Problem Solving

gmattesttaker2 wrote:Hello,

I came across the following question that was posted in this forum some time ago:

Richard has 3 green, 2 red, and 3 blue balls in a bag.He randomly picks 5 from the bag without replacement.What is the probability that of the 5 drawn balls, Richard has picked 1red, 2 green, and 2 blue balls?

These were the answer choices:

8/28 9/28 10/28 10/18 11/18

The correct answer was given as 9/28.

The way I am trying to solve this is as follows:

( P(E) = Probability for the event to occur
n(E) = Number of ways the event could occur
n(S) = Total number of possible outcomes )


P(E) = P(E1). P(E2). P(E3)
= (n(E1)/n(S1)). (n(E2)/n(S2)). (n(E3)/n(S3))
= (2C1/8C1).(3C2/8C2).(3C2/8C2)

However, I am not sure if this approach is correct since I am not getting 9/28.

I was wondering if you can please explain why I am going wrong here? Thanks for your help.

Best Regards,
Sri

Brent@GMATPrepNow wrote:

dhonu121 wrote:Hi Brent,
In your solution,
1.How do you accomodate the fact that the balls are taken out without replacement ?
2.What have you done in your solution above, if int the question, balls were taken out with replacemnet ?

1. In the denominator, I'm looking for the number of ways to select 5 of the 8 balls. This does not allow for replacement. The same applies to the numerator.

2. This would require several steps. If you want to post it as a separate question, I'd be happy to answer it in a new thread. For the curious out there, if we allow for replacement each time, the probability = (30)(1/4)(3/8)(3/8)(3/8)(3/8)

Cheers,
Brent

gmat_and_me wrote:This approach doesn't seem correct. First of all this is a problem
without replacement. That means that every time you pick a ball,
the denominator for calculation of probability has to decrease by 1.

So, _as per your method_, the probability will have to be

2/8 * 3/7 * 2/6 * 3/5 * 2/4 = 3/280 (in the order R, G, G, B, B).
But that is only one arrangement. The resultant probability is probability
of getting R, G, G, B & B in that order. What about getting R, G, B, G, B ?
Again P is same. But there is one more arrangement to the original.
So there are more such possibilities. In fact, if you fix position of
R as 1, then G, G, B, B can be arranged in (4 * 3 * 2 * 1)/ (2 * 2) = 6
ways. Now R can move from positions 1 to 5 and that means that there are
30 such arrangements. So probability will be 30 * 3/280 = 9/28

Another way to look at the problem:

Total possibilities = 8 * 7 * 6 * 5 * 4

Now if there were only 5 balls, number of ways of selecting them
will be 5 * 4 * 3 * 2 * 1 (i.e R, G, G, B, B). For each of these
possibilities, you have one more duplicate because there are 2
reds. Hence the result now becomes 2 * (5 * 4 * 3 * 2 * 1). We
can select 2 greens in 3c2 ways => 3 more ways. Similarly for
Blue, there are 3 more ways (3c2).

So P = (2 * 3 * 3)(5 * 4 * 3 * 2 * 1)/(8 * 7 * 6 * 5 * 4)

HTH

gmattesttaker2 wrote:Hello,

I came across the following question that was posted in this forum some time ago:

Richard has 3 green, 2 red, and 3 blue balls in a bag.He randomly picks 5 from the bag without replacement.What is the probability that of the 5 drawn balls, Richard has picked 1red, 2 green, and 2 blue balls?

These were the answer choices:

8/28 9/28 10/28 10/18 11/18

The correct answer was given as 9/28.

The way I am trying to solve this is as follows:

( P(E) = Probability for the event to occur
n(E) = Number of ways the event could occur
n(S) = Total number of possible outcomes )


P(E) = P(E1). P(E2). P(E3)
= (n(E1)/n(S1)). (n(E2)/n(S2)). (n(E3)/n(S3))
= (2C1/8C1).(3C2/8C2).(3C2/8C2)

However, I am not sure if this approach is correct since I am not getting 9/28.

I was wondering if you can please explain why I am going wrong here? Thanks for your help.

Best Regards,
Sri

Brent@GMATPrepNow wrote:

gmattesttaker2 wrote: Richard has 3 green, 2 red, and 3 blue balls in a bag.He randomly picks 5 from the bag without replacement.What is the probability that of the 5 drawn balls, Richard has picked 1red, 2 green, and 2 blue balls?

