2015 regional secondary no. 2 probability

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Md. Rifat uddin
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2015 regional secondary no. 2 probability

Unread post by Md. Rifat uddin » Sat Jan 07, 2017 9:41 pm

There are 9 seats for passengers in a water bus. Probability of ticket buyer of the water bus remains absent on the day of journey is 50%. If the owners of the water bus sell 11 tickets then the probability of every passenger getting a seat is a/b,where a and b are co-prime. b-a = ? :!: :!:

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Thanic Nur Samin
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Re: 2015 regional secondary no. 2 probability

Unread post by Thanic Nur Samin » Fri Jan 13, 2017 1:28 am

The only cases where every passenger doesn't get a seat are when $10$ and $11$ passenger comes. The probability of these happening are simultaneously $\dfrac{11}{2^{10}}$ and $\dfrac{1}{2^{11}}$

So, the probability of everyone getting a seat would be $1-\dfrac{11}{2^{10}}-\dfrac{1}{2^{11}}=\dfrac{2^{11}-23}{2^{11}}$. So the answer is clearly $\boxed{23}$
Hammer with tact.

Because destroying everything mindlessly isn't cool enough.

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ahmedittihad
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Re: 2015 regional secondary no. 2 probability

Unread post by ahmedittihad » Sat Jan 14, 2017 9:08 pm

Erm, I'm getting a different result. The probability of $10$ passengers coming is $11/2^{11}$. The probability of $11$ passengers coming is $1/2^{11}$.

So the probability of everyone getting a seat would be $1- 11/2^{11} - 1/2^{11}=( 2^{11}-12)/2^{11}$. So the answer is $3$.
Frankly, my dear, I don't give a damn.

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Thanic Nur Samin
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Re: 2015 regional secondary no. 2 probability

Unread post by Thanic Nur Samin » Sat Jan 14, 2017 10:40 pm

ahmedittihad wrote:Erm, I'm getting a different result. The probability of $10$ passengers coming is $11/2^{11}$.
:oops: :oops: Such a silly mistake by me. I guess this is what happens when you solve divisional problems after midnight without tea :roll:
Hammer with tact.

Because destroying everything mindlessly isn't cool enough.

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