Question

Three number are chosen at random without replacement from $\{1,2,3,...,10\}$. The probability that minimum of the chosen number is 3 or their maximum is 7, cannot exceed

• A. $11/30$
• B. $11/40$
• C. $1/7$
• D. $1/8$

Answer 1

Answer: (A), (B)

$11/30$
$11/40$

Let $A$ and $B$ denote the following events:

$A$: minimum of the chosen number is $3$.

$B$: maximum of the chose number is $7$.

We have $P\left(A\right)=P$ (choosing 3 and two other numbers from 4 to 10)

$=\frac{{}^7{C}_2}{{}^{10}{C}_3}=\frac{7\times 6}{2}\times \frac{3\times 2}{10\times 9\times 8}=\frac{7}{40}$

$P\left(B\right)=P$ (choosing 7 and two other numbers from 1 to 6)

$=\frac{{}^6{C}_2}{{}^{10}{C}_3}=\frac{6\times 5}{2}\times \frac{3\times 2}{10\times 9\times 8}=\frac{1}{8}$

$P(A\cap B)=P$ (choosing 3 and 7 and one other number from 4 to 6).

$=\frac{3}{{}^{10}{C}_3}=\frac{3\times 3\times 2}{10\times 9\times 8}=\frac{1}{40}$.

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

$=\frac{7}{40}+\frac{1}{8}-\frac{1}{40}=\frac{11}{40}<\frac{11}{30}$.