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Perms and combination way to solve probability question

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Perms and combination way to solve probability question

by Joepc » Wed Oct 12, 2016 9:37 pm
HiAll,

Please help me in sorting my confusion

Question There are 5 green balls, 4 red balls and 6 blue balls
What is the probability of getting all three are different colour, when the three balls are picked one after the other without replacement

Answer

Method 1 :-Perms and comms way

Total ways of selecting 3 Balls one after the other is 15*14*13

No of ways of selecting 3 different colour balls are 5*4*6

Is it not 5*4*6 / 15 *14*13 ?

But in an article i saw it is (5*4*6 /15*14*13) *6
why a 6 is multiplied?

Method 2 :- Probability Way

But When i solve using probability approach I am getting a reason for 6

Probability of selecting Green ball without replacement is 5/15
Probability of selecting Red ball without replacement is 4/14
Probability of selecting blue ball without replacement is 6/13

hence for
Green , Red ,Blue it is 5/15 * 4/14 * 6/13
similarly it goes for various other 5 arrangement hence it 6 *(5/15 * 4/14 * 6/13).

Getting confused, which one is the right answer?
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by DavidG@VeritasPrep » Thu Oct 13, 2016 7:03 am
Joepc wrote:HiAll,


Is it not 5*4*6 / 15 *14*13 ?

But in an article i saw it is (5*4*6 /15*14*13) *6
why a 6 is multiplied?

The 3! or 6 encompasses all the different-color scenarios you'd have if order mattered. You could pick:

GRB
GBR
BRG
BGR
RGB
RBG

(And because order doesn't matter, you'd have to divide 5*4*6 by 6 to ensure that you aren't counting each of those scenarios as a discrete group)

Note that you'd also divide 15*14*13 by 6 as order won't matter for this group either. The 6's will cancel out.
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by Matt@VeritasPrep » Fri Oct 14, 2016 12:40 am
You could also do it combinatorially like so:

((5 choices of Green) * (4 choices of Red) * (6 choices of Blue)) / ((15 choices for first) * (14 choices for second) * (13 choices for third))

then multiply that by the six ways of arranging GRB.
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by Matt@VeritasPrep » Fri Oct 14, 2016 12:42 am
Your two methods are basically the same: the issue is that you're ignoring the order in which these things could happen in your first approach, but probability has to account for this.

For instance, suppose I have a red die and a blue die (each fair and with six sides), and I want to know the probability that when I roll them their faces give a sum of 3. I could get a 1 and a 2 or a 1 and a 2, so I've got

(1/6)*(1/6) + (1/6)*(1/6)

or two arrangements, since there are two distinct ways this could happen.

With the green, red, and blue marbles we've got similar conditions: there are a number of ways in which this could happen in some order, and you have to account for all of them.
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