Sometimes, when working with math, the exercises are not well defined. Since the following phrase is not preceded by an exactly or an at least and because it does not specify the number of men, it could take on multiple meanings.

7 of the committee will be women

Other than the previous interpretation and answer, it is possible to have

7,

8,

9,

or

10

of the committee be women and still say there are

7

women on the committee. To find the probability we can use the terms of the binomial from where we started above to the end.

k = 7 \sum 10 \frac{1 0 !}{k ! ( 1 0 - k ) !} m^{10 - k} w^{k}

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Evaluate

WriteSum

Write as a sum

\frac{1 0 !}{7 ! ( 1 0 - 7 ) !} m^{10 - 7} w^{7} + \frac{1 0 !}{8 ! ( 1 0 - 8 ) !} m^{10 - 8} w^{8} f + \frac{1 0 !}{9 ! ( 1 0 - 9 ) !} m^{10 - 9} w^{9} + \frac{1 0 !}{1 0 ! ( 1 0 - 1 0 ) !} m^{10 - 10} w^{10}

SubTerms

Subtract terms

\frac{1 0 !}{7 ! \cdot 3 !} m^{3} w^{7} + \frac{1 0 !}{8 ! \cdot 2 !} m^{2} w^{8} f + \frac{1 0 !}{9 ! \cdot 1 !} m^{1} w^{9} + \frac{1 0 !}{1 0 ! \cdot 0 !} m^{0} w^{10}

Write as a product

\frac{1 0 \cdot 9 \cdot 8 \cdot 7 !}{7 ! \cdot 3 \cdot 2 \cdot 1} m^{3} w^{7} + \frac{1 0 \cdot 9 \cdot 8 !}{8 ! \cdot 2 \cdot 1} m^{2} w^{8} f + \frac{1 0 \cdot 9 !}{9 ! ( 1 )} m^{1} w^{9} + \frac{1 0 !}{1 0 ! ( 1 )} m^{0} w^{10}

CalcQuot

Calculate quotient

120 m^{3} w^{7} + 45 m^{2} w^{8} + 10 m^{1} w^{9} + m^{0} w^{1} 0

Simplify power

120 m^{3} w^{7} + 45 m^{2} w^{8} + 10 m w^{9} + w^{1} 0

Now, from the other solution, we know the denominator is

1024 .

\frac{1 2 0}{1 0 2 4} + \frac{4 5}{1 0 2 4} + \frac{1 0}{1 0 2 4} + \frac{1}{1 0 2 4}

AddFrac

Add fractions

\frac{1 7 6}{1 0 2 4}

ReduceFrac

$\frac{a}{b} = \frac{a / 1 6}{b / 1 6}$

\frac{1 1}{6 4}

To have at least

7

women on the committee, the probability is

\frac{1 1}{6 4} .

Exercise