Old math tests... Argh!

• May 27th 2007, 07:57 PM
3deltat
Old math tests... Argh!
So its time! The Final Exam is but two days away!
I am studying old math tests... but, alas, I did not correct my errors!

There are 4 problems that i did and I answered incorrectly. I will give my work... Ive looked over them a couple times and (since im rusty from going over them a half year ago) Im not 100% sure what the real thing is.

1) A basketball team of 10 players - 4guards, 3forwards, 3centers - is electing two captains. Find the probability that a guard And a center will be elected

so.... 4Guards
3Forwards
3Centers
with 10 in total

i did {(4C1) X (6C0) / 10C1 } X {(3C1) X (7C0) / 10C1}
(so it was the Probability for Guards times the probability for Centers )

and it = .12
where did i go wrong? :confused:

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

P(senior) + P(male) - P(Senior and Male)

so we have 18 juniors with 6M and 12F
and 10 seniors with 6males and 4 females

(10/18) + (12/28) - (16/10) = .384126

3) Find the probability that three cards drawn from a dec without replacement contain at least two hearts

a deck of card has 52 cards and 13 hearts

I had 13C2 X 39C1 / 52C3

Im guessing though, that this one, it was supposed to be
3C2 X (13/52)^2 X (39/52)^1 since it is WITHOUT replacement, right?

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

I had 12C6 + 12C6 = 1848
but im guessing it should have been 12C6 X 12C6 correct?

Thank you very very much for your time!
• May 27th 2007, 08:54 PM
Soroban
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: .$\displaystyle P(\text{guard and center}) \:=\:\frac{12}{45} \:=\:\frac{4}{15}$

Quote:

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.

$\displaystyle \begin{array}{ccccccc} & & M & & F & & Total \\ \hline Juniors & | & 6 &| & 12 & | & 18 \\ \hline Seniors & | & 6 & | & 4 & | & 10 \\ \hline Total & | & 12 & | & 16 & | & 28 \\ \hline\end{array}$

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

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

Quote:

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

Quote:

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!

• May 28th 2007, 09:25 AM
3deltat
ah alright. I was looking at your answer for the Without replacement question and was compairing it to a different problem and I think i understand now. What i did the first time, on the test, I made it so that the problem would be trying to find the probability of finding 2 hearts, not "at least" two hearts. Thanks for clearing that up!