# Thread: stuck with simple question

1. ## stuck with simple question

each of three balls are randomly places into one of three bowls. find the mean and standard deviation for Y= the number of empty bowls. what is the probability that the value of Y falls within 2 standard deviations of the mean?

my working:

let X = the fill bowl.

E(X) = 1/3 x 1/ 3 = 1/9

then im stuck..

2. If i understand the question correctly, i think that solution is this:

Y......................0..........1..........2.... ......Sum
p(y)..................1/7.......3/7.......3/7........1
y.P(y)................0..........3/7.......6/7........9/7
$[p(y)-yP(y)]^2$...81/49.....42/49....21/49

So, probability that the value of Y falls within 2 standard deviations is

Y=1 and Y=2, P(Y=1)+P(Y=2)=3/7+3/7=6/7

3. Originally Posted by alexandrabel90
each of three balls are randomly places into one of three bowls. find the mean and standard deviation for Y= the number of empty bowls. what is the probability that the value of Y falls within 2 standard deviations of the mean?

my working:

let X = the fill bowl.

E(X) = 1/3 x 1/ 3 = 1/9

then im stuck..
Let $X_i$ be the bowl in which the ith ball is placed. Then all the triples $(X_1, X_2, X_3)$ where $X_i = 1, 2, 3$ are equally likely. The straightforward way to solve the problem is to simply list all 27 triples and count the number of empty bowls in each case. This will allow you to find the distribution of Y, then the mean and standard deviation will be easy to compute.

4. I have counted only 10 triplets:

300,030,003 - two empty bowls
210,201,120,102,021,012 - one empty bowl
111 - zero empty bowls

Did I miss something?

5. Originally Posted by losm1
I have counted only 10 triplets:

300,030,003 - two empty bowls
210,201,120,102,021,012 - one empty bowl
111 - zero empty bowls

Did I miss something?
You are counting the wrong kind of triplets. Count triplets of the form $(X_1, X_2, X_3)$ where $X_i = 1, 2, 3$ is the bowl in which the ith ball is placed.
(1,1,1) -- 2 empty bowls
(1,1,2) -- 1 empty bowl
(1,1,3) -- 1 empty bowl
(1,2,1) -- 1 empty bowl
etc.

The reason that this is important is that all the triplets $(X_1, X_2, X_3)$ are equally likely. If you count according to the number of balls in each bowl, i.e. (3 balls, 0 balls, 0 balls), (2 balls, 1 ball, 0 balls), etc., then the arrangements are NOT all equally likely. (It's a very easy mistake to make, and I myself have made it many times.)

For example, suppose you had only two balls and two bowls. You might think that (2 balls, 0 balls), (0 balls, 2 balls), and (1 ball, 1 ball) are equally likely, but they are not.
To get (2 balls, 0 balls) you must have ball 1 in bowl 1 and ball 2 in bowl 1, with probability
(1/2) x (1/2) = (1/4).
But you get (1 ball, 1 ball) in two ways: (1) Ball 1 in bowl 1 and ball 2 in bowl 2, and (2) ball 2 in bowl 1 and ball 1 in bowl 2, with probability
(1/2) x (1/2) + (1/2) x (1/2) = 1/2.

I hope you see what I mean.

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# Each of three balls are randomly placed into one of three bowls. Let X be the number of empty bowls. i. ii. iii. Define the sample space S.

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