I am trying to determine the probability of a "mixed panel" assignment (i.e., a panel of judges w/at least 1 woman).

Consider the following: A court has a total of 20 judges, 8 of whom are women. Panels of 5 judges are randomly drawn to decide any given case. What is the probability of drawing a panel on which at least 1 of the judges is female?

I realize this is a "simple" question, but it is beyond me. I would greatly appreciate a discussion of the steps required for its solution. Thank you in advance.


1 Answer 1


There is a large and rich branch of mathematics, combinatorics, devoted to solving such problems. The most important step used here is to recognize that "at least one female" is more simply characterized as "not all males." Details follow.

The name for the number of distinct 5-member panels from a pool of 20 judges is the "binomial coefficient," $\binom{20}{5}$. We will worry later about how to compute this. All probabilities involving 5-member panels of these judges will be fractions with this number in the denominator. Computing a probability is a matter of counting which panels are described by an event; that will go in the numerator.

The all-male panels are drawn from a smaller pool of just 20-8 = 12 judges. There are therefore $\binom{12}{5}$ of them. The remainder of the possible panels, equal in number to $\binom{20}{5} - \binom{12}{5}$, have at least one female. Therefore the desired probability is

$$\Pr(\text{Panel with a female judge}) = \frac{\binom{20}{5} - \binom{12}{5}}{\binom{20}{5}}.$$

To compute the binomial coefficients, consider that the number of ordered sequences of 5 people out of 20 equals $20 \cdot 19 \cdot 18 \cdot 17 \cdot 16$ because there are 20 ways to pick the first in the panel, 19 remaining people from whom to choose the second, and so on. Any panel of 5 people determines $5! = 5\cdot 4 \cdots 2 \cdot 1$ such orderings. Therefore $\binom{20}{5} = 20\cdots 16 / (5 \cdots 1) = 15504.$ Likewise $\binom{12}{5} = 792$. The desired probability is

$$\frac{\binom{20}{5} - \binom{12}{5}}{\binom{20}{5}} = \frac{15504 - 792}{15504} = \frac{613}{646}.$$

  • 2
    $\begingroup$ "not all males" rather than "no males"? $\endgroup$
    – mark999
    Jun 12, 2011 at 3:41

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