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1. ## Few probability questions

I'm having trouble with a couple of problems.

Let a chip be taken at random from a bowl that contains 6 white chips, 3 red chips, and 1 blue chip. Let the random variable X = 1 if the outcome is a white chip; let X = 5 if the outcome is a red chip; and let X = 10 if the outcome is a blue chip.
a. Find the p.m.f. of X
b. Find the mean and standard deviation of X.

For a I need a function where you can plug in X=1 and get 6/10, X=5 and get 3/10, etc..

Once I have that I can find the mean and standard deviation. I've been trying to figure out that equation.
I have (x^2+5)/10x
It gives the correct x and y values for the 1st 2 x's but not the 3rd.

Also

Suppose X is a random variable such that E(X+4)=10 and E((X+4)^2)=116

How do I find the mean and variance?
Any help would greatly be appreciated.

Thanks

2. Originally Posted by sebastianzx6r
I'm having trouble with a couple of problems.

Let a chip be taken at random from a bowl that contains 6 white chips, 3 red chips, and 1 blue chip. Let the random variable X = 1 if the outcome is a white chip; let X = 5 if the outcome is a red chip; and let X = 10 if the outcome is a blue chip.
a. Find the p.m.f. of X
b. Find the mean and standard deviation of X.

For a I need a function where you can plug in X=1 and get 6/10, X=5 and get 3/10, etc..

Once I have that I can find the mean and standard deviation. I've been trying to figure out that equation.
I have (x^2+5)/10x
It gives the correct x and y values for the 1st 2 x's but not the 3rd.

Also

Suppose X is a random variable such that E(X+4)=10 and E((X+4)^2)=116

How do I find the mean and variance?
Any help would greatly be appreciated.

Thanks
Why don't we try:
P(X=x) = 6/10, x = 1
3/10, x = 5
1/10, x = 10

Also, the fact that it is a probability mass function of a discrete (as opposed to a continuous) distribution implies that over all x, the sum of the probabilities must equal 1. This certainly is true for our P(X).

For the second question, since the expected value is a linear operator and the expected value of a constant is the constant itself, E(X+4)=E(X)+4=10-->E(X)=6.

Also, E((X+4)^2)=E(X^2+8X+16)=E(X^2)+E(8X)+E(16)
=E(X^2)+8*E(X)+16=E(X^2)+8*6+16=116-->E(X^2)=52
And, Var(X)=E(X^2)-E(X)^2=52-6^2=16.

Help?

3. I'd really appreciate if you don't make duplicate threads next time. You can post anywhere and I will find it. I see I just wasted time answering your question in your duplicate thread.

4. Originally Posted by Ken
I'd really appreciate if you don't make duplicate threads next time. You can post anywhere and I will find it. I see I just wasted time answering your question in your duplicate thread.