A committee of $6$ is chosen at random from $7$ men and $5$ women.
$i)$ find the probability that $4$ men and $2$ women are chosen.
$ii)$ find the probability that fewer women than men are chosen.
My working:
$i)$ The part $a$ answer I have found to be: $Pr$($4$ men and $2$ women being chosen) = $\frac{{7 \choose 4}{5 \choose 2}}{12\choose 6}.$ Is my working Okay?
$ii)$ For second part I need help. I can't the write mathematical expression or equation of the statement: fewer women than men. How do approach it. any help will be appreciated. Thanks
Best Answer
Hint:
The probability fewer women than men are chosen is