An basket contains 3 black balls and 2 red balls

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late4thing: Junior | Next Rank: 30 Posts; Posts: 11; Joined: Thu Oct 15, 2015 2:14 pm

by late4thing » Wed Oct 05, 2016 5:17 pm

Treating this as 2 questions:
An urn contains 3 black balls and 2 red balls. Balls are selected one at a time without replacement.

Q1: What is the probability that the second ball is red?

Q2: What is the probability that the second ball is red, given that the first ball is black?

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crackverbal: Master | Next Rank: 500 Posts; Posts: 157; Joined: Mon Aug 16, 2010 7:30 pmFollowed by: 3 members; Location: India; Thanked: 65 times

by crackverbal » Wed Oct 05, 2016 11:37 pm

Hi late4thing,

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 same 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).

Coming to the question you have asked, we have an urn that contains 3 black and 2 red balls and we are dealing with the case of without replacement.

1. Probability that the second ball is red : There are two cases possible here.
1. P(BR) = 3/5 * 2/4 * 1 (since the order is specified) ----> 3/10
or 2. P(RR) = 2/5 * 1/4 * 1 ----> 1/10
Total probability = 3/10 + 1/10 = 2/5

2. Probability that the second ball is red given the first ball is black

P(BR) = 3/5 * 2/4 * 1 (since the order is specified) ----> 3/10

Hope this helps!

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MBA Challengers: Senior | Next Rank: 100 Posts; Posts: 51; Joined: Fri Sep 09, 2016 11:59 pm; Thanked: 20 times

by MBA Challengers » Thu Oct 06, 2016 1:02 am

Hi late4thing,

An easy way to solve is to remember that here we are talking about a pool of 5 balls - 3 black and 2 red.

Coming to the first question, the probability of the second ball being red, where the balls are being drawn without replacement.
There will be 2 scenarios in this question, which will also end up covering your second question:

Scenario 1: The first ball drawn is black - this implies drawing one of the 3 balls from a total pool of 5. Thus, that would mean 3/5.

Now, after this, there would be 4 remaining balls of which the second ball must be red (2 out of the 4 remaining). This would mean 2/4
So, scenario 1 can be summarised as 3/5 * 2/4 which simplifies to 3/10. Incidentally, this answers the 2nd question.

Scenario 2: The first ball drawn is red - this implies drawing one of the 2 balls from a total pool of 5., This would mean 2/5.
Now, after this, there would be 4 remaining balls of which the second ball must be red (1 out of the 4 remaining as one of them has already been drawn out). This would mean 1/4
So, scenario 2 can be summarised as 2/5 * 1/4 which simplifies to 1/10.

To combine the 2 scenarios, the 2 ratios need to be added as in case of a probability, when we are looking at 2 option instances or scenarios, the individual probabilities are added up.

So, the total probability of the 2nd ball drawn being red (without replacement) = 3/10 + 1/10 = 4/10. This can be further simplified to 2/5.

Question 2: As explained in question 1, this will be 3/10

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late4thing: Junior | Next Rank: 30 Posts; Posts: 11; Joined: Thu Oct 15, 2015 2:14 pm

by late4thing » Thu Oct 06, 2016 6:44 am

Thank you both, this is very helpful. So just to confirm I understood this, let's say if I want to find the probability that the first ball is black, given that second ball is black as well.

P(BB) = 2/5(Since two black balls) * 3/4(Since second is definitely black). = 3/10?

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regor60: Master | Next Rank: 500 Posts; Posts: 416; Joined: Thu Oct 15, 2009 11:52 am; Thanked: 27 times

by regor60 » Thu Oct 06, 2016 8:19 am

On question 2, let's lay out the scenarios:

RR first and second balls both red: Probability 2/5 x 1/4 =.1

BR first ball black, second red: Probability 3/5 x 2/4 = .3

RB first ball red, second black: Probability 2/5 x 3/4 =.3

BB both balls black " 3/5 x 2/4 =.3

Notice the total probability equals 1, so we haven't missed a scenario

The question asks for the probability of whether the second ball is Red given that the first ball is black. So, we're being told to focus on only the scenarios where the first ball is black.

