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Probability Approach to solve a Probability question

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Probability Approach to solve a Probability question

by Joepc » Wed Oct 12, 2016 8:52 pm
Question :- There are 5 green balls, 4 red balls and 6 blue balls
What is the probablility of getting all three are different colour, when the three balls are picked simulatneously

Using Combinatorial approach I can bring the answer to

5C1*4C1*6C1 / 15C3

How to solve this in probability way?
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by crackverbal » Wed Oct 12, 2016 11:52 pm
Hi Joepc.

Let me first give you a brief summary of how to tackle probability questions of this kind and then let me work out the question as an example.

By definition, Probability = (Number of favorable outcomes)/(total number of outcomes)

Consider that we have a bag containing 2 blue, 3 green and 4 yellow balls and have to select one ball from the bag then the total outcomes will be 2 + 3 + 4 = 9. If we need to find the probability of selecting one blue ball P(B) from the bag, then P(B) = 2/9, the favorable outcomes here being the 2 blue balls and the total outcomes being all the 9 balls. Now this procedure holds good when we select one ball at a time (selecting one object at a time is referred to as a single event), but things change when we start selecting more than one ball (selecting more than one object at a time is referred to as a complex event).

Lets consider that we now select two balls at a time (complex event). Now this selection can be done in any one of three ways
1. With Replacement : Involves picking the first ball, replacing it, and then picking the second
2. Without Replacement : Involves picking the first ball, NOT replacing it, and then picking the second
3. Simultaneous : Involves picking both balls at once

For the above mentioned 3 cases the universal formula that you can use is

P(complex) = P(product of individual events) * arrangement

Now considering the example of a bag with 2 blue, 3 green and 4 yellow balls, if we need to find the probability of getting the first ball blue and the second ball green when the balls are selected one by one with replacement then

P(BG) = P(B) * P(G) * arrangement ----> P(BG) = 2/9 * 3/9 * 1 = 2/27.

Here the arrangement is considered as 1 since the order here is specified i.e the first ball needs to be blue and the second green.

If the same question were rephrased as find the probability of getting one blue and one green when the balls are selected one by one with replacement then

P(BG) = P(B) * P(G) * arrangement ----> P(BG) = 2/9 * 3/9 * 2!

The 2! here represents the arrangement of the word BG. If we have a word MISSISSIPPI then the arrangements will be 11!/(4!4!2!) where 11! represents the number of alphabets, 4! represents the number of I's, 4! represents the number of S's and 2! represents the number of P's.

The arrangement here takes care of the two possible cases of getting a blue and a green i.e. P(BG) and P(GB).

If the same question were rephrased as find the probability of getting one blue and one green when the balls are selected one by one without replacement then

P(BG) = P(B) * P(G) * arrangement----> P(BG) = 2/9 * 3/8 * 2!

When we deal with a simultaneous selection, the probability way of solving is to treat a simultaneous selection as one by one without replacement and then use the above illustrated procedure.

Question :- There are 5 green balls, 4 red balls and 6 blue balls
What is the probability of getting all three are different color, when the three balls are picked simultaneously

Here we just need to find P(GRB)

P(GRB) = P(G)*P(R)*P(B) * arrangement ----> 5/15 * 4/14 * 6/13 * 3! which is exactly the same answer as what you got using the combination approach.

Hope this helps!

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by GMATGuruNY » Thu Oct 13, 2016 2:52 am
Joepc wrote:Question :- There are 5 green balls, 4 red balls and 6 blue balls
What is the probablility of getting all three are different colour, when the three balls are picked simulatneously
Let G = green, R = red, B = blue.
Here, a good outcome consists of EXACTLY 1 G, EXACTLY 1 R, and EXACTLY 1 B.

P(exactly n times) = P(one way) * total possible ways.

P(one way):
One way to get a good outcome is GRB.
P(1st marble is G) = 5/15. (15 marbles, 5 of them G)
P(2nd marble is R) = 4/14. (14 marbles left, 4 of them R)
P(3rd marble is B) = 6/13. (13 marbles left, 6 of them B)
Since we want all of these events to happen together, we multiply the fractions:
P(GRB) = 5/15 * 4/14 * 6/13 = 4/91.

Total possible ways:
Any arrangement of GRB will yield exactly 1 G, 1 R and 1 B.
Thus, the result above must be multiplied by the number of ways to arrange the 3 elements GRB.
Number ways to arrange GRB = 3! = 6.

Thus, P(1 of each color) = 4/91 * 6 = [spoiler]24/91[/spoiler].

More practice:
https://www.beatthegmat.com/marbles-t88714.html
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https://www.beatthegmat.com/probability- ... 14250.html
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https://www.beatthegmat.com/probability-t227448.html
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by Matt@VeritasPrep » Fri Oct 14, 2016 1:08 am
Visually, we could also think of it this way:

First pick: I need any color, let's say G. This would be G/15.

Second pick: I need any new color, let's say R. This would be R/14.

Third pick: I need the final color, here B. This would be B/13.

The probability of MY arrangement is thus

(G/15) * (R/14) * (B/13)

But I could've done GRB in any order, and there are 6 of these. So the answer is

6 * (G * R * B) / (15 * 14 * 13)

Replace G, R, and B with the number of green, red, and blue marbles, respectively, and you're done!
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