PLEASE EXPLAIN PROBABILITY WITH AN WITHOUT REPLACENT!!!

[tsotnemir]

Want to clear this question:

Jar contains 3 green, 2 red and 3 blue balls. If we randomly pick 5 balls WITH replacement. What is the probability that of the 5 drawn balls picked 1 would be red, 2 green, and 2 blue balls?

My approach:
(5!/2!*2!)*(2/8*3/8*3/8*3/8*3/8)

AND:

Jar contains 3 green, 2 red and 3 blue balls. If we randomly pick 5 balls WITHOUT replacement. What is the probability that of the 5 drawn balls picked 1 would be red, 2 green, and 2 blue balls?

My approach:
(5!/2!*2!)*(2/8*3/7*2/6*3/5*2/4)

Is this correct?

Please advise.

Thanks.

P.S. This is just a question to clear my doubts, it's not from any source.

[tsotnemir]

Anybody?

Instructors... Moderators... Anyone... Please.

[esledge]

I get the same result you do for the WITH replacement question.

But for the WITHOUT replacement question, I think your probability is too low. An alternative way to think about these questions is to think "What's left in the jar after all the selections have been made?" (This simplifies things, for me at least, especially when more balls are selected than are left behind.)

Here's the question again:

Jar contains 3 green, 2 red and 3 blue balls. If we randomly pick 5 balls WITHOUT replacement. What is the probability that of the 5 drawn balls picked 1 would be red, 2 green, and 2 blue balls?

In other words, what is the probability that the 3 balls that remain in the jar are 1 each of R,G,B?

First, ignoring color altogether, there are 8!/3!5! = (8)(7)(6)/(3)(2) = 56 ways to select 5 balls and leave 3 in the jar.

How many ways to leave 1 R in the jar? 2!/1!1! = 2
How many ways to leave 1 G in the jar? 3!/1!2! = 3
How many ways to leave 1 B in the jar? 3!/1!2! = 3

Thus, the total # of ways to leave 1 each of R,G,B = 2*3*3 = 18. (Note: order doesn't matter--in the jar is in the jar.)

Probability = 18/56 = 9/28.

[tsotnemir]

I get the same result you do for the WITH replacement question.

But for the WITHOUT replacement question, I think your probability is too low. An alternative way to think about these questions is to think "What's left in the jar after all the selections have been made?" (This simplifies things, for me at least, especially when more balls are selected than are left behind.)

Here's the question again:

Jar contains 3 green, 2 red and 3 blue balls. If we randomly pick 5 balls WITHOUT replacement. What is the probability that of the 5 drawn balls picked 1 would be red, 2 green, and 2 blue balls?

In other words, what is the probability that the 3 balls that remain in the jar are 1 each of R,G,B?

First, ignoring color altogether, there are 8!/3!5! = (8)(7)(6)/(3)(2) = 56 ways to select 5 balls and leave 3 in the jar.

How many ways to leave 1 R in the jar? 2!/1!1! = 2
How many ways to leave 1 G in the jar? 3!/1!2! = 3
How many ways to leave 1 B in the jar? 3!/1!2! = 3

Thus, the total # of ways to leave 1 each of R,G,B = 2*3*3 = 18. (Note: order doesn't matter--in the jar is in the jar.)

Probability = 18/56 = 9/28.

Thanks Emily,

Your approach is clear for. It's funny but my approach also gives the same result for the WITHOUT replacement: 9/28.

I guess as both got the same result mine is also right.

[esledge]

Ah, sorry about that--I did the computation of your resulting probability product incorrectly! Yes, it is a good sign that we got the same answer by two different methods.