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

Probability of Picking 1 green (3/10)*probability of orange (4/9)*probability of another orange (3/8)=1/20. There are three ways this can happen goo
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1/20*3=3/20

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

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

Thx

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.

Thanks a lot
Makes sense, just thought it the wrong way around

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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

10C3 = 120

3C1 = 3

4C2 = 6

6 * 3 = 18

18/120 = 6/40 = 3/20

Answer is E.

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