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March 21st, 2017, 09:26 AM   #1
Joined: Mar 2017
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Compound Probability

Alicia writes the number $1$ to $45$ on separate cards. She then randomly chooses three of the cards. What is the probability that the $2nd$ and $3rd$ cards will include the digit $9$ in the number.

Using a permutation solution method, I find the total number of possible outcomes ($ _{45}P_3$, $45$ cards taken $3$ at a time) which gives $85140$ different arrangements of the $3$ numbers.

I also use permutations to find the total number of ways the 4 numbers ($9, 19, 29, and 39$) can be arranged taken two at a time ($ _{4}P_2$) to get $12$ different arrangements. (These are the arrangements of the LAST $2$ numbers with a $9$). For each of the $12$ arrangements, there are $43$ different numbers that can come first in the $3$ letter group, so $12*43=516$ total arrangements (desired outcomes).

Using the basic probability model I divide the number of desired outcomes by the number of possible outcomes.


*I have even made lists to check that this is correct.*

However, the answer in the book uses the multiplication rule (which seems valid) to find the solution.

P(any card the fist time) * P(9 the 2nd time) * P(9 the 3rd time)

=$\cfrac{45}{45} * \cfrac{4}{44} * \cfrac{3}{43} = .00634 = .634%$

This doesn't agree with the permutation method (which I trust more unless you can find a flaw).

However, I can get the same solution by multiplication if I start with 45 in the denominator.

$\cfrac{4}{45} * \cfrac{3}{44} = .00606 = .606%$

But I can't find any logic at all in why you would do this.
Can you shed any light on this for me?


Last edited by skipjack; March 21st, 2017 at 09:31 AM.
canaanbowman is offline  
March 21st, 2017, 11:29 AM   #2
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The book is incorrect. I'm surprised such a mistake made it past the editor.

If you select a number with a digit 9 on the first selection then the probabilities of selecting them on the 2nd and 3rd selections will be $\dfrac {3}{44}$ and $\dfrac{2}{43}$

the correct answer is

$p = \dfrac{41\cdot 4 \cdot 3 + 4 \cdot 3 \cdot 2}{45 \cdot 44 \cdot 43} =\dfrac{516}{ 85140} = \dfrac {1}{165}$

which is what you came up with.
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