Family Tree

Family Tree

About Me

My photo
Kathmandu, Bagmati Zone, Nepal
I am Basan Shrestha from Kathmandu, Nepal. I use the term 'BASAN' as 'Balancing Actions for Sustainable Agriculture and Natural Resources'. I am a Design, Monitoring & Evaluation professional. I hold 1) MSc in Regional and Rural Development Planning, Asian Institute of Technology, Thailand, 2002; 2) MSc in Statistics, Tribhuvan University (TU), Kathmandu, Nepal, 1995; and 3) MA in Sociology, TU, 1997. I have more than 10 years of professional experience in socio-economic research, monitoring and documentation on agricultural and natural resource management. I had worked in Lumle Agricultural Research Centre, western Nepal from Nov. 1997 to Dec. 2000; CARE Nepal, mid-western Nepal from Mar. 2003 to June 2006 and WTLCP in far-western Nepal from June 2006 to Jan. 2011, Training Institute for Technical Instruction (TITI) from July to Sep 2011, UN Women Nepal from Sep to Dec 2011 and Mercy Corps Nepal from 24 Jan 2012 to 14 August 2016 and CAMRIS International in Nepal commencing 1 February 2017. I have published articles to my credit.

Tuesday, June 12, 2018

Permutation of Outcomes Without Replacement and Tree Diagram, Statistical Note 13

A bag contains four balls of different colors (A of Red color, B Blue, C Green and D yellow). Select two balls randomly one at a time and do not replace back to the bag after selecting. Calculate the total number of permutation of possible outcomes containing two balls selected without replacement.

Calculating the total number of outcomes using some objects selected randomly one at a time without replacement from the total number of objects is important to calculate the probability of events. Because, in sampling some sampling units are selected without replacement from the population. Also, important is a mathematical concept of the ‘Permutation’ used in identifying the total number of possible outcomes given the condition and calculating the probability of events. I show the permutation of favorable outcomes using the tree diagram 1.






















Diagram 1: Permutation of Outcomes in Selecting Two Out of Four Balls Without Replacement

There will be two stages required to select or arrange two out of four balls. At the first stage or draw, there are four possibilities of randomly selecting the first ball. The first ball could be one of red (A), blue (B), green (C) or yellow (D) balls shown by boxes of those letters and colors in Diagram 1. Once the first ball, red (A), is drawn and not replaced back, there will be three balls left in the bag, B, C and D for the second stage or draw. If the second ball drawn is B, the first outcome constituting two balls is AB which is numbered 1 on the right most side of the diagram 1.

If the first ball selected is B and not replaced back, there will be three balls left in the bag, A, C and D. At the second stage or draw, there are three possibilities, one of B, C or D balls for second stage or draw. If the second ball drawn is A, the outcome constituting two balls is BA which is numbered 4 on the right most side of the diagram 1. Because they are different outcomes as the position of the ball first or second is meaningful. Thus, there are 12 permutations each with two balls selected randomly without replacement from four balls.

Now, I will let you brainstorm how many will be the permutation of outcomes without replacement if there are six balls out of which two balls are drawn one at a time without replacement? Could you use a diagram to show all possible outcomes? Perhaps, difficult to show.  Although the diagrammatic logic is important to understand, as the number of balls or object increases, the diagram becomes complicated and it will be less possible to show all outcomes in the diagram. One may need to use the formula. Could you calculate using a formula? If not, wait for my other statistical notes.

No comments:

Post a Comment