# Math Help - Help, SVP.

1. ## Help, SVP.

Hey... if you could answer/explain any of the following (even one would be awesome), I'd be forever thankful.

1. A six-member working group to plan a student lounge is to be selected from five teachers and nine students. If the working group is randomly selected, what is the probability of the committee including at least two teachers.

2. Cha’tima has five white and six grey huskies in her kennel. If a wilderness expedition chooses a team of six sled dogs from Cha’tima’s kennel, what is the probability that the team will consist of:
all white huskies?
all grey huskies?
three of each colour?

3. A survey at a school asked students if they were ill with a cold or the flu during last months.
The results were as follows. None of the students had both a cold and the flu.
Cold Flu Healthy
Females 32 18 47
Males 25 19 38
Use these results to estimate the probability that:
a randomly selected student had a cold last month.
a randomly selected females students was healthy last month.
a randomly selected student who had the flu last month is male.
a randomly selected student had either a cold or the flu last month.

4. To get out of jail free in the board game MonopolyÒ, you have to roll doubles with a pair of dice. Determine the odds in favour of getting out of jail on your first or second roll.

5. If a survey on teenage reads of popular magazines shows that 38% subscribe to Teen Life and 47% subscribe to Cool Teen and 35% subscribe to neither magazine, what is the probability that a randomly selected teenager,
subscribes to both magazines?
subscribes to either one magazine or both?
subscribes to only one of the two magazines?

2. I got this for 2,3, and 4.

2)

Prob is 0 b/c only 5 whites are there
P(all grey) = 1/(11 6) = 1/462, ~ 0.0022
P(3 white & 3 grey) ((5 3) (6 3))/(11 6), 200/462, ~ 0.433

3)

P(cold)=57/179, =0.3184, ~ 31.8%
P(health | female) (47/179)/(97/179) = 47/97 = 0.4845, ~ 48.5%
P(male | flu) 19/37 = 0.5135, ~51.3%
44/82 = 0.5366, ~53.7%

Double on first roll = 6/36 = 1/6, ergo not rollin = 5/6
Ergo prob of rolling doubles on the second roll: 5/6 * 1/6 = 5/36
Prob of rolling double on first or second: 1/6 + 5/6 = 11/36
Ergo odds of getting out of jail on first or second roll are 11:25

Confirm?

And help with 1 and 5?

3. Originally Posted by acc
Confirm?

And help with 1 and 5?
I didn't get a calculator out to confirm final answers, but I agree with your work for 2,3,4 except for the following:

for 3b, I would simply write 47/97 without the initial (47/179)/(97/179) step.

for 3d, I get 94/179. Your answer is for a randomly chosen male student.

for 4, there's an assumption we have to make, and that is that we don't want to count the probability of getting doubles on both the first and second rolls (because presumably, rolling a double on the first roll will get us out of jail and we won't make the second roll). Assuming that, I agree with your answer. (But you made a typo by writing 5/6 in one spot where you meant 5/36.)

For problem 1, I would start with a "1" and subtract the following probabilities from it

P(0 teachers and 6 students)
P(1 teacher and 5 students)

Problem 5, you could draw a Venn diagram. Or using formal notation,

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

(Edit: fixed typo.)

Edit 2: The part in gray is incorrect. What you wrote for 4 is fine the way it is, other than the typo.

4. Hello, acc!

2. Cha’tima has five white and six grey huskies in her kennel.
If a wilderness expedition chooses a team of six sled dogs from Cha’tima’s kennel,
what is the probability that the team will consist of:

(a) all white huskies?
(b) all grey huskies?
(c) three of each colour?
There are: . ${11\choose6} \:=\:462$ possible selections.

$(a)\;P(\text{6 white}) \;=\;0$

$(b)\;P(\text{6 grey}) \;=\;\frac{{6\choose6}}{462} \;=\;\frac{1}{462}$

$(c)\;P(\text{3 white, 3 grey}) \;=\;\frac{{5\choose3}{6\choose3}}{462} \;=\;\frac{10\cdot20}{462} \;=\;\frac{100}{231}$

3. A survey at a school asked students
if they were ill with a cold or the flu during last month.

The results were as follows.
None of the students had both a cold and the flu.

. . $\begin{array}{c||c|c|c||c|}
& \text{Cold} & \text{Flu} & \text{Healthy } & \text{Total} \\ \hline\hline \text{Females} & 32 & 18 & 47 & 97 \\ \hline
\text{Males} & 25 & 19 & 38 & 82 \\ \hline \hline
\text{Total} & 57 & 37 & 85 & 179 \\ \hline \end{array}$

Use these results to estimate the probability that:

(a) a randomly selected student had a cold last month.
(b) a randomly selected female student was healthy last month.
(c) a randomly selected student who had the flu last month is male.
(d) a randomly selected student had either a cold or the flu last month.

$(a)\;P(\text{Cold}) \:=\:\frac{57}{179}$

$(b)\;P(\text{Female}\,\wedge\,\text{Healthy}) \;=\;\frac{47}{179}$

$(c)\;P(\text{Male}|\text{Flu}) \;=\;\frac{19}{37}$

$(d)\;P(\text{Cold} \vee\text{Flu}) \;=\;P(\text{Cold}) + P(\text{Flu}) \;=\;\frac{57}{179} + \frac{37}{170} \;=\;\frac{94}{179}$

5. Originally Posted by Soroban

$(a)\;P(\text{Cold}) \:=\:\frac{57}{179}$

$(b)\;P(\text{Female}\,\wedge\,\text{Healthy}) \;=\;\frac{47}{179}$

$(c)\;P(\text{Male}|\text{Flu}) \;=\;\frac{19}{37}$

$(d)\;P(\text{Cold} \vee\text{Flu}) \;=\;P(\text{Cold}) + P(\text{Flu}) \;=\;\frac{57}{179} + \frac{37}{170} \;=\;\frac{94}{179}$

Wow, I completely misread (c) and thought it said "a randomly selected student had the flu last month."

For (b), I'm wondering if it is a matter of interpretation. To get the probability you found, wouldn't it have better been stated "a randomly selected student is female and had the flu last month"?

Edit: I must be tired because I didn't even address 3c, I'll stop posting now and get some rest.

6. Originally Posted by acc
Hey... if you could answer/explain any of the following (even one would be awesome), I'd be forever thankful.

1. A six-member working group to plan a student lounge is to be selected from five teachers and nine students. If the working group is randomly selected, what is the probability of the committee including at least two teachers.

[snip]

5. If a survey on teenage reads of popular magazines shows that 38% subscribe to Teen Life and 47% subscribe to Cool Teen and 35% subscribe to neither magazine, what is the probability that a randomly selected teenager,
subscribes to both magazines?
subscribes to either one magazine or both?
subscribes to only one of the two magazines?
Duplicate posted (and answered) here: http://www.mathhelpforum.com/math-he...tml#post511545