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

combinatoricsprobability

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

  • the probability that $4$ men and $2$ women are chosen (as you have correctly found), plus
  • the probability that $5$ men and $1$ woman are chosen (find similarly), plus
  • the probability that $6$ men and $0$ women are chosen (find similarly).
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