Still not quite there: choosing student council positions

heidi18

New member
Joined
May 30, 2008
Messages
4
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

Super Moderator
Staff member
Joined
Sep 28, 2005
Messages
7,216
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: \(\displaystyle \frac{P(4,3)}{2370}=\frac{4}{455}\)

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

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

Add them up and get \(\displaystyle \frac{34}{455}\)
 

soroban

Elite Member
Joined
Jan 28, 2005
Messages
5,588
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: \(\displaystyle _{15}P_3 \:= \:2,730\) 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:
. . \(\displaystyle \begin{array}{ccccc}\text{P(all seniors)} & = & \frac{6}{15}\cdot\frac{5}{14}\cdot\frac{4}{13} &=& \frac{20}{455} \\ \\[-3mm] \text{P(all juniors)} &=& \frac{5}{15}\cdot\frac{4}{14}\cdot\frac{3}{13} &=& \frac{10}{455} \\ \\[-3mm] \text{P(all sophs)} &=& \frac{4}{15}\cdot\frac{3}{14}\cdot\frac{3}{13} &=& \frac{4}{455} \end{array}\) . . . . 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) . . .

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

. . which matches galactus' answer.

 
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