probability

Question

For this problem, assume the balls in the box are numbered 1 through 7, and that an experiment consists of randomly selecting 2 balls one after another without replacement.

 

What probability should be assigned to the event that at least one ball has an odd number?

 

 

note: i believe the answer is 17/21 however i am being told this is wrong

 

Answer 1

Let E be the event that at least one of two numbers is odd.

 

The only way that E doesn't happen is if you draw two even numbers.

 

There are a total of C(7,2) = (7·6)/2! = 21 pairs possible and of those, C(3,2) = 3 pairs where both numbers are even.

 

Thus P(E) = 1 - P(E^c) = 1 - 3/21 = 18/21 = 6/7

 

The probability you calculated is for the event of drawing at least one even number, since there are 4 odd numbers in the box.

Answer 2

One way to get the answer is to note that the probability of getting at least one odd number is equal to

  1 - probability of drawing two even numbers in a row.  

 

   This latter probability is   3/7 x  2/6  =   1/7

   The factor of 3/7 is the probability of drawing even on first draw,

    The factor of 2/6 is the probability of drawing one of the two remaining even numbers from the six remaining

 

   1 -  1/7  =   6/7      which is the final answer.