D01-37

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!

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

I am not sure what am i doing wrong in the problem? my approach is :
1 red and one green or both green
thus, (6/11*5/10) or (5/11*4/10)
=5/11

please help me know what i am missing in my approach

Hi..

Where you are going wrong is that you are taking two cases RG and GR as one case..
You can pick green then red OR pick red then green OR pick both green..
So 6/11*5/10+5/12*6/10+5/11*4/10

You will get your answer

Can anyone explain ( Experts in particular)

Why has the order not matter here? Is it because of the word ( selection) used in Question stem which hints for a Combination question?

Atleast one green will be interpreted as ( Green, Green ) Or ( Green , Red) [ Now this can also be taken as ) --> (Red, Green)

But here we have only take these two ( Green, Green ) Or ( Green , Red) as possibilities and left out the third one. Kindly clarify.

Hi..

Here the order matters and you have to take all 3 cases.
See solutions above

Hello,

Can I solve it as (5C2 + 6C1 * 5C1)/11C2?

Jayesh24

Can you explain reasoning behind each term and how you arrived at this?

May be then someone can help you!

Hello, Jayesh24. The short answer is yes. You will arrive at the same answer by solving the problem in the proposed manner. That is,

From 5 green hats, you could select 2 in successive selections: 5C2 (which comes to 10) or
From two successive selections, you could select 1 red hat and 1 green hat: 6C1 * 5C1 (in either order, which will come to 6 * 5, or 30)
Since there are two possibilities to select at least 1 green hat, you create an or condition, meaning you add the independent probabilities: (5C2) + (6C1 * 5C1) (or 10 + 30)
Finally, out of 11 total hats, you are selecting 2: 11C2 (which comes to 55) and
since probability is given by a desired outcome divided by all possible outcomes,
\(\frac{(5C2 + 6C1 * 5C1)}{11C2}\)
is correct. Or, if you prefer,
\(\frac{(10 + 6 * 5)}{(55)}\), which becomes \(\frac{40}{55}\), or \(\frac{8}{11}\) when reduced.
This is a roundabout way of getting the answer, but a valid method nonetheless. Well done.

- Andrew

11 hats, 6 red, 5 green.

P(at least one green) = 1 - P(no green) = \(1-\frac{6}{11}.\frac{5}{10} = \frac{8}{11}\)

if I say "the probability of the first hat is 5/11. and we don't pick the green hat, then the next probability of picking a green hat for the second pick is 5/10.
5/11 + 5/10 = 8/11.

can someone tell me why this approach is not discussed?

5/11 + 5/10 = 21/22, not 8/11. Therefore, that solution is not incorrect.

The correct solution should be:

P(at least 1 green hat) = P(2 green hats) + P(1 green hat and 1 red hat) =

= 5/11 * 4/10 + 5/11 * 6/10 * 2 = 8/11 (we multiply 5/11 * 6/10 by 2 because 1 green hat and 1 red hat can occur in two different ways: either the first hat is green and the second hat is red OR the first hat is red and the second hat is green, each with a probability of 5/11 * 6/10).

I hope this helps.

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.