Question

A basketball player has a 50% chance of making each free throw. What is the probability that the player makes at most nine out of eleven free throws?A. 15/16B. 397/2048C. 509/512D. 193/512

Answer 1

`\begin{gathered} P(X\le9)\text{ probability of at most }9\text{ successes} \\ P(X\le9)=P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)+P(9) \\ \text{ We find the individual probability by} \\ P(X)=\binom{n}{X}\cdot p^X\cdot(1-p)^(n-X) \\ \text{ Find }P(0) \\ P(0)=(11!)/(0!(11-0)!)\cdot0.5^0\cdot(1-0.5)^(11-0) \\ P\mleft(0\mright)=0.00048828125 \\ \text{ Do this for the remaining probabilities }P(1)\text{ up to }P(9)\text{ and we get} \\ P\mleft(1\mright)=0.00537109375 \\ P\mleft(2\mright)=0.02685546875 \\ P\mleft(3\mright)=0.08056640625 \\ P\mleft(4\mright)=0.1611328125 \\ P\mleft(5\mright)=0.2255859375 \\ P\mleft(6\mright)=0.2255859375 \\ P\mleft(7\mright)=0.1611328125 \\ P\mleft(8\mright)=0.08056640625 \\ P\mleft(9\mright)=0.02685546875 \\ \text{Add them all together from }P(0)\text{ up to }P(9) \\ P(X\le9)=P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)+P(9) \\ P(X\le9)=0.994140625 \\ P(X\le9)=(509)/(512) \end{gathered}`