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Updated on: 17 Apr 2014, 00:52
Originally posted by tauchmeister on 16 Apr 2014, 15:55.
Last edited by Bunuel on 17 Apr 2014, 00:52, edited 1 time in total.
Last edited by Bunuel on 17 Apr 2014, 00:52, edited 1 time in total.
Edited the question
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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
A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 2/5
16 Apr 2014, 19:17
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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
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
General Discussion
20 Apr 2015, 03:06
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tauchmeister
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
