My Math Forum Probability problem

 Probability and Statistics Basic Probability and Statistics Math Forum

 May 12th, 2013, 03:12 AM #1 Member   Joined: Feb 2013 Posts: 44 Thanks: 0 Probability problem OK so there's 80 light bulbs 6 are defective, which means 74 are not defective. If I select 2 at random, what are the odds at least one is defective. the answer key says. 1- C(74,2)/ C(80,2) = 1- 74x73/80x79 = .145 I don't understand the answer key at all. I thought to use a stem-graph ( a line with more lines protruding from that line, shwoing the varous probabilites of otcomes.)) can someone explain please?
May 12th, 2013, 04:51 AM   #2
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 408

Re: Probability problem

Hello, hamburgertime!

Quote:
 There are 80 light bulbs: 6 are Defective, 74 are Good. If I select 2 at random, what is the probability at least one is Defective. The answer key says:[color=beige] .[/color]$1\,-\, \frac{C(74,2)}{C(80,2)} \:=\: 1\,-\,\frac{(74)(73)}{(80)(79)} \:=\: 0.145$ Can someone explain please?

The opposite of "at least one defective? is "NO defective".

What is the probability of no Defective (both are Good)?

There are:[color=beige] .[/color]$C(80,2)$ possible pairs of bulbs.
There are:[color=beige] .[/color]$C(74,2)$ pairs of Good bulbs.

$P(\text{both Good}) \:=\:\frac{C(74,2)}{C(80,2)} \:=\: \frac{\frac{74!}{2!\,72!}} {\frac{80!}{2!\,78!}} \:=\:\frac{(74)(73)}{(80)(79)} \:=\:0.854746835$

$\text{Therefore: }\:P(\text{at least one Defective}) \;=\;1\,-\,0.854745835 \;\approx\;0.145$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

It could be solved like this . . .

There are three possible scenarios:

$\;\;\;\begin{array}{ccccccccc}(1)& \text{1st is D, 2nd is G.} && P(DG) &=& \left(\frac{6}{80}\right)\left(\frac{74}{79}\right ) &=& \frac{444}{6320} \\ \\ \\
(2) & \text{1st is G, 2nd is D.} && P(GD) &=& \left(\frac{74}{80}\right)\left(\frac{6}{79}\right ) &=& \frac{444}{6320} \\ \\ \\
(3) & \text{Both are D.} && P(DD) &=& \left(\frac{6}{80}\right)\left(\frac{5}{70}\right) &=& \frac{30}{6320} \end{array}$

$\text{Therefore:}
\;\;\;P(\text{at least one D}) \:=\:\frac{444}{6320}\,+\,\frac{444}{6320}\,+\,\fr ac{30}{6320} \:=\:\frac{918}{6320} \:\approx\:0.145$

 Tags probability, problem

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post bsjmath Advanced Statistics 2 March 1st, 2014 02:50 PM ystiang Advanced Statistics 1 February 13th, 2014 04:27 PM halloweengrl23 Advanced Statistics 3 March 23rd, 2012 02:27 AM adiptadatta Probability and Statistics 1 May 11th, 2010 06:46 AM indian Algebra 0 February 21st, 2008 04:10 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top