A bag contains 3 blue disks, 3 green disks and 4 orange disks. If thre

Question

A bag contains 3 blue disks, 3 green disks and 4 orange disks. If three disks are selected random from the bag, what is the probability that 1 of the disks will be green and the other 2 of the disks will be orange?

A) 1/30
B) 1/20
C) 3/40
D) 1/10
E) 3/20

A) 1/30
B) 1/20
C) 3/40
D) 1/10
E) 3/20

My solution:

Total disks: 10
The prob. of selecting a green=3/10
Once a green disk is selected, we are left with 9 total disks
Then the probability of selecting an orange disk = 4/9
Once one orange is selected, we are left with 8 total disks
Then the probability of selecting the second orange disk = 3/8

Total Probability = 3/10 * 4/9 * 3/8 ==> 1/20

However thats not the correct answer. What am I missing?

Appreciate your help!

Thanks

mikemcgarry · Accepted Answer

Dear HarveyKlaus , I'm happy to respond. :-)

One very tricky thing about probability is that, when a question asked about the probability of a certain result, we have to be careful not to assume that there is only one way to get that result. For example, you approach explicitly assumes that the green disk was selected first. What if it wasn't? You calculated the probability of getting to that result by a specific route, but this excludes other ways to get to that same result. One way to approach this would be a kind of tree approach--ultimately, the choices, in order, could be GOO or OGO or OOG, and all three of those would have to be calculated and added. P(GOO) = 1/20 (what you calculated) P(OGO) = (4/10)*(3/9)*(3/8) = 1/20 P(OOG) = (4/10)*(3/9)*(3/8) = 1/20 Add P(total) = 3/20 Another approach would be to apply counting techniques. See: GMAT Probability and Counting Techniques Think of it this way. Suppose the discs are give letters, A - J. Discs A, B, & C are green. Discs D-G are orange, and rest are blue. Question #1 : how many different sets of three altogether could we pick? 10C3 = \frac{10!}{(7!)(3!)} = \frac{10*9*8}{3*2*1} = 10*3*4 = 120 . That's the denominator. Question #2 : how many different sets of three involve just one green and two oranges? Three ways to choose the single green member, and 4C2 = 6 ways to choose the two orange members. By the Fundamental Counting Principle , number of ways = 3*6 = 18 . That's the numerator. Probability = 18/120 = 3/20 Here are a couple more blogs you may find helpful: GMAT Data Sufficiency Practice Questions on Probability GMAT Advanced Probability Problems Does this make sense? Mike :-)

RMD007 · Answer

Probability of complex events = Individual Events * their arrangements.

In this case - it is equal to -- 
Probability of 1 green - 3/10
Probability of 1 orange - 4/9 (Since 1 green has already selected.)
Probability of 2nd orange - 3/8

Their arrangements = 3!/2!*1! = 3

Probability = 3/10*4/9*3/8*3 = 3/20

GMATforSeaCliff · Answer

Could you explain the part with " Their arrangements = 3!/2!*1! = 3 " ?

Thx

RMD007 · Answer

Sure. I meant the arrangements of the selected events i.e how many ways these events can arrange themselves. Total we have 3 events so the arrangement is 3! and out of them 2 are same i.e. selecting orange disk hence 3!/2!

Hope its clear.

ScottTargetTestPrep · Answer

We are given that in a bag there are 3 blue disks, 3 green disks, and 4 orange disks. We need to determine the probability that from 3 selected disks, 1 will be green and 2 will be orange.

Let’s first determine the total number of ways to select 1 green disk from 3.
 
3C1 = 3

Next, let’s determine the number of ways to select 2 orange disks from 4.

4C2 = (4 x 3)/2! = (4 x 3)/(2 x 1) = 6 ways

So, there are 3 x 6 = 18 ways to select 1 green and 2 orange disks.

Now let’s determine the number of ways to select 3 disks from 10:

10C3 = (10 x 9 x 8)/3! = (10 x 9 x 8)/(3 x 2) = 5 x 3 x 8 = 120

Thus, the probability of selecting 1 green disk and 2 orange disks is 18/120 = 3/20.

Alternate Solution:

Selecting 3 disks at once is equivalent to selecting them one at a time without replacement. Remember, we are determining the probability of selecting 1 green disk and 2 orange disks from 10 total disks.

If we select the green disk first, since there are 3 green disks and 10 total disks, there is a 3/10 chance that the green disk will be selected. Next, since there are 4 orange disks and 9 disks left, there is a 4/9 chance an orange disk will be selected. Similarly, for the third disk chosen, there is a 3/8 chance another orange disk will be selected. However, there are 3 different ways to select the 1 green disk and 2 orange disks:

G - O - O

O - G - O

O - O - G

Each of these 3 ways has the same probability of occurring. Thus, the total probability is:

3 x (3/10 x 4/9 x 3/8) = 3/20

Answer: E

RastogiSarthak99 · Answer

Either one has to solve it by 1/20 * 3 or use combination

Such as:

total ways = 10c3
1 Blue = 3c1
2 Orange = 4c2

(3c1 * 4C2)/10c3

= 3/20

E

adroitpensador · Answer

Bunuel   ScottTargetTestPrep

Fundamental Gap:

Could you help me differentiate between Combinations (3C1*4C2)/(10C3) method and Division method (4/10*3/9*2/9*3!/2!)? Or if you can share an article to read more on it.
I use these interchangeably. But during 1-off instance, I ended up including 3!/2! (essentially the arrangements) along with the Combinations method.

Bunuel · Answer

The key here is maintaining consistency between the numerator and denominator, and understanding what multiplying by 3!/2! actually does.

We multiply 4/10 * 3/9 * 3/8 by 3!/2! because the GOO outcome can occur in 3 different ways:

GOO (first disk is green, second is orange, third is orange) 
 OGO (first disk is orange, second is green, third is orange) 
 OOG (first disk is orange, second is orange, third is green)

These are three different scenarios, and we need to account for all of them when calculating the probability.

Alternatively, we could calculate:

P(GOO) + P(OGO) + P(OOG) =

= (3/10 * 4/9 * 3/8) + (4/10 * 3/9 * 3/8) + (4/10 * 3/9 * 3/8) =

= (3/10 * 4/9 * 3/8) * 3

Notice that both approaches yield the same result: (3/10 * 4/9 * 3/8) * 3

Now, when using the combinations approach, we calculate 3C1 * 4C2 / 10C3, where the numerator represents the number of different groups of three disks where 1 is green and 2 are orange, and the denominator represents the total number of ways to select 3 disks from 10. It's crucial to observe that both the numerator and the denominator give unordered groups of three disks (meaning XYZ is counted only once, without considering XZY, YXZ, and other arrangements). This maintains consistency between them.

Therefore, multiplying the numerator by 3!/2!, which accounts for arrangements, would break the consistency. In the numerator, you’d have the number of groups of three disks considering arrangement, while in the denominator, it would still be without it. Thus, multiplying by 3!/2! does not make sense.

Hope that clears things up!

adroitpensador · Answer

Thanks, this does help.

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