mestrec wrote:

Hi All!

Could someone please explain how I could solve this exercise without doing the opposite?

Should not the solution be 5/11 X 4/10?

Regards!

Bunuel wrote:

Official Solution:

A box contains 11 hats, out of which 6 are red hats and 5 are green hats. If two hats are to be selected at random without replacement, what is the probability that at least one green hat will be selected?

A. \(\frac{10}{11}\)

B. \(\frac{8}{11}\)

C. \(\frac{7}{12}\)

D. \(\frac{5}{13}\)

E. \(\frac{2}{7}\)

It's easier to find the probability of the opposite event and subtract it from 1. The opposite event would be if we select zero green hats, or which is the same if we select 2 red hats, so: \(P(G \ge 1)=1-P(RR)=1-\frac{6}{11}*\frac{5}{10}=\frac{8}{11}\).

Answer: B

Hi,

although you should solve by looking at the prob of opposite event, since it is less time consuming and less error prone..

the normal way would be..there are 5 green and 6 red ...

so way you can pick up atleast one of red is..

both green=\(\frac{5}{11}*\frac{4}{10}\)..

one red and one green=\(\frac{6}{11}*\frac{5}{10}\) and \(\frac{5}{11}*\frac{6}{10}\)

(2 ways one of each can be picked up)total prob=\(\frac{5}{11}*\frac{4}{10} + \frac{6}{11} * \frac{5}{10} *2= \frac{(20+60)}{110}= \frac{8}{11}\)

hope it helps you

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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372

2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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