Question

Three numbers are chosen at random without replacement from {1, 2, ......, 15}. Let $E_1$ be the event that minimum of the chosen numbers is 5 and $E_2$ be that their maximum is 10 then

• A. $P\left(E_1\right)=\frac{9}{91}$
  
• B. $P\left(E_2\right)=\frac{36}{455}$
  
• C. $P(E_1\cap E_2)=\frac{4}{455}$
  
• D. $P\left(E_1/E_2\right)=\frac{1}{9}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A

$P\left(E_1\right)=\frac{9}{91}$

B

$P\left(E_2\right)=\frac{36}{455}$

C

$P(E_1\cap E_2)=\frac{4}{455}$

D

$P\left(E_1/E_2\right)=\frac{1}{9}$

$P\left(E_1\right)$ =P(Choosing 5 and two tickets from 6 to 15)
=$\frac{{10}_{C_2}}{{15}_{C_3}}=\frac{10\times 9}{2}\times \frac{3\times 2}{15\times 14\times 13}=\frac{9}{91}$
$P\left(E_2\right)$ =P(Choosing 10 and two tickets from 1 to 9)
$\frac{9_{C_2}}{{15}_{C_3}}=\frac{9\times 8}{2}\times \frac{3\times 2}{15\times 14\times 13}=\frac{36}{455}$
$P(E_1\cap E_2)$=P(Choosing 5 and 10 and one ticket from 6 to 9)
$=\frac{4_{C_1}}{{15}_{C_3}}=4\times \frac{3\times 2}{15\times 14\times 13}=\frac{4}{455}P\left(E_1/E_2\right)=\frac{1}{9}$