This Valentine’s Day, Drake Martinet proposed to Stacy Green, Mashable’s Vice President of Marketing and Communications, via an infographic on Mashable.
Here’s the part of it that reminded me of a GRE problem:
How did we get from 1,340,000 women to 0.48 of a woman?
This answer came about because the writer took 3% of 1,340,000, then 2% of that, then 6% of that, then 1% of that.
Here’s how you do that:
1,340,000(.03)(.02)(.06)(.01) = .48
The writer goes on to conclude, rather sweetly, that since probability has it that there should be less than one-half a woman on earth who is perfect for him, he is very lucky to find one entire such woman.
Note that you can do many consecutive percents in a row without calculating some kind of intermediate answer. Order doesn’t even matter when it’s all multiplication!
The actual math for this situation is a bit more complicated, though.
There is a big difference between “3% of women could keep Martinet laughing and 2% are gorgeous inside and out” and “3% of women could keep Martinet laughing and 2% OF THOSE are gorgeous inside and out.”
Let’s try a separate example.
Imagine the difference between:
Scenario 1: “50% of the cookies are chocolate chip and 50% have nuts”
Scenario 2: “50% of the cookies are chocolate chip and 50% OF THOSE have nuts”
In Scenario 1, half of the cookies are chocolate chip and half have nuts. Since a cookie can certainly have both chocolate chips and nuts, we would need some information about the overlap to calculate any further.
For example, there could be ZERO overlap: half of the cookies could have chocolate chips (no nuts) and half could have nuts (no chocolate chips). Or there could be TOTAL overlap: half of the cookies have both nuts and chocolate chips, and the other half could be something else entirely.
In Scenario 2 — the one that matches the kind of math Martinet did — half of the cookies have chocolate chips and half OF THOSE have nuts, meaning that we don’t know anything at all about the other half of the cookies (other than that they don’t have chocolate chips).
For Martinet’s words to match up with his math, he should have written, “Only 3% of people could keep me laughing, and only 2% of those are gorgeous inside and out, and only 6% of those could love a huge geek, and only 1% of those could bring total joy to my life.”
If it is instead simply the case that the 3%, 2%, 6%, and 1% are percents of the original total (1,340,000), then we would need information about the overlap among the groups to do any calculating.
For instance, if I told you that, among a group of 100 people, 5% are straight-A students and 6% have read The Lord of the Rings, we have no way to figure out how how many straight-A students have read The Lord of the Rings. We can say that 5 people are straight-A students and 6 people have read LoTR. Maybe we’re talking about 11 different people here, or maybe we’re talking about 6 people who have all read LoTR, and 5 of whom are also straight-A students.
Want to test your understanding of this topic? Answer these two questions:
1. In a group of 200 high school girls, 8% have a puppy and 25% of those have a boyfriend. How many have both a puppy and a boyfriend?
2. In a group of 200 high school girls, 8% have a puppy and 25% have a boyfriend. If x students have both a puppy and a boyfriend, what is the range of possible values for x?
Ready for the answers?
Answer to #1: 200(.08)(.25) = 4 students.
Answer to #2: 200(.08) = 16 students have a puppy. 200(.25) = 50 students have a boyfriend. Are the 16 contained within the 50? We have no idea. What is the minimum number of students who could have a puppy and a boyfriend? Well … zero. There are 200 total girls — there could be 16 girls with puppies, 50 separate girls with boyfriends, and 144 other girls. Or, the 16 girls with puppies could ALL also have boyfriends. So, the answer is 0 ≥ x ≥ 16.
Er… congratulations to the happy couple!