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March 21st, 2017, 09:26 AM  #1 
Newbie Joined: Mar 2017 From: Tennessee Posts: 1 Thanks: 0  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. $\cfrac{516}{85140}=.00606=.606%$ *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? THANKS! Last edited by skipjack; March 21st, 2017 at 09:31 AM. 
March 21st, 2017, 11:29 AM  #2 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,493 Thanks: 752 
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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compound, compound probability, multiplication rule, permutations, probability 
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