A) 8/28
B) 9/28
C) 10/28
D) 10/18
E) 11/18

We can also solve this question using a combination of probability rules and counting methods:

Let's examine one possible outcome: RGGBB (the first ball is red, the second ball is green, the third ball is green, etc.)

What is the probability of this particular outcome?

Well, P(RGGBB) = P(red on 1st AND green on 2nd AND green on 3rd AND blue on 4th AND blue on 5th)
= P(red on 1st) x P(green on 2nd) x P(green on 3rd) x P(blue on 4th) x P(blue on 5th)
= (2/8)(3/7)(2/6)(3/5)(2/4)
= 3/280

Of course this is just the probability of one possible outcome. We need to consider others. For example, the outcome BGRGB also meets the criteria.

So, P(BGRGB) = P(blue on 1st AND green on 2nd AND red on 3rd AND green on 4th AND blue on 5th)
= P(blue on 1st) x P(green on 2nd) x P(red on 3rd) x P(green on 4th) x P(blue on 5th)
= (3/8)(3/7)(2/6)(2/5)(2/4)
= 3/280

Important: The probability of each acceptable outcome will equal 3/280. So the question now becomes, "In how many ways can we draw 1 red, 2 green and 2 blue balls?" In other words, in how many different ways can we arrange the letters in RGGBB?

This is what is known as a MISSISSIPI question, where we are arranging objects where some of the objects are identical.

Here we have 5 objects (letters) and we have 2 identical letters, and another 2 identical letters. So, we can arrange these 5 objects is 5!/2!2! ways. This is equal to 30

So there are 30 different ways to draw 1 red, 2 green and 2 blue balls, and the probability that each one occurs = 3/28

So, P(drawing 1 red, 2 green and 2 blue) = [spoiler](30)(3/28) = 9/28 = B[/spoiler]

Cheers,
Brent

Think of the 8 balls as G1, G2, G3, R1, R2, B1, B2, and B3

# of ways to select 5 balls (in any order)
= 8C5
= 56

# of ways to select 1 red, 2 green and 2 blue
- # of ways to select 1 red ball from 2 red balls (R1 and R2) = 2C1 = 2
- # of ways to select 2 green balls from 3 green balls = 3C2 = 3
- # of ways to select 2 blue balls from 3 blue balls = 3C2 = 3
So, total # of ways to select 1 red, 2 green and 2 blue = 2x3x3 = 18

P(select 1 red, 2 green and 2 blue) = 18/56 = 9/28

Brent@GMATPrepNow wrote:The first method I used was primarily an application of the Fundamental Counting Principle (FCP). . . along with some combinations.

Aside: People also refer to the application of the FCP as the "slot method"


The second method I used would be best described as using the rules of probability.

If you're interested, you can watch a free video about the FCP here: https://www.gmatprepnow.com/module/gmat-counting?id=775

Cheers,
Brent

Richard has 3 green, 2 red, and 3 blue balls in a bag.He randomly picks 5 from the bag without replacement.What is the probability that of the 5 drawn balls, Richard has picked 1red, 2 green, and 2 blue balls?

These were the answer choices:

8/28 9/28 10/28 10/18 11/18

Brent@GMATPrepNow wrote:

gmattesttaker2 wrote: Richard has 3 green, 2 red, and 3 blue balls in a bag.He randomly picks 5 from the bag without replacement.What is the probability that of the 5 drawn balls, Richard has picked 1red, 2 green, and 2 blue balls?

8/28 9/28 10/28 10/18 11/18

Think of the 8 balls as G1, G2, G3, R1, R2, B1, B2, and B3

# of ways to select 5 balls (in any order)
= 8C5
= 56

# of ways to select 1 red, 2 green and 2 blue
- # of ways to select 1 red ball from 2 red balls (R1 and R2) = 2C1 = 2
- # of ways to select 2 green balls from 3 green balls = 3C2 = 3
- # of ways to select 2 blue balls from 3 blue balls = 3C2 = 3
So, total # of ways to select 1 red, 2 green and 2 blue = 2x3x3 = 18

P(select 1 red, 2 green and 2 blue) = [spoiler]18/56 = 9/28[/spoiler]

Cheers,
Brent