
Probability and Statistics Basic Probability and Statistics Math Forum 
 LinkBack  Thread Tools  Display Modes 
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: USA Posts: 1,977 Thanks: 1026 
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. 

Tags 
compound, compound probability, multiplication rule, permutations, probability 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Compound Angle  justusphung  Algebra  3  February 28th, 2013 11:24 PM 
Compound Angles  bilano99  Algebra  3  October 31st, 2012 09:29 AM 
Compound Angles  bilano99  Algebra  2  October 30th, 2012 09:30 AM 
compound inequality  HellBunny  Algebra  14  April 3rd, 2012 07:59 PM 
Calculating Compound  lilwayne  Calculus  2  November 27th, 2010 02:03 AM 