If a number is selected at random from a set of 6 distinct integers

Question

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} .

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nick1816 · Answer

Statement 1- 
There is 5 even numbers and 1 odd number

There can be 0,1 or 2 prime numbers

Insufficient

Statement 2-
There are 2 prime numbers; we will have at least 1 odd number

Insufficient

Combining both statements
We have 2 prime numbers (one odd and one even) and we have only 1 odd number, which is prime too.

Sufficient

exc4libur · Answer

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

odd/6=1/6, one odd and five even: {1,2,4,6,8,10}; (odd,prime)/6 = 1/6
odd/6=1/6, one odd and five even: {2,3,4,6,8,10}; (odd,prime)/6 = 2/6

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

prime/6=1/3: two primes {1,2,3,6,8,10}; (odd,prime)/6 = {1,2,3}/6 = 3/6
prime/6=1/3: two primes: {2,3,4,6,8,10}; (odd,prime)/6 = {2,3}/6

(1&2)   sufic

Only even prime number is 2, the others are odd; 
If there is only one odd, and two primes, then there is one odd prime and one even prime;
{odd,prime}/6=2/6=1/3

Ans (C)

luukav · Answer

As we don't know which distinct numbers are in the set, to know the possibility of the number being odd or prime we need to know more about which numbers are in the set i.e. the range.

Statement 1:
The probability of the number being odd is 1/6.
This doesn't tell us anything about the probability of the number being prime. There are many odd numbers that are prime.

Statement 2:
The probability of the number being prime is 1/3.
This tell us that most of the numbers are odd, except the number two which is prime and even. So there are still multiple outcomes.

Both statements together gives us the information we need to solve the probability for odd or prime.

C IMO

MHIKER · Answer

(1) The set may prime, composite, odd, even. We are given the probability of an odd number only. Insufficient.

(2) The set may prime, composite, odd, even. We are given the probability of Prime number only. Insufficient.

Considering both:
The question asked for odd or prime. An Odd number can be prime and Odd too. So, the events are mutually inclusive. Thus we will multiply each probability.  \frac{1}{6} * \frac{1}{3} ; Sufficient.

The answer C.

CEdward · Answer

P(odd or prime) = P(odd) + P(prime)

St1 - P(odd) = 1/6
Case 1: 1,4,6,8,10,12
Case 2: 2,3,6,8,10,12

P(odd or prime) = 1/6 in Case 1
P(odd or prime) = 1/3 in Case 2

Insufficient.

St2 - P(prime) = 1/3 
Case 1: 1,3,6 ---> P(odd or prime) = 2/3
Case 2: 3,27,6 ---> P(odd or prime) = 1/3

Combined: Sufficient.
Answer is C.

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If a number is selected at random from a set of 6 distinct integers

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If a number is selected at random from a set of 6 distinct integers

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