Vemuri wrote:A bag contains 4 $5 notes, 7 $2 notes and 9 $1 notes. If the 3 notes are drawn at random from the bag, then find the odds against drawing the minimum possible amount

A) 99:88
B) 7:88
C) 88:95
D) 88:7
E) 90:7

My answer is (C)
Probability of drawing the minimum possible amount = C(3,9)/C(3,20)=7/95,
Thus the odds is (95-7)/95=88/95

billzhao wrote:

Vemuri wrote:A bag contains 4 $5 notes, 7 $2 notes and 9 $1 notes. If the 3 notes are drawn at random from the bag, then find the odds against drawing the minimum possible amount

A) 99:88
B) 7:88
C) 88:95
D) 88:7
E) 90:7

My answer is (C)
Probability of drawing the minimum possible amount = C(3,9)/C(3,20)=7/95,
Thus the odds is (95-7)/95=88/95

OA is D

Any idea how to solve this problem?

what I do understand is if we find the odds of drawing the minimum amt and subrtact that from 1, we'll find the answer:

The min. amt that can be drawn is $3 in 1 ways:
i) All the 3 notes are $1 each.

But I don't understand how to put it into the Prob. terms and then solve it.

total notes=20. so (9C3)/(20C3). subtract the outcome of this from 1. I spent quite a lot of time on this, so I hope someone provides a simple and sweet soln.