While taking the pre-test for the GMAT I ran across a problem that must have a simple solution, but despite educational studies at the Calculus II level, I can’t figure it out. I come humbly to the masses for help.
75 percent of the class got the first question right. 55 percent of the class got the second question right. 20 percent got neither question right. What percent got both questions right?
How did you get the answer? What’s the answer?
Pick a random student in the class. There is a 75% probability he got the first question right. There is a 55% probability he got the second question right. Stick both probabilities together, and you get a 41.25% probability that any given person got both questions right.
There is a 63% probability that I got this question wrong.
Out of a class of 100:
20 got both wrong.
75 got the first right.
55 got the second right.
20 got the first right but the second wrong.
5 got the first wrong but the second right.
So 20 got both wrong and 25 got only one right, leaving 55 to get both right.
That doesn’t sound right.
Lessee,
If 75% got the first right, 20% got both wrong then
100-75-20= 5% got the first wrong and second right
Since 55% got the second right, with 5% getting the first wrong then
50% got them both right
I must respectfully disagree with friedo. If you use his logic the other way around you don’t get 20% getting both questions wrong.
View it as a diagram with four boxes – Column 1 is Question #1 Right, column #2 is question #1 wrong. The two rows are Question #2 Right and Question #2 Wrong. Label the boxes A, B, C, and D. (A = both questions right, B = #1 Wrong and #2 right, C = #1 Right and #2 Wrong, D = both wrong)
View the probabilities as fractions rather than as percentages. A + B + C + D = 1
They tell you that D = 0.20
They also tell you that A + B = 0.55 and A + C = 0.75
From the above you can use algebra to work out that A = 0.50, B = 0.05 (easy question, I guess), and C = 0.25. The probability that you get both questions right is 0.50, or 50%
We know that 80% got one or both questions correct. Since 75% got the first question correct, that leaves 5% that got the first question wrong, but answered the second correctly. Similarly, 25% (55% - 80%) got the second correct and missed the first. We now have four disjoint sets: 20% who missed both, 5% who missed the second and got the first, 25% who got the second and missed the first, and the remainder (100% - 20% - 5% - 25%) 50% who got both correct. It helps to draw a Vinn diagram.
I shouldn’t have to ask this, but what’s the flaw in friedo’s reasoning?
friedo is assuming that the 75% probability that a student got the first question is independent of the 55% probability that a student got the second question correct.
So, how many of us got the first question correct and what is the second question?
Freido may be mucking up the works with probability. My guess would be that a person who got the first one right would be more likely to get the second right, which would bring his answer back in line.
I found some of the other ones confusing, maybe it will help to outline the reasoning that leads me to the 50% answer:
IF: 75 got #1 right
THEN: 25 got #1 wrong
IF: 55 got #2 right
THEN: 45 got #2 wrong
IF 20 got both wrong THEN:
5 got only #1 wrong (25-20)
and 25 got only #2 wrong (45-20)
SO:
20 (both wrong)
plus 25 (only #1 wrong)
plus 5 (only #2 wrong)
equals 50.
The remaining 50 must have gotten both right.
[To the slow go the nits]That’s Venn diagram. And yeah, independence is a really strong assumption, tempting as it may be to multiply probabilities to get intersections.
Dr. Matrix, that echo you hear is me…me…me…
2 flaws:
He answered a question that wasn’t asked. Specifically, he focused on the probability of a given student passed both and used it to extrapolate for the group as a whole. Even if there were no additional information available, this would be an incorrect answer, because it gives only a likely answer, while the question asked for an exact and definite number
He ignored the additional information that 20% got both wrong. This gave the ability to determine the exact answer, as noted by several posters.
The question did not ask “what are the chances that X number of people got both questions right”, but simply “how many got both questions right”, suggesting that it is possible to calculate an actual number, not just generate probabilities. Can probability be used to yield a factual answer?
I flipped a coin. Sure, there is a 50% probability that it landed heads, but the OP is asking what side did the coin land on?
By guessing that, aren’t you working on probability?
I say probably.
No,No, it used to be Vinn, after Vinnie Descarte, notorious gambler, brother of Rene, and cardcutter. He came up with this while playing craps. They changed it to Venn to to obscure the true source.
The standard way to approach this would be to use the Principle of Inclusion-Exclusion, a simple version of which can be found here.
With that knowledge, it’s simply 75 + 55 - 80 = 50%.
This’d be the second question then. In this case, there’s only one way that the probabilities sum to 1, so there’s no guessing involved. The probabilities are known proportions, not forward-looking estimates.
As Cal showed,it can worked out through a Venn diagram since it’s based on set theory and not probability.(of course probability also uses set theory but not the other way round)
Actually Cal just mentioned diagrams. picmr was the first to mention Venn diagrams. It would have been me if I could spell worth a damn. I was the first to mention Vinn diagrams, though.
Not to nitpick, but a Venn is overkill. Just draw a pie and see what happens.
slide rule? 