Question

From a group of 4 men and 3 women, 2 persons are selected at random, Find the probability at least one man is selected.

• A. $\frac{6}{7}$
• B. $\frac{7}{6}$
• C. $\frac{5}{6}$
• D. $\frac{11}{21}$
• E. $\frac{12}{21}$

Answer 1

The correct option is A $\frac{6}{7}$

${}^7{C}_2=\frac{7\times 6}{2}=21$
${}^4{C}_1{\times }^3{C}_1{+}^4{C}_2=4\times 3+\frac{(4\times 3)}{2}=12+6=18P=\frac{18}{21}=\frac{6}{7}$