Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks

Why do you have 5!/2!2!? Does order matter? No...

Let's clear out this one:

CASE I.
Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 5 from the bag WITH replacement. What is the probability that of the 5 drawn balls, Rich has picked 1 red, 2 green, and 2 blue balls?

The probability would be: \(\frac{2}{8}*(\frac{3}{8}*\frac{3}{8})*(\frac{3}{8}*\frac{3}{8})*\frac{5!}{2!2!}=14.8%\)

CASE II.
If the problem was WITHOUT replacement:

Then the probability would be: \(\frac{2}{8}*(\frac{3}{7}*\frac{2}{6})*(\frac{3}{5}*\frac{2}{4})*\frac{5!}{2!*2!}=\frac{9}{28}=32.14%\)

OR different way of counting: \(\frac{2C1*3C2*3C2}{8C5}=\frac{9}{28}\)


The only difference we have that in FIRST case we are always drawing from 8 balls and the initial number of balls in bag remains the same, as we are putting withdrawn balls back (we have replacement).

In the SECOND case after each withdrawal we are reducing total number of balls by 1 AND reducing the withdrawn balls number by 1.

We are then multiplying this, in BOTH CASES, by \(\frac{5!}{2!2!}\), because we want to calculate number of ways this combination is possible (1 red, 2 green, 2 blue), total combination of 5 balls is 5!, BUT as among these 5 balls we have 2 green and 2 red, we should divide 5! by 2!2! to get rid of repeated combinations.

The second way of calculating "WITHOUT replacement case" just counts the number of combinations of picking desired number of each ball from its total number (1 red out of two, 2 green out of three, and 2 blue out of three), multiplies them and divides this by the number of ways we can pick total of 5 balls out of 8.

Hope now it's clear.

"The second way of calculating "WITHOUT replacement case" just counts the number of combinations of picking desired number of each ball from its total number (1 red out of two, 2 green out of three, and 2 blue out of three), multiplies them and divides this by the number of ways we can pick total of 5 balls out of 8."

Bunuel,
About the above statement, can you explain the way of calculating Case I this way - using the counting approach. How would the calculation be different with the WITH case.

Also as a general doubt when one says 5 balls are drawn randomly WITH replacement - drawing 5 balls randomly suggests I grab 5 directly, however with the WITH replacement clause, it seems to indicate that I draw one then put it back, then draw another and then replace it, essentially drawing balls 1 by 1.. Is this understanding correct?

Let's clear out this one:

CASE I.
Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 5 from the bag WITH replacement. What is the probability that of the 5 drawn balls, Rich has picked 1 red, 2 green, and 2 blue balls?

The probability would be: \(\frac{2}{8}*(\frac{3}{8}*\frac{3}{8})*(\frac{3}{8}*\frac{3}{8})*\frac{5!}{2!2!}=14.8%\)

CASE II.
If the problem was WITHOUT replacement:

Then the probability would be: \(\frac{2}{8}*(\frac{3}{7}*\frac{2}{6})*(\frac{3}{5}*\frac{2}{4})*\frac{5!}{2!*2!}=\frac{9}{28}=32.14%\)

OR different way of counting: \(\frac{2C1*3C2*3C2}{8C5}=\frac{9}{28}\)


The only difference we have that in FIRST case we are always drawing from 8 balls and the initial number of balls in bag remains the same, as we are putting withdrawn balls back (we have replacement).

In the SECOND case after each withdrawal we are reducing total number of balls by 1 AND reducing the withdrawn balls number by 1.

We are then multiplying this, in BOTH CASES, by \(\frac{5!}{2!2!}\), because we want to calculate number of ways this combination is possible (1 red, 2 green, 2 blue), total combination of 5 balls is 5!, BUT as among these 5 balls we have 2 green and 2 red, we should divide 5! by 2!2! to get rid of repeated combinations.

The second way of calculating "WITHOUT replacement case" just counts the number of combinations of picking desired number of each ball from its total number (1 red out of two, 2 green out of three, and 2 blue out of three), multiplies them and divides this by the number of ways we can pick total of 5 balls out of 8.

Hope now it's clear.

Does the order matter in this question??? I dont think so.. as the question says... wat is the probability of getting 1 red, 2 green and 2 blue... and says nothing of the arrangement. Please comment.

Responding to a pm:
himanshu

Why are we multiplying by \(\frac{5!}{2!2!}\)? This is the equivalent of 2 in the previous question. There we picked 2 balls so we had 2 different cases RW and WR
Here we are picking 5 balls RGGBB. How many different cases can we have?
RGGBB
RGBGB
RBBGG
BBGGR
BGBGR
and many more...
The total number of cases is 5!/2!*2! (We divide by 2!s because there are 2 blues and 2 greens.)

NOte that we multiply by this number in both with and without replacement. The only thing that changes in without replacement case is the probability as discussed before.

I couldn figure out yet why dont we find simply the probability to select. For example selectin 2 green balls of 10 balls just 2/10. what is the difference here? dont it is asked simply to find prob of selecting these balls?

Let's clear out this one:

CASE I.
Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 5 from the bag WITH replacement. What is the probability that of the 5 drawn balls, Rich has picked 1 red, 2 green, and 2 blue balls?

The probability would be: \(\frac{2}{8}*(\frac{3}{8}*\frac{3}{8})*(\frac{3}{8}*\frac{3}{8})*\frac{5!}{2!2!}=14.8%\)

CASE II.
If the problem was WITHOUT replacement:

Then the probability would be: \(\frac{2}{8}*(\frac{3}{7}*\frac{2}{6})*(\frac{3}{5}*\frac{2}{4})*\frac{5!}{2!*2!}=\frac{9}{28}=32.14%\)

OR different way of counting: \(\frac{2C1*3C2*3C2}{8C5}=\frac{9}{28}\)


The only difference we have that in FIRST case we are always drawing from 8 balls and the initial number of balls in bag remains the same, as we are putting withdrawn balls back (we have replacement).

In the SECOND case after each withdrawal we are reducing total number of balls by 1 AND reducing the withdrawn balls number by 1.

We are then multiplying this, in BOTH CASES, by \(\frac{5!}{2!2!}\), because we want to calculate number of ways this combination is possible (1 red, 2 green, 2 blue), total combination of 5 balls is 5!, BUT as among these 5 balls we have 2 green and 2 red, we should divide 5! by 2!2! to get rid of repeated combinations.

The second way of calculating "WITHOUT replacement case" just counts the number of combinations of picking desired number of each ball from its total number (1 red out of two, 2 green out of three, and 2 blue out of three), multiplies them and divides this by the number of ways we can pick total of 5 balls out of 8.

Hope now it's clear.