Question

Three numbers are chosen at random without replacement from $\{1,2,3,...14\}$.
Let $A=\left\{\text{minimum of chosen numbers is 5}\right\}$
$B=\left\{\text{maximum of chosen numbers is 11}\right\}$ and $P(A\cup B)=\frac{K+11}{91}$, the $K=$ ____.

Answer 1

3 no.s are chosen without replacement.

$A=\left\{\text{minimum of chosen numbers is 5}\right\}$

$\therefore$ 2 no.s can be selected from 6 to 14.

$\therefore P\left(A\right)=\frac{1}{14}\times \frac{9}{13}\times \frac{8}{12}$

$B=\left\{\text{maximum of chosen numbers is 11}\right\}$

$\therefore$ 2 numbers can be selected from 1 to 10.

$\therefore P\left(B\right)=\frac{1}{14}\times \frac{10}{13}\times \frac{9}{12}$

$A\cap B=\left\{\text{minimum number is 5 and maximum number is 11}\right\}$

$\therefore$ 1 no. can be chosen from 6 to 10.

$\therefore P(A\cap B)=\frac{1}{14}\times \frac{1}{13}\times \frac{5}{12}$

$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$

$\frac{K+11}{91}=\frac{72+90-5}{14\times 13\times 12}$

$K+11=\frac{157}{24}$

$24K=157-264$

$24K=-107$

$K=\frac{-107}{24}$