Set S is the set of all prime integers between 0 and 20. If three numbers are chosen randomly from set S and each number can be chosen only once, what is the positive difference between the probability that the product of these three numbers is a number less than 31 and the probability that the sum of these three numbers is odd?
(A) 1/336
(B) 1/2
(C) 17/28
(D) 3/4
(E) 301/336

• answer is c i.e 17/28
  
  there are 8 prime numbers in s that are 2,3,5,7,11,13,17,19
  now the three numbers whose product is less than 31 are 2,3,5
  so the number of ways of taking these numbers in a group of 3 is 1
  total number of ways of taking 3 numbers from 8 numbers is 8c3 = 56
  so probability that a product of three numbers is less than 31 is 1/56
  let it be p1
  p1= 1/56
  
  now for sum to be odd we have to take all three numbers as odd that is we should not take 2
  this can be found by finding the ways in which 2 appears and subtracting it from total number of ways
  now if 2 occurs then remaining two numbers can be selected in 7c2 ways = 21 ways
  so number of ways in which 2 not occurs is 8c3-7c2 = 56-21 = 35
  so probability that sum is odd is 35/56
  let it be p2
  p2= 35/56
  difference between these two probabilities is 35/56-1/56 = 34/56 = 17/28
  so the answer is 17/28
• 10 years agoHelpfull: Yes(5) No(0)