Question

Three numbers are chosen at random without replacement from $\{1,2,3,\cdots ,10\}$. The probability that, the smallest of the chosen number is $3$ or the greatest one is $7,$ is equal to

• A. $\frac{7}{40}$
• B. $\frac{11}{40}$
• C. $\frac{9}{40}$
• D. $\frac{13}{40}$

Answer 1

The correct option is B $\frac{11}{40}$

Let $E_1$ be the event:
$\text{getting the smallest number}3$ and
$E_2$ be the event:
$\text{getting the greatest number}7$
Then, $P\left(E_1\right)=\text{P (getting one number 3 and other two from 4 to 10)}$
$=\frac{{}^1{C}_1{\times }^7{C}_2}{{}^{10}{C}_3}=\frac{7}{40}$
Similarly, $P\left(E_2\right)=\text{P(getting one number 7 and other two number from 1 to 6)}$
$=\frac{{}^1{C}_1{\times }^6{C}_2}{{}^{10}{C}_3}=\frac{1}{8}$
$P(E_1\cap E_2)=\text{P(getting one number 3 and second number 7 and third number from 4 to 6)}$
$=\frac{{}^1{C}_1{\times }^1{C}_1{\times }^3{C}_1}{{}^{10}{C}_3}=\frac{1}{40}$
$\therefore P(E_1\cup E_2)=P\left(E_1\right)+P\left(E_2\right)-P(E_1\cap E_2)$
$=\frac{7}{40}+\frac{1}{8}-\frac{1}{40}=\frac{7+5-1}{40}=\frac{11}{40}$