Consulting the table of scenarios above where the first ball is black, we have BB and BR. Each has a probability of 0.3. So the probability of a Red ball being picked after a Black ball is first picked is (.3+.3)/.6 or 1/2.

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crackverbal: Master | Next Rank: 500 Posts; Posts: 157; Joined: Mon Aug 16, 2010 7:30 pmFollowed by: 3 members; Location: India; Thanked: 65 times

by crackverbal » Fri Oct 07, 2016 10:34 am

Hi late4thing,

Perfect! Looks like you got the hang of this

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regor60: Master | Next Rank: 500 Posts; Posts: 416; Joined: Thu Oct 15, 2009 11:52 am; Thanked: 27 times

by regor60 » Mon Oct 10, 2016 10:05 am

late4thing wrote:Thank you both, this is very helpful. So just to confirm I understood this, let's say if I want to find the probability that the first ball is black, given that second ball is black as well.

P(BB) = 2/5(Since two black balls) * 3/4(Since second is definitely black). = 3/10?

No, this is not correct.

Your wording asks for the probability that the first ball is black, given that two balls have been drawn and the second ball is black.

This is a posterior probability question. You can look this concept up.

This question is solved by identifying the two scenarios in which the second ball is black and their associated probabilities. They are both 0.3.

Therefore, the probability that the first ball is black, given that the second is black, is 0.3/(0.3+0.3), or 0.5.

So, obviously, the probability that the first ball is red is also 0.5

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Matt@VeritasPrep: GMAT Instructor; Posts: 2630; Joined: Wed Sep 12, 2012 3:32 pmFollowed by: 119 members; Location: East Bay all the way; Thanked: 625 times; GMAT Score:780

by Matt@VeritasPrep » Fri Oct 14, 2016 2:10 am

regor60 wrote:
This is a posterior probability question. You can look this concept up.

IMO it's not really a good idea to answer a simple question with a more complex set of ideas, but if the OP or anyone else is actually keen on reading a lot more on this, the term to Google is Bayesian probability. We don't need to delve too far into it, though, since the basic idea is the same. We find our probability by doing

(Target probability) / (Sum of all possible probabilities)

In our case, the target is B first AND B second, and the possible probabilities are ANY that have B second.

In our case, GIVEN THAT the second ball is black, we have two ways of getting there:

Way 1: The first ball was black

In this case, we have B * B, or (3/5) * (2/4) = 3/10.

Way 2: The first ball was red

In this case, we have R * B, or (2/5) * (3/4) = 3/10.

We want P(B first, B second) given B second, so we do

(Target Scenario) / (All Possible Scenarios) =>

P(B, then B) / (P(B, then B) + P(R, then B)) =>

3/10 / (3/10 + 3/10) =>

1/2

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Matt@VeritasPrep: GMAT Instructor; Posts: 2630; Joined: Wed Sep 12, 2012 3:32 pmFollowed by: 119 members; Location: East Bay all the way; Thanked: 625 times; GMAT Score:780

by Matt@VeritasPrep » Fri Oct 14, 2016 2:15 am

This is overkill, but if we use Bayes' rule, this is pretty quick too:

P(Black first, given Black second) = P(Black second, given Black first) * P(Black second) / P(Black first)

=>

P(B1 given B2) * P(B1) = P(B2 given B1) * P(B2)

Since P(B2 given B1) and P(B1) are easy to find (they're 3/5 and 1/2), we can substitute them in =>

P(B1 given B2) * 3/5 = 1/2 * P(B2)

P(B1 given B2) = 5/6 * P(B2)

P(B2) = P(R1,B2) + P(B1,B2) = (2/5)*(3/4) + (3/4)*(2/5) = 3/5

so

P(B1 given B2) = 5/6 * (3/5) = 1/2

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