A bag contains 5 blue, 3 black and 10 red balls. If three balls are dr

Question

A bag contains 5 blue, 3 black and 10 red balls. If three balls are drawn at random, without replacement. What is the probability that one is blue, one is black and one is red ?

A. 25/136

B. 4/21

C. 1/2

D. 111/136

E. 4/5

AbidHasan2002 · Answer

I guess, answer is A
3 balls can picked in 3! Or 6 ways
 So without replacing,
Getting blue, black and red ball is
 (5/18)×(3/17)×(10/16)
 =25/816
Finally,
 (25/816)×6= 25/136
I hope it's the answer 😃

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MBAHOUSE · Answer

Again I used partially the additive principle and partially the multiplication principle.
It’s very necessary at the GMAT you be able to be fast in simplifying the fractions.

ThatDudeKnows · Answer

What's the probability that we are going to get exactly Blue, Black, Red in that order?
5/18 * 3/17 * 10/16

But that's not the only order; we need to account for the probabilities of other orders. How many ways are there to order the three selections? We have three choices for which color we draw first, two choices for which color we draw second, and one choice for which color we draw third, so 3*2*1=6 possible orders. In each case, our denominators will be 18, 17, and 16 and our numerators will be 5, 3, and 10 in some order.

So the overall probability is 6 * 5/18 * 3/17 * 10/16 = 50/272 = 25/136.

Answer choice A.

Regor60 · Answer

Denominator: number of ways to pull 3 balls from 18:

18!/3!15! = 3*17*16

Numerator: ways to pull a blue ball*ways to pull black*ways to pull red:

5*3*10

Probability: 5*3*10/3*17*16 =

5*5/17*8 = 25/136

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Suruchim12 · Answer

Experts :  Bunuel  /  GMATNinja

Can you please help explain how we can arrive at the total outcomes using the nCr formula. 
In other similar problems we have used nCr to determine the total no. of outcomes where order does not matter but here all explanations I see are using the counting method.

TIA!

Sroy1 · Answer

Total number of ways to choose 3 coloured balls from total 18 balls are =  18P3=18*17*16 (as order of the drawn coloured balls matters)
Now our desired case that we are going to get exactly one Blue, one Black, one Red balls once  5*3*10*6! .
( 6! Is coming from the fact that those 3 drawn balls can arrange themselves in 6! Ways.).

Thus required probability: 
(5*3*10*6!)/(18*17*16)=25/136

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A bag contains 5 blue, 3 black and 10 red balls. If three balls are dr

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A bag contains 5 blue, 3 black and 10 red balls. If three balls are dr

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