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### Two Probability Questions

```
Date: 06/08/99 at 00:30:29
From: prashant
Subject: Probability

Dear sir:

I am a student at University of Mississippi medical center. I have a
great interest in probability, so I keep solving questions I get from
texts and friends. I have two questions I could not solve. Kindly help
me solve them.

Q1) Assume it is known (from past experience, surveys, etc.) that 40%
of buyers of Fatcow products prefer unsalted butter and the remaining
60% prefer salted butter. A store manager expects to sell no more than
20 packages of Fatcow butter a day, so only 8 packages of unsalted
butter and 12 packages of salted butter are put on the shelf. If 15
people buy Fatcow butter on a particular day (each person buying one
package) what is the probability that all buyers will find the kind of
butter they want?

Q2) Suppose that on average one in every 100 passengers does not show
up for an airplane flight. An airline has sold 250 tickets for a
flight serviced by an airplane that has 247 seats. What is the
probability that every person who shows up for the flight will get a
seat? [This I have done using Poisson approximation to the binomial
distribution. If you can do this using that procedure, please let me
know.]
```

```
Date: 06/08/99 at 07:50:50
From: Doctor Anthony
Subject: Re: Probability

>what is the probability that all buyers will
>find the kind of butter they want?

The probabilities where everyone gets what he or she wants are:

Unsalted    Salted            Probability
--------------------        ------------------
8          7        C(15,8) x 0.4^8 x 0.6^7   =  0.118055
7          8        C(15,7) x 0.4^7 x 0.6^8   =  0.177084
6          9        C(15,6) x 0.4^6 x 0.6^9   =  0.206597
5         10        C(15,5) x 0.4^5 x 0.6^10  =  0.185938
4         11        C(15,4) x 0.4^4 x 0.6^11  =  0.126776
3         12        C(15,3) x 0.4^3 x 0.6^12  =  0.063388
--------------
Total  =  0.877837

>What is the probability that every person who shows up
>for the flight will get a seat?

The Poisson or Normal approximations could be used.

If you use the Poisson, the expected number not turning up is 250 x
1/100 = 2.5

Everyone gets a seat if 3 or more fail to turn up.

So we calculate  1 - [P(0) + P(1) + P(2)]

P(0) =     e^(-2.5)     =  0.082085
P(1) = 2.5/1 x P(0)     =  0.205212
P(2) = 2.5/2 x P(1)     =  0.256516
---------------
Total =  0.543813

So the probability that everyone will get a seat = 1 - 0.543813 =
0.456187

If you use the Normal approximation we have
np = 2.5   npq = 2.5 x 0.99 = 2.475   s.d. = 1.5732

Using continuity correction, the x value is 2.5

2.5 - 2.5
z =  ----------- = 0   A(z) = 0.5
1.5732

So using the Normal approximation the probability that 3 or more will
fail to turn up is 0.5.

If you use the Binomial distribution with n = 250, p = 0.01, q = 0.99
we get

P(0) =    0.99^250                = 0.0810585
P(1) = C(250,1) x 0.01 x .99^249  = 0.2046932
P(2) = C(250,2) x .01^2 x .99^248 = 0.2574172
---------------
Total = 0.5431689

So the probability that everyone will get a seat is 0.456831

We can see that the Poisson approximation in this example is much
closer to the true probability than the Normal approximation. This
will generally be true if p is very small and n is large.

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Probability

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