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 April 8th, 2011, 04:06 PM #1 Senior Member   Joined: Apr 2009 Posts: 106 Thanks: 0 Probability Word Problem Involving Inequalities Leah is asked to get a red sleeping bag out of the back seat of her mother's car. There are 7 sleeping bags in the car, only 2 of which are red. All the bags are in black, zippered cases, so there is no way to tell which sleeping bag is which without opening the cases. Leah picks s cases, with s < 5, and discovers that not one of these cases contains a red sleeping bag. She tosses each of these cases into the front seat of the car. In terms of s, what is the probability that the next case Leah chooses from the back seat will contain a red sleeping bag? I have never seen a probability problem like this before. It seems to me that the answer is 2/(2
 April 8th, 2011, 06:06 PM #2 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Probability Word Problem Involving Inequalities If she removed s bags, there must be 7-s bags remaining, of which 2 are red. So the answer is 2/(7-s).
April 8th, 2011, 07:10 PM   #3
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Re: Probability Word Problem Involving Inequalities

Hello, julian21!

I solved this at another site, but here is my explanation.

There is no formula for this type of problem.
We must make an exhaustive list . . . carefully.

Quote:
 Leah is asked to get a red sleeping bag out of the back seat of her mother's car. There are 7 sleeping bags in the car, only 2 of which are red. All the bags are in black zippered cases, so there is no way to tell their colors. Leah picks $\,s$ cases, with $\displaystyle s < 5$, opens the cases, and discovers [color=beige]. . [/color]that not one of these cases contains a red sleeping bag. She tosses each of these cases into the front seat of the car. In terms of $\,s$, what is the probability that the next case Leah chooses from the back seat will contain a red sleeping bag?

There are 2 Red sleeping bags and 5 Others.

And there are four cases to consider . . .

$[1]\;s= 1:\:\text{ Leah gets 1 Other.}$
[color=beige]. . [/color]$P(\text{1 Other}) \:=\:\frac{5}{7}$

$\text{There are 2 Reds and 4 Others.}$
[color=beige]. . [/color]$P(\text{Red}) \:=\:\frac{2}{6} \:=\:\frac{1}{3}$
$\text{Hence: }\:P(s= 1\,\wedge\,\text{Red}) \:=\:\frac{5}{7}\cdot\frac{1}{3} \:=\:\dfrac{5}{21}$

$[2]\;s= 2\;\text{ Leah gets 2 Others.}$
[color=beige]. . [/color]$P(\text{2 Others}) \:=\:\frac{{5\choose2}}{{7\choose2}} \:=\:\frac{10}{21}$

$\text{There are 2 Reds and 3 Others.}$
[color=beige]. . [/color]$P(\text{Red}) \:=\:\frac{2}{5}$
$\text{Hence: }\:P(s= 2\,\wedge\,\text{Red}) \:=\:\frac{10}{21}\cdot\frac{2}{5} \:=\:\dfrac{4}{21}$

$[3]\;s= 3:\;\text{ Leah gets 3 Others.}$
[color=beige]. . [/color]$P(\text{3 Others}) \:=\:\frac{{5\choose3}}{{7\choose2}} \:=\:\frac{2}{7}$

$\text{There are 2 Reds and 2 Others.}$
[color=beige]. . [/color]$P(\text{Red}) \:=\:\frac{2}{4} \:=\:\frac{1}{2}$
$\text{Hence: }\:P(s= 3\,\wedge\,\text{Red}) \:=\:\frac{2}{7}\cdot\frac{1}{2} \:=\:\frac{1}{7} \:=\:\dfrac{3}{21}$

$[4]\;s= 4:\;\text{ Leah gets 4 Others.}$
[color=beige]. . [/color]$(\text{4 Others}) \:=\:\frac{{5\choose4}}{{7\choose2}} \:=\:\frac{1}{7}$

$\text{There are 2 Reds and 1 Other.}$
[color=beige]. . [/color]$P(\text{Red}) \:=\:\frac{2}{3}$
$\text{Hence: }\:P(s= 3\,\wedge\,\text{Red}) \:=\:\frac{1}{7}\cdot\frac{2}{3} \:=\:\dfrac{2}{21}$

$\text{Therefore: }\:P(s\text{ Others}\,\wedge\,\text{Red}) \;=\;\dfrac{6-s}{21}$

 April 8th, 2011, 08:51 PM #4 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Probability Word Problem Involving Inequalities Soroban's answer assumes that Leah has only just been asked to get a red sleeping bag, and that s is known beforehand. My answer assumes that Leah has already taken out the s sleeping bags and found that none were red. This seems to me more appropriate given that we asked about the "next" sleeping bag.

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