, not a Venn diagram. Truth to tell, I don’t know how I’d do this on a Venn diagram (or a Vinnie diagram).
And yeah, I use a Slide Rule – got a collection of 'em at home, and intend to teach MilliCalm how to use one. But you don’t need it for this problem.
I’ll try a Vinnie diagram, using fractions rather than percentages:
___________________________
| __________ |
| _____/______ \B |
| / | \ | |
| | | | | |
| A | | AnB | | |
| | \______|___/ |
| \____________/ 0.2 |
|___________________________|
Box A contains the people who got Q1 right.
Box B contains the people who got Q2 right.
The probability of being in box A, P(A) is 0.75.
The probability of being in box B, P(B) is 0.55.
Now, since 20% got neither right, 80% must have got Q1, Q2 or both right. So we can say P(AUB)*=0.8.
My Maths book tells me P(AUB) = P(A) + P(B) - P(AnB)**
so P(AnB)= P(A) + P(B) - P(AUB)
= 0.75 + 0.55 - 0.8
= 0.5
So 50% got them both right.
Damn that was hard. A lot of people have got to that answer so I think it’s right.
[sub]*That U is supposed be ‘union’, if you know about that sort of thing. If not it just means the probability of being in box A and/or box B.
**The n is supposed to be the intersection symbol. It just means where the boxes overlap ie people who got both questions right.[/sub]
First, forget percentages, and pretend the class has 100 students.
20 of them answered neither question correctly, so forget about them[sup]1[/sup]. That leaves 80 who answered either or both.
If we add the 75 who answered the first question correctly to the 55 who answered the second question correctly, we get 130, Of course there’s only 80 people, so we subtract 80 from 150 to get the 50 people who answered both.
It’s interesting how this question was cast as a probability question, when probability really isn’t an issue.
[sub]1. As Herbert Kornfeld would say, they be no-skeelz havin’ wackasses who can’t EXECUTE!
I’m a little late, but I’ll post my reasoning anyway
Assume class of 100 people
#1 right = 75 #1 wrong = 25
#2 right = 55 #2 wrong = 45
That gives a total of (45+25) 70 wrong answers on the two questions. However, 20 people answered both wrong. Subtract the (20*2) 40 wrong answers they contributed from the total of 70 to get (70-40) 30 wrong answers. We already accounted for those idiots who missed both, so the 30 wrong answers left represent 30 unique people that each missed one question.
100 people - 20 missed both - 30 missed only one = 50 that missed none.
50 people got both right.
javaman needs a slide rule or something 
Just use algebra…
A = number who got only Q1 right, C = number who got only Q2 right, B = number who got both right (assume 100 students)
A + B + C = 80 (number of people who got a question right)
A + B = 75 (number who got first one right)
B + C = 55 (number who got the second question right)
Solve for B.
Surely probability is an issue, it is just this question can be solved by simple algebra. Funky McDuck’s theory is the best way of solving this kinda problems.
PIE CHART, DAMMIT. You’ll see the problem was probably designed to be solved by a pie chart. It took me about 6 seconds to solve it with one.
I’m curious–how did you solve this problem with a pie chart?
Probablity is not mentioned in the OP, and appealing to probablity theorems is not necessary for solving this. But, it happens that the algebra needed to solve it matches with concepts commonly used in probability, and is an alternate way of approaching it.
You could change it into a probability problem by changing just a few words, and would come up with the same answer.
What kind of pie?
I thought in math you used pi 