DAT Math Destroyer 2018 Test 3 Q#12

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Jul 25, 2017
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It is found that 14% of a population has blue eyes, while 25% are left handed, and 8% are both. What is the probability that a person randomly chosen from the population is either blue eyed or left handed?

(The answer says it is 25+14-8=32)

I was wondering if this would be incorrect because wouldn't you need to subtract 8 from both the 25 and the 14 and then add them together for a probability of 23%?

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You are correct in that you subtract 8% from each— this accounts for people who only match one of the two criteria. So you have 17% and 6%. If you add those together, you get 23%. However, since the 8% you subtracted accounts for individuals with both criteria, you add 8% to 23% to get 31% (answer choice E). Hopefully that helps! Let me know if doesn’t, and I can do my best to help you work through the problem.
Subtract the "both" from each of the topic individuals and then add all three values together.
(Blue eyed - both) + (left handed -both) + both
= (14-8) + (25-8) + 8 = 31.
Most easily seen if you do a Venn diagram! The left circle will have 6 blue eyed people, the right will have 17 left handed people, and the crossover will have 8 people that are both blue-eyed and left handed. That means a total of 8+17+6 people out of the assumed 100 people (since percentages are given) will be your answer. 31%

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