An urn contains six balls numbered 1 to 6. One after another 3 balls

Question

An urn contains six balls numbered 1 to 6. One after another 3 balls are drawn from the urn and put aside. What is the probability that the number on the 1st ball is less than the number on the 2nd ball which in turn is less than the number on the 3rd ball?

A. 1/12
B. 1/8
C. 1/6
D. 1/4
E. 1/3

A. 1/12
B. 1/8
C. 1/6
D. 1/4
E. 1/3

Regor60 · Accepted Answer

Since the balls are being drawn without replacement, we know they each will have a different number.😉

There is only one order of those 3 numbers that will satisfy the question.

The 3 numbers can be ordered in 3! = 6 ways.

Probability: 1/6

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ThatDudeKnows · Answer

We need to find the number of ways to "win" and divide by the total number of ways.

For the total number of ways:
There are 6 ways to draw the first ball. Once we've done that, there are 5 ways to draw the second ball. Once we've done that, there are 4 ways to draw the third ball. So 6*5*4=120 total ways.

For the number of ways to win:
We could learn and deploy real math, or we could just use simple observation and be 100% sure we aren't using the wrong formula.
If the first ball is 6 or 5, there would be no way to win.
If the first ball is 4, the only way to win is for the second ball to be 5 and the third ball to be 6. So, one way to win.
If the first ball is 3, we could win with the following combinations of second and third balls: 4/5, 4/6, 5/6. So, three ways to win.
If the first ball is 2, we could win with the following combinations of second and third balls: 3/4, 3/5, 3/6, 4/5, 4/6, 5/6. So, six ways to win.
If the first ball is 1, we could win with the following combinations of second and third balls: 2/3, 2/4, 2/5, 2/6, 3/4, 3/5, 3/6, 4/5, 4/6, 5/6. So, ten ways to win.
Add those up and there are 20 ways to win.

20/120 = 1/6

Answer choice C

ThatDudeKnowsProbability

sanjitscorps18 · Answer

The first call has to be <=4. So it can be 1,2, 3, 4
a < b < c

First ball number 4, we have = 2C1 x 1C1 /2 = 1 (Dividing by 2 for symmetric values removal as we would also get the triplet 4,6,5 that does not fit the criteria)
it is 4,5,6

First ball number 3, we have = 3C2 x 2C1 / 2 = 6/2 = 3
3,4,5 or 3,5,6 or 3,4,6

First ball number 2, we have = 4C3 x 3C2 / 2 = 4 x 3 / 2 = 6
2,3,4 or 2,3,5 or 2,3,6
2,4,5 or 2,4,6
2,5,6

First ball number 1, we have = 5C4 x 4C3 / 2 = 4 x 3 / 2 = 10
1,2,3 or 1,2,4 or 1,2,5 or 1,3,6
1,3,4 or 1,3,5 or 1,3,6
1,4,5 or 1,4,6
1,5,6

Hence total valid ways is 1 + 3 + 6 + 10 = 20
Total ways = 6 x 5 x 4 = 120

Probability = 20/120 = 1/6
Option C

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An urn contains six balls numbered 1 to 6. One after another 3 balls

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An urn contains six balls numbered 1 to 6. One after another 3 balls

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