Kinshook
If 2 prime numbers are chosen less than 100 without repetition, then what is the probability that their sum will be an odd number?
\(A. \frac{6}{75}\)
\(B. \frac{3}{75}\)
\(C. \frac{1}{12}\)
\(D. \frac{276}{300}\)
\(E. \frac{273}{300}\)
The prime numbers less than 100 are:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97There are 25 primes in total, and only 1 is even.
For this question, I would probably use counting techniques, but we can also saw that using probability rules quite quickly.
The key here is to recognize that, in order for the sum to be odd, one number must be ODD, and the other must be EVEN. So,....
P(sum is odd) = P(1st number is odd
AND 2nd number is even
OR 1st number is even
AND 2nd number is odd)
= [P(1st number is odd)
x P(2nd number is even)]
+ [P(1st number is even)
x P(2nd number is odd)]
= [24/25
x 1/24]
+ [1/25
x 24/24]
= 1/25
+ 1/25
= 2/25
Check answer choices.... not there!
However, 6/75 is equivalent to 2/25, which means the correct answer is A