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heidi18
06-01-2008, 05:15 PM
Hi, thanks for all that you guys do, lets see if I get this problem

A student council is made up of 6 seniors, 5 juniors, and 4 sophmores. How many ways can this student council select its president, vice-president, and treasurer? What is the probability that the officers will be either all seniors, or all juniors, or all sophmores?

For the first part of the question I have 15 nPr 3 = 2,730 ways they can be chosen, is this correct?

The second part gets a little hazy for me. I have the probability being... probability of all seniors + probability of all juniors + probability of all somphmores since the word OR was used. Maybe I am not reading the question right...

Then I broke it down as the senior probability equals (6/15)(5/14)(4/13)=
junior probability equals (5/15)(4/14)(3/13)=
sophmore probabilty eqauls (4/15)(3/14)(2/13)=

I feel like I need to divide something by the number of ways it can be done. What am I doing wrong???

galactus
06-01-2008, 05:36 PM
You are correct for the first part.

For the second question, all seniors or all juniors or all sophomores.

Choose 3 from the 4 sophomores and none from the others: \frac{P(4,3)}{2370}=\frac{4}{455}

Choose 3 from the 5 juniors: \frac{P(5,3)}{2730}=\frac{2}{91}

Choose 3 from the seniors: \frac{P(6,3)}{2730}=\frac{4}{91}

Add them up and get \frac{34}{455}

soroban
06-02-2008, 02:32 AM
Hello, Heidi!


A student council is made up of 6 seniors, 5 juniors, and 4 sophmores.
(a) How many ways can this student council select its president, vice-president, and treasurer?
(b) What is the probability that the officers will be either all seniors, or all juniors, or all sophmores?

For (a) I have: _{15}P_3 \:= \:2,730 ways they can be chosen, is this correct? . . . . Yes!

Part (b) gets a little hazy for me.
I have the probability being: P(all seniors) + P(all juniors) + P(all sophs) since the word OR was used.
Maybe I am not reading the question right. . . . . You're doing fine!

Then I broke it down as:
. . \begin{array}{ccccc}\text{P(all seniors)} & = & \frac{6}{15}\cdot\frac{5}{14}\cdot\frac{4}{13} &=& \frac{20}{455} \\ \\[-3mm] \text{P(all juniors)} &=& \frac{5}{15}\cdot\frac{4}{14}\cdot\frac{3}{13} &=& \frac{10}{455} \\ \\[-3mm] \text{P(all sophs)} &=& \frac{4}{15}\cdot\frac{3}{14}\cdot\frac{3}{13} &=& \frac{4}{455} \end{array} . . . . Good!

I feel like I need to divide something by the number of ways it can be done.
No ... you've already included the denominators in your calculations.
Now just add them (as you suggested) . . .

\text{P(all seniors }\vee\text{ all juniors }\vee\text{ all sophs)} \;=\;\frac{20}{455} + \frac{10}{455} + \frac{4}{455} \;=\;\frac{34}{455}

. . which matches galactus' answer.