Problem of the Day


If a number is selected at random from a set of 6 distinct integers, what is the probability that the number is odd or prime?

(1) The probability of the number being odd is \frac{1}{6}.
(2) The probability of the number being prime is \frac{1}{3}.

 

Reveal Answer

Answer

C. Both statements together are sufficient, but neither statement alone is sufficient.

See the Solution

Solution

[latexpage]

P(O)=probability that the number is odd.
P(P)=probability that the number is prime.

Basic probability theory tells us that P(O or P)=P(O)$+$P(P)$-$P(O and P).

Statement #1: INSUFFICIENT. This essentially tells us that exactly one out of the six numbers is odd. However, it tells us nothing about how many are prime. Thus, it is insufficient by itself.

Statement #2: INSUFFICIENT. This essentially tells us that exactly two out of the six numbers are prime. However, it tells us nothing about how many of the numbers are odd. Thus, it is insufficient by itself.

Statements #1 and #2 together: SUFFICIENT. Statement 1 tells us that exactly one number is odd. Statement 2 tells us that exactly two numbers are prime. Remember that 2 is the only even prime number. Thus, if two distinct integers are prime, then they are either both odd, or one is odd and one is even. They can’t both be odd, because that would violate statement 1. Therefore, one of the primes is even and one of the primes is odd. In other words, exactly one of the six integers is both odd and prime, which means that P(O and P)=$\frac{1}{6}$. Going back to our original equation:

P(O or P)=P(O)$+$P(P)$-$P(O and P)
P(O or P)=$\frac{1}{6}+\frac{1}{3}-\frac{1}{6}$
P(O or P)=$\frac{1}{3}$

 


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