1. ## Portraits

Set of portraits consists of 6 boys and 4 girls. Little brother drew mustache on three of these set portraits. What is the probability that a randomly selected image will be a portrait of a girl without a mustache?

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2. Hello speculus

Welcome to Math Help Forum!
Originally Posted by speculus
Photo fiber consists of 6 friends to 4 friends portraits. Little brother 3 nuotraukse drew people mustache. What is the probability that a randomly selected image will be intact companion portrait?

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Sorry. Your question does not make sense in English.

3. excuse me for wrong explanation

I've changed question, see above

4. Little brother will select from none to three girls. That is the partition of the space.
Probability of selecting a girl without a mustache when he is done is $\displaystyle \sum\limits_{k = 0}^3 {\left[ {\frac{{\binom{4}{k}\binom{6}{3-k}}}{\binom{10}{3}} \cdot \frac{{4 - k}}{{10}}} \right]}$

5. thank you

6. Hello, speculus!

Here's an explanation of Plato's excellent solution.

There are portraits of 6 boys and 4 girls.
Little brother drew mustaches on three of these portraits.

What is the probability that a randomly selected portrait
will be of a girl without a mustache?

Assume that little brother randomly chose three portraits to deface.
. . He has: .$\displaystyle {10\choose3} \,=\,120$ possible choices.

He has 4 choices for the genders of the subjects:

. . $\displaystyle \begin{array}{cccccccccc} (1) & \text{3 boys:} & {6\choose3} &=& 20\text{ ways.} & P(\text{3B}) &=& \dfrac{20}{120} &=& \dfrac{5}{30} \\ \\ (2) & \text{2 boys, 1 girl:} & {6\choose2}{4\choose1} &=& 60\text{ ways.} & P(\text{ 2B, 1G}) &=& \dfrac{60}{120} &=& \dfrac{15}{30} \end{array}$

. . $\displaystyle \begin{array}{cccccccccc} (3) & \text{1 boy, 2 girls:} & {6\choose1}{4\choose2} &=& 36\text{ ways.} & P(\text{1B, 2G}) &=& \dfrac{36}{120} &=& \dfrac{9}{30} \\ \\ (4) & \text{3 girls:} & {4\choose3} &=& 4\text{ ways.} & P(\text{3G}) &=& \dfrac{4}{120} &= &\dfrac{1}{30} \end{array}$

For each choice, there is a probability of choosing a girl without a mustache.

. . $\displaystyle \begin{array}{cccc} \text{Mustaches on:} & P(\text{girl w/o mustache}) \\ \hline\ \\[-3mm] \text{3 boys} & \dfrac{4}{10} \\ \\[-3mm] \text{2 boys, 1 girl} & \dfrac{3}{10} \\ \\[-3mm] \text{1 boy, 2 girls} & \dfrac{2}{10} \\ \\[-3mm] \text{3 girls} & \dfrac{1}{10} \end{array}$

Therefore: .$\displaystyle P(\text{girl w/o mustache}) \;=\; \left(\frac{5}{30}\right)\left(\frac{4}{10}\right) + \left(\frac{15}{30}\right)\left(\frac{3}{10}\right ) +$ $\displaystyle \left(\frac{9}{30}\right)\left(\frac{2}{10}\right) + \left(\frac{1}{30}\right)\left(\frac{1}{10}\right)$

. . . . . . . . . . . . . . . . . . . . . .$\displaystyle =\;\ \frac{20}{300} + \frac{45}{300} + \frac{18}{300} + \frac{1}{300} \;\;=\;\;\frac{84}{300} \;\;=\;\;\boxed{\frac{7}{25}}$

7. Hello Soroban,

thank you very much for a detailed solution.