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orbs in a sack

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orbs in a sack

by sanju09 » Sat Apr 24, 2010 12:45 am
Bernard has 3 green, 2 red, and 3 blue orbs in a sack. He aimlessly elects to choose 5 orbs out of the sack, without replacement. What is the probability that of the 5 drawn orbs, he has picked 1 red, 2 green, and 2 blue orbs?
(A) 3/280
(B) 9/280
(C) 3/28
(D) 9/28
(E) 27/28
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by pradeepkaushal9518 » Sat Apr 24, 2010 1:21 am
what i understand

5 out of 8 is 8C5= 8*7*6

1 red out of 2 red = 2C1=2
2 green out of 3 = 3C2= 3
2 blue out of 3= 3C2=3
probability= possible outcomes/total outcomes

= 2*3*3/8*7*6
= 3/56

which i dont find in option or i am doing any mistake
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by sanju09 » Sat Apr 24, 2010 1:53 am
pradeepkaushal9518 wrote:what i understand

5 out of 8 is 8C5= 8*7*6

1 red out of 2 red = 2C1=2
2 green out of 3 = 3C2= 3
2 blue out of 3= 3C2=3
probability= possible outcomes/total outcomes

= 2*3*3/8*7*6
= 3/56

which i dont find in option or i am doing any mistake
We can get 1 Red-2 Green-2 Blue in many different ways (as the order does not matter).
The mind is everything. What you think you become. -Lord Buddha



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by rockeyb » Sat Apr 24, 2010 2:32 am
We can select 5 orbs out of 8 in 8 C 5 ways = 8!/ (5!x2!) = 56 .

So Number of possible ways = 56 .

Now we can select a red orb in 2 ways as there are only 2 red orbs .

2 green in 3C2 ways = 3 ways .

2 blue in 3C2 ways = 3 ways .


Number of desired out come = 2 x 3 x 3 = 18 .

Number of possible out come = 56 .

Probability = Number of desired out come / Number of possible out come .

Probability= 18 / 56 [spoiler]= 9 / 28 that is D .[/spoiler]
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by Fiver » Sat Apr 24, 2010 6:54 am
Agree with D.
Here's another approach.
Total no. of items is 8
P (1R, 2G & 2B) without replacement = 2/8 * 3/7 * 3/6 = 3/56
And since the order does not matter there are 3! possible orders.
Hence our answer is (3/56) * 6 = 9/28
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