May I know how to expand this??

Given that

x=2,

y=3,

Lambda1 = 1.5

Lambda2 = 1.0

Lambda3 = 0.05

What's the answer?

Thanx in advance(Nod)

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- Sep 30th 2008, 09:50 PMphishboaneSummation Series... How to solve?? Plz help...
May I know how to expand this??

Given that

x=2,

y=3,

Lambda1 = 1.5

Lambda2 = 1.0

Lambda3 = 0.05

What's the answer?

Thanx in advance(Nod) - Oct 1st 2008, 09:55 AMSoroban
Hello, phishboane!

It seems to be simple substitution . . . then arithmetic.

Quote:

Given that: .$\displaystyle x=2,\;y=3, \; \lambda_1 = 1.5,\;\lambda_22 = 1.0,\;\lambda_3 = 0.05$

Evaluate: .$\displaystyle \sum^{\min(x,y)}_{i=0} {x\choose i}{y\choose i}(i!)\left(\frac{\lambda_3}{\lambda_1\lambda_2}\r ight)^i $

We see that: .$\displaystyle \min(x,y) \:=\:\min(2,3) \:=\:2$

. . and: .$\displaystyle \frac{\lambda_3}{\lambda_1\lambda_2} \:=\:\frac{0.05}{(1.5)(1.0)} \:=\:\frac{1}{30}$

And we have: .$\displaystyle \sum^2_{i=0}{2\choose i}{3\choose i}(i!)\left(\frac{1}{30}\right)^i $

. . $\displaystyle = \;{2\choose0}{3\choose0}(0!)\left(\frac{1}{30}\rig ht)^0 + {2\choose1}{3\choose1}(1!)\left(\frac{1}{30}\right )^1 + {2\choose2}{3\choose2}(2!)\left(\frac{1}{30}\right )^2$

. . $\displaystyle = \;1\cdot1\cdot1\cdot1 \;+\; 2\cdot3\cdot1\cdot\frac{1}{30} \;+ \;1\cdot3\cdot2\cdot\frac{1}{30^2} $

. . $\displaystyle = \;1 + \frac{1}{5} + \frac{1}{150}$

. . $\displaystyle = \;\frac{181}{150}$

- Oct 2nd 2008, 04:06 PMphishboaneThanx Soroban.
Hi Soroban,

By looking at ur reply, ur solution steps. I know that I've made a correct decision to join this forum, n perhaps, I should have join earlier.

I appreciate ur kind approach, n ur detailed solutions. Thanx a lot.

The way u solved it was exactly the way i tried before but didn't get it right. So I think, the min(x,y) not as easy as we thought...(Doh)

Do u have some other solutions?

Im really glad that u've replied my thread. TQ very much n nice to know u.(Hi)