There are $9$ seats for passengers in a bus. Probability of ticket buyer of the bus remains absent of the day of the journey is $50\%$. If the owner of the bus sell $11$ tickets then find the probability of every passenger getting a seat.
I am really newbie in probability. So can't find way to start.
Source: BdMO 2015 Cox Bazar Regional Higher Secondary question no. 6.
In order for every (present) passenger to get a seat, there must at least $2$ absentees. We can determine this directly, but it's easier to get the complement: in order for a passenger to not get a seat, there must be $0$ or $1$ absentees.
Individual attendence is composed of independent bernoulli trials of equal probability, so we can use the Binomial Distribution.
What is the probability that there are $0$ absentees? What is the probability that there is $1$ absentee? Add these together, and subtract from $1$, to get the final probability.
The only ways that somebody would be left without a seat would be if either 10 or 11 people showed up. you should use the binomial distribution with $p=q=\frac 12$ $$ P = 1 - \binom {11 }{10} \left(\frac12 \right)^{10} \left(\frac12 \right)^{1} - \binom {11 }{11} \left(\frac12 \right)^{11} \left(\frac12 \right)^{0}$$
$$ = 1 - 12 \left(\frac12 \right)^{11} $$