Question

Three numbers are chosen at random without replacement from $1,2,....10$ . The probability that the minimum of the chosen numbers is $3$, or their maximum is $7$ is ..............

• A. $7/24$
• B. $11/40$
• C. $17/60$
• D. $1/4$

Answer 1

The correct option is B $11/40$

Let A and B denote the following events:
A:minimum of the chosen number is 3
B: maximum of the chosen 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 $P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)=\frac{7}{40}+\frac{1}{8}-\frac{1}{40}=\frac{11}{40}$