The box contains 4 odds and three evens. The first draw the probability of drawing an odd is 4/7 (times 100)
if an even has been drawn the box now contains 2 evens and 6 balls in total. Now the probability of drawing an odd is 4/6
The probability of getting at least one odd combining two draws is:
[4/7 + (3/7)(4/6)] times 100
The 4/7 is drawing an odd on the first draw. The subsequent draws don't matter we already have an odd. The 3/7 is the odds of drawing an even on the first draw. This is multiplied by the odds of drawing an odd at this point. After drawing an even the box now contains 4 odds and 6 balls in total. These two fractions are multiplied.
But it there are three draws. The probability of getting at least one odd is:
[4/7
+ (3/7)*(4/6)
+ (3/7)*(2/6)*(4/5)] times 100
[4/7 -- first draw drew one of the 4 odds. There were 7 balls in the box
+ [(3/7) -- first draw drew one of the evens there were 7 balls in the box
*(4/6)] -- second draw drew one of the four odds only 6 balls left in the box
+ [(3/7) -- first draw drew an even
*(2/6) -- second draw another even
*(4/5)] -- third draw finally got one of the odds
Then the fractional number is multiplied by 100 to make it a percentage.
=97%
A different " tricky" way to consider this problem is if we draw all 3 evens in the first three draws. This is the opposite of what we want.
(3/7)(2/6)(1/5) = 1/35
since this is the opposite of what is wanted subtract this from one.
Prob = (1 - (1/35)) * 100
= (34/35) * 100
