problem of porobability

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AntiviruShahriar
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problem of porobability

Unread post by AntiviruShahriar » Mon Dec 26, 2011 11:49 am

টেবিল টেনিস বোর্ডে A,B,C,D চারটি ভাগ আছে। আরমান টেবিল টেনিস বল বোর্ডের উপর ছুড়ে দিল।
বলটি সাতবার ড্রপ খেয়ে থেমে গেল। [বলটি যেকোন সময় যেকোন দিকে যেতে পারে তবে বোর্ডের বাইরে নয়]
বোর্ডের B অংশে বল্টির চারতি ড্রপ পড়ার সম্ভাবনা কত?

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bristy1588
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Re: problem of porobability

Unread post by bristy1588 » Mon Dec 26, 2011 6:17 pm

Tomar Uttor Koto, Shahriar??
Bristy Sikder

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Labib
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Re: problem of porobability

Unread post by Labib » Mon Dec 26, 2011 7:47 pm

Let (B) and (P) denote the event of the ball dropping on B and other regions respectively.
The other three incidents can be chosen in $3^3$ ways. They can happen in $5P3$ ways.
So total way, $3^3*5P3$.
So my solution was, $\frac{3^3*5P3}{4^7}$.
Last edited by Labib on Mon Dec 26, 2011 10:12 pm, edited 1 time in total.
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MATHPRITOM
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Re: problem of porobability

Unread post by MATHPRITOM » Mon Dec 26, 2011 9:36 pm

JUST WE CAN CHOOSE 4 DROPS IN C(7,4) WAYS. THEN, 1/4 IS PROBABILITY FOR THE BALL DROPPING IN B.SO,THE ANSWER IS $C(7,4)*(1/4)^3*(3/4)^3$.

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