BdMO Online Forum

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 = ?

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 $\boxed{23}$

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 $3$.

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

Such a silly mistake by me. I guess this is what happens when you solve divisional problems after midnight without tea