Probability mean, variance, uniform

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Hi there ,

I have some simple (on 1st look) questions. I just need some help.

http://img59.imageshack.us/img59/3018/5hby.png

1st question.
I think the expected value should be 150 since the probability of getting a head is 1/2 .. 300x1/2 = 150 so the mean should also be 150.
the variance i think is 1/2x1/2x300 = 75
2nd question.
Here I have something for 6 eggs with 2 bad and 4 good.. but when I use the same logic for 600 eggs total, 20 bad I get very big number. So my logic for 6 eggs 4 good 2 bad is:
Let X be the number of bad eggs picked. If X=0, we have 4/6x3/5= 6/15 =2/5
P(X-1)= 4/6x2/5 + 2/6x4/5 =8/15
P(X=2) = 1/15 so the EX is 0x6/15+1x8/15+2x1/15 = 2/3
But how I can use this logic for the bigger problem?
3rd question
Why do we have to care about the 1st line? Since the probability of picking one of the 2 coins is 1/2.
a) 2 heads P(HHT)+P(THH)+P(HTH)=3/8?
b)if the 1st one is heads so we need one head out of 2 flips so isnt it just 1/2?
or I am missing something here?
4th question
I think it will be better to start from part b)
b) I think the length function is f(x)=min(x,1-x) as x is on the interval [0,0.5] and 1-x is on the interval [0.5,1], We have to find E(f(X)) which is equal to:

http://img707.imageshack.us/img707/1160/3taz.jpg

a) from part b) since the average size is 1/4 it can be either left or right segment. So the left can be either 1/4 or 3/4
c) don't have idea about this question.. intuitively think something about 2/3
but cannot explain why Very Happy

Thanks

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@gl0ck,

1) mean = 300 * 1/2 = 150, variance =300 * 1/2 * 1/2 = 75

2) P(n bad eggs) = C(20,n)*C(580,20-n)/C(600,20)
The sum of the products is 2/3.

3a) P(2 heads with coin 1) = 3 * 1/8 = 3/8
P(2 heads with coin 2) = 3 * 4/27 = 4/9
P(2 heads) = (1/2 * 3/8) + (1/2 * 4/9) = 59/144

3b) P(you have the first coin) = (1/2) / (1/2 + 2/3) = 3/7
P(you have the second coin) = (2/3) / (1/2 + 2/3) = 4/7
If you have the first coin, P(1 more head) = 1/2
If you have the second coin, P(1 more head) = 4/9
P(2 heads) = (3/7 * 1/2) + (4/7 * 4/9) = 59/126

4a) 1/2 (must be the same as the expected length of the right segment)
4b) yep, 1/4
4c) 2/3 - the break can occur in these regions: (0-1/3] and [2/3, 1)

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@markr,

Thanks for the reply, but can you explain a bit more the 2nd question? It should be trivial but it seems that I cannot see it

1 Reply

@gl0ck,

P(n bad eggs) = C(20,n)*C(580,20-n)/C(600,20)

There are C(20,n) ways to choose n bad eggs. There are C(580,20-n) ways to choose such that the rest of the eggs are good. Multiply these to get the number of ways to choose exactly n bad eggs. Divide by C(600,20) to get the probability of choosing exactly n bad eggs.

To compute the expected number of bad eggs:
sum(n = 0 to 20, n*P(n))

1 Reply

@markr,

thanks man, you saved me again Smile

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Probability mean, variance, uniform

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