Summation Series... How to solve?? Plz help...

**phishboane** · September 30th 2008, 09:50 PM

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

**Soroban** · October 1st 2008, 09:55 AM

Hello, phishboane!

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

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

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

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

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

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

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

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

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

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

**phishboane** · October 2nd 2008, 04:06 PM

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...

Do u have some other solutions?
Im really glad that u've replied my thread. TQ very much n nice to know u.

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Summation Series... How to solve?? Plz help...

Thanx Soroban.

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