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1) In a population of men, the probability that a man’s left eye is of brown colour is p, and the probability that a man’s right eye colour is brown is also p. Therefore the probability that a man has at least one eye of brown colour is: Pr(left eye brown or right eye brown) = Pr(left eye brown) + P(right eye brown) = 2p.


To be eligible for a certain type of manual work in the petroleum industry, a male must be above both a certain minimum height and a certain minimum weight. The separate probabilities that these are satisfied are ph (for height) and pw (for weight). Therefore for a man selected at random, the probability that he meets both criteria for the job is: Pr( Meets height and weight criteria) = Pr(Meets height criterion) x Pr(Meets weight criterion) = ph x pw


Suppose a student guesses the answers to three true/false questions. Whether the guess for any question is correct is independent of the guess for any other question being correct, and the probability of any guess being correct is 1=2. Overall, there are four possibilities for the number of questions the student guesses correctly, namely the student may guess 0; 1; 2; or 3 answers correctly. Therefore, because all four of these possibilities are equally likely, the probability that the student guesses all three questions correctly is 1=4.

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First Approach: You are counting the case where someone has two brown eyes twice. I think this principle is called Inclusion-Exclusion. You should have $2p-p^2$.

Second Approach: The height and weight of someone aren't mutually exclusive. If they were, your approach would be perfectly fine.

Third Approach: Getting one question right is not the same as getting all three questions right. The probability that you get all questions right is $({\frac1 2})^3=\frac1 8$. The probability that you get two questions right is $\frac3 8$.

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  1. Incomplete formula. $P(\text{A or B}) = P(A) + P(B) - P(\text{A and B}).$ I've never met a person who had $2$ different colored eyes. I think it is a reasonable simplification to say that given that a person has a brown left eye, they also have a brown right eye. Since $P(A \text{ and } B) = P(A) * P(B \text{ given A)}.$ I would say $P(\text{ A and B }) = P(A). $In other words, I think you should get $'p'$ as your answer rather than $'2p'$
  2. Variables height and weight aren't independent of each other. Tall people tend to weigh more than short people. I need to know the probability that he meets the weight requirement after we know he met the height requirement.
  3. Let $X$ be the random variable of how many answers he gets right out of those $3.$ It isn't a uniform random variable.

If he gets $0$ questions right, the probability of that is $(\frac{1}{2})^{3}= \frac{1}{8}$. If he gets $1$ question right (the first, second, or third), the probability is $(\frac{1}{2})^{3} * 3 = \frac{3}{8}$ If he gets $2$ questions right (the first and second, the first and third, or the second and third), the probability is $(\frac{1}{2})^3 *3 = \frac{3}{8}.$ If he gets all questions right, the probability is $(\frac{1}{2})^3 = \frac{1}{8}.$

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  • $\begingroup$ A cursory Google search says the probability of two different colored eyes in humans is approximately $0.006$; however, since two different shades of brown would be included in this probability, the probability of one brown and one non-brown would be even lower than that. $\endgroup$ – GFauxPas Sep 18 '16 at 15:47

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