- Wed Jul 07, 2010 2:06 am
- Forum: Statistics
- Topic: Let X, Y, be ind. Poisson rand. vars; mean 1; find P(X+Y=4),
- Replies:
**7** - Views:
**9631**

I couldnt understand how to evaluate P(A+B=4) as is it P(A+B=4) = P(A) + P(B) = 4 or is it some other way If X is poisson 1 and Y is poisson 1 then X+Y is Poisson 2 but how should i represent it in my solution P(A+B) = 1+1 = 2 but what about the '4' I am having a hard time how to formulate this ? s...

- Tue Jul 06, 2010 2:53 am
- Forum: Statistics
- Topic: Need to know which formulae which has been used
- Replies:
**2** - Views:
**5193**

Thanks!!If you're not sure what formula this is, try working backwards from the "sample" formulas they've given you. Which one uses in the numerator? Which one also uses in the denominator?

- Tue Jul 06, 2010 2:50 am
- Forum: Statistics
- Topic: Let X, Y, be ind. Poisson rand. vars; mean 1; find P(X+Y=4),
- Replies:
**7** - Views:
**9631**

Start with this... http://en.wikipedia.org/wiki/Poisson_distribution#Properties Thanks so \mu = \lambda = 1 P(A+B=4) = ? stuck here !!! From the link I provided look at "Sums of Poisson-distributed random variables" I couldnt understand how to evaluate P(A+B=4) as is it P(A+B=4) = P(A) + P(B) = 4 o...

- Mon Jul 05, 2010 4:04 pm
- Forum: Statistics
- Topic: Let X, Y, be ind. Poisson rand. vars; mean 1; find P(X+Y=4),
- Replies:
**7** - Views:
**9631**

ThanksStart with this...http://en.wikipedia.org/wiki/Poisson_di ... PropertiesQuestion :

Suppose X and Y are independent Poisson random variable , each with the mean 1 , obtain

i) P(X+Y=4)

ii) E[(X+Y)^{2}]

so = = 1

P(A+B=4) = ?

- Mon Jul 05, 2010 11:41 am
- Forum: Statistics
- Topic: Let X, Y, be ind. Poisson rand. vars; mean 1; find P(X+Y=4),
- Replies:
**7** - Views:
**9631**

Question :

Suppose X and Y are independent Poisson random variable , each with the mean 1 , obtain

i) P(X+Y=4)

ii) E[(X+Y)^{2}]

Suppose X and Y are independent Poisson random variable , each with the mean 1 , obtain

i) P(X+Y=4)

ii) E[(X+Y)

- Mon Jul 05, 2010 3:44 am
- Forum: Statistics
- Topic: prob. of 3 red, 2 white, 1 blue (is the answer correct)
- Replies:
**5** - Views:
**6189**

Sorry silly mistakes ....Assuming the decimal points stand for multiplication, could you please show the steps in your arithmetic with the fractions?

P(R).P(R).P(R) + P(W).P(W) + P(B) = 0.4289

- Sun Jul 04, 2010 4:08 pm
- Forum: Statistics
- Topic: What is the probability that he really knows the answer
- Replies:
**4** - Views:
**5679**

If u know the answer can u please let me know ?That number seems a little low.then p(knows|correct)=p(correct|knows)*p(knows)/p(correct)

=1*1/2*1/(p(correct|knows)p(knows)+p(correct|guess)p(guess)+p(correct|copy)p(copy))

=.5*1/(p(knows)+p(guess)+1/48)

=24/41

- Sun Jul 04, 2010 4:03 pm
- Forum: Statistics
- Topic: random var. X is -1, 0, 1 w/ equal probability (check ans)
- Replies:
**4** - Views:
**5746**

Please can u tell what is wrong

My work ........

E(X) = 0

_{1}= 0

_{2}= 2/3

^{2}= 2/3

var(X) =_{2}-^{2}= 2/3 - 2/3 = 0

Thanks !!!

- Sun Jul 04, 2010 4:00 pm
- Forum: Statistics
- Topic: prob. of 3 red, 2 white, 1 blue (is the answer correct)
- Replies:
**5** - Views:
**6189**

Could u please help me solve this problem by pointing out the mistake12 is not a probability

- Sun Jul 04, 2010 10:27 am
- Forum: Statistics
- Topic: prob. of 3 red, 2 white, 1 blue (is the answer correct)
- Replies:
**5** - Views:
**6189**

Question : A box contains 5 red balls, 4 white balls and 3 blue balls . A ball is selected at random from the box . the colour of the ball is noted and then replaced back . Find the probability that out of 6 balls selected in this manner , 3 are red, 2 are white , 1 is blue. Solution : Let P(R) be t...