Hello, 3deltat!

Quote:

1) A basketball team of 10 players - 4 guards, 3 forwards, 3 centers -

is electing two captains.

Find the probability that a guard and a center will be elected.

Choosing 2 captains from 10 players, there are: /$\displaystyle _{10}C_2 \,=\,45$ possible choices.

There are 4 choices for a guard and 3 choices for a center: .$\displaystyle 4 \times3 \:=\:12$ choices.

Therefore: .$\displaystyle P(\text{guard and center}) \:=\:\frac{12}{45} \:=\:\frac{4}{15}$

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2) There are 18 juniors and 10 seniors in a class.

If 6 of the seniors are males and 12 of the juniors are females,

find the probability of selecting a senior or a male.

$\displaystyle \begin{array}{ccccccc} & & M & & F & & Total \\ \hline

Juniors & | & 6 &| & 12 & | & 18 \\ \hline

Seniors & | & 6 & | & 4 & | & 10 \\ \hline

Total & | & 12 & | & 16 & | & 28 \\ \hline\end{array}$

Your formula was correct:

. . $\displaystyle P(\text{senior or male}) \:=\:P(\text{senior}) + P(\text{male}) - P(\text{senior and male})$

But the values are: .$\displaystyle P(\text{senior or male}) \;=\;\frac{10}{28} + \frac{12}{28} - \frac{6}{28} \;=\;\frac{16}{28} \;=\;\frac{4}{7}$

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3) Find the probability that three cards drawn from a deck

without replacement contain at least two hearts

There are: .$\displaystyle _{52}C_3 = 22,\!100$ possible 3-card hands.

To get exactly three hearts, there are: .$\displaystyle _{13}C_3 = 286$ ways.

To get exactly two hearts, there are: .$\displaystyle \left(_{13}C_2\right)\left(_{39}C_1\right) = 3042$ ways.

Hence, there are: .$\displaystyle 286 + 3042 \:=\:3328$ ways to get at least two hearts.

Therefore: .$\displaystyle P(\text{at least 2 hearts}) \;=\;\frac{3328}{22,\!100} \;=\;\frac{64}{425}$

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4) How many ways can a jury of 6 men and 6 women be selected from 12 men and 12 women?

I had: .$\displaystyle _{12}C_6 + _{12}C_6 \:= \:1848$

but I'm guessing it should have been: .$\displaystyle _{12}C_6 \times\, _{12}C_6$, correct? .Right!