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# In a group of 10 girls, only 4 have blue eyes. If 3 girls ar

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Intern
Joined: 15 Jun 2010
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In a group of 10 girls, only 4 have blue eyes. If 3 girls ar  [#permalink]

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Updated on: 16 Apr 2014, 23:52
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Difficulty:

55% (hard)

Question Stats:

66% (02:26) correct 34% (02:39) wrong based on 130 sessions

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In a group of 10 girls, only 4 have blue eyes. If 3 girls are to be selected at random, without replacement, what is the probability that at least 2 girls will have blue eyes?

A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 2/5

Originally posted by tauchmeister on 16 Apr 2014, 14:55.
Last edited by Bunuel on 16 Apr 2014, 23:52, edited 1 time in total.
Edited the question
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Re: In a group of 10 girls, only 4 have blue eyes.  [#permalink]

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16 Apr 2014, 18:17
atleast 2 girls means 2 girls (or) more than 2 girls.

Total = 10;
Having blue eyes =4;
3 girls are selected at random without replacement
probability that at least 2 girls will have blue eyes is 4c2*6c1(exacltly 2 girls)/10c3 + 4c3*6c0(more than 2 girls)/10c3
= 3/10 + 1/30 = 1/3;

Press kudos if this helps you
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In a group of 10 girls, only 4 have blue eyes. If 3 girls ar  [#permalink]

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20 Apr 2015, 02:06
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tauchmeister wrote:
In a group of 10 girls, only 4 have blue eyes. If 3 girls are to be selected at random, without replacement, what is the probability that at least 2 girls will have blue eyes?

A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 2/5

The Event here can happen in 2 ways:
i) 2 of the 3 girls have blue eyes and the 3rd girl has non-blue eyes
ii) All 3 of the 3 girls have blue eyes

Now, the important point that students must know in PnC and Probability questions is that OR scenarios imply ADDITION, and AND scenarios imply MULTIPLICATION.

In the current question, either Case (i) will happen OR Case (ii) will happen. The two cases cannot happen at the same time (You cannot have 2 girls out of the 3 girls having blue eyes AND 3 girls out of the 3 girls having blue eyes at the same time)

Since this is an OR scenario, we will ADD the probabilities of Case (i) and Case (ii).

So, we'll write: P(at least 2 blue-eyed girls) = P(2 blue-eyed girls) + P(3 blue-eyed girls) . . . (1)

Let's now find P(2 blue-eyed girls)

Here the favorable scenario is: 2 girls are blue-eyed AND 1 girl is non-blue-eyed

Since this is an AND scenario, the number of favorable cases will be obtained by multiplying the number of ways in which 2 blue-eyed girls can be selected (out of the 4 blue-eyed girls) and the number of ways in which 1 non-blue-eyed girl can be selected (out of the 6 non-blue-eyed girls)

So, P(2 blue-eyed girls) = (4C2) X(6C1)/(10C3) = 3/10 . . . (2)

P(3 blue-eyed girls) = (4C3)/(10C3) = 1/30. . . (3)

Subsituting (2) and (3) in (1), we get:

P(at least 2 blue-eyed girls) = $$\frac{3}{10}+\frac{1}{30}$$ = 1/3

Hope this helped!

- Japinder
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Re: In a group of 10 girls, only 4 have blue eyes. If 3 girls ar  [#permalink]

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16 Jan 2017, 03:52
tauchmeister wrote:
In a group of 10 girls, only 4 have blue eyes. If 3 girls are to be selected at random, without replacement, what is the probability that at least 2 girls will have blue eyes?

A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 2/5

Total selection = 10C3 = 10 * 3 * 4
Exactly 1 girl has blue eyes = 4C1 = 4
Exactly 3 girls don't have blue eyes = 6C3 = 5 * 4

At least 1 girl has blue eyes = $$1 - [(4 * 5 * 4)/ 10 * 3 * 4]$$ = $$1 - 2/3$$ =$$1/3$$
Ans.: B
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Joined: 04 Mar 2011
Posts: 2799
Re: In a group of 10 girls, only 4 have blue eyes. If 3 girls ar  [#permalink]

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18 Jan 2017, 10:01
tauchmeister wrote:
In a group of 10 girls, only 4 have blue eyes. If 3 girls are to be selected at random, without replacement, what is the probability that at least 2 girls will have blue eyes?

A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 2/5

We are given that we have a group of 10 girls; 4 girls have blue eyes and 6 girls do not have blue eyes. We need to determine the probability, when 3 girls are selected, that at least 2 girls will have blue eyes. There are two scenarios in which we can select at least 2 girls with blue eyes.

Scenario 1: Blue-Blue-Not Blue

Scenario 2: Blue-Blue-Blue

Before calculating each scenario, let’s first determine the total number of ways to select 3 girls from a group of 10 girls.

The number of ways to select 3 girls from a group of 10 is:

10C3 = (10 x 9 x 8)/3! = (10 x 9 x 8)/(3 x 2 x 1) = 5 x 3 x 8 = 120

Next we need to determine the number of ways to select the blue-eyed girls.

First, let’s calculate scenario 1: Blue-Blue-Not Blue

The number of ways to select 2 girls with blue eyes from 4 is:

4C2 = 4!/2!2! = (4 x 3)/2! = 6 ways

The number of ways to select 1 girl with no blue eyes from 6 is:

6C1 = 6 ways

Thus, the probability of selecting 2 girls with blue eyes and 1 with no blue eyes is:

(6 x 6)/120 = 36/120

Next, let’s calculate scenario 2: Blue-Blue-Blue

The number of ways to select 3 girls with blue eyes from 4 is:

4C3 = (4 x 3 x 2)/3! = 4 ways

Thus, the probability of selecting 3 girls with blue eyes is:

4/120

Finally, we can determine the probability of selecting at least 2 girls with blue eyes:

36/120 + 4/120 = 40/120 = 1/3

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Re: In a group of 10 girls, only 4 have blue eyes. If 3 girls ar   [#permalink] 18 Jan 2017, 10:01