# Probability Word Problem involving Inequalities

• Apr 8th 2011, 04:01 PM
thamathkid1729
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<s<7)?
• Apr 8th 2011, 05:39 PM
NOX Andrew
After Leah tosses each of the $\displaystyle s$ cases into the front seat of the car, there are $\displaystyle 7 - s$ remaining cases, 2 of which are red. Therefore, the probability the next case Leah chooses from the back seat will contain a red sleeping bag is $\displaystyle \dfrac{2}{7 - s}$.
• Apr 8th 2011, 06:49 PM
Soroban
Hello, thamathkid1729!

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 $\displaystyle \,s$ cases, with $\displaystyle s < 5$, opens the cases, 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 $\displaystyle \,s$, 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 . . .

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

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

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

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

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

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

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

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

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