classical approach word problem help

david langstaff · #1 $Old$ October 31st, 2009, 06:13 AM

I an having a some troubel with a word problem.
A customer randomly selects 6 doughnuts form 58 doughnuts of which 8 contain jelly. What is the probalility that she selects none of those containing jelly?
Here is what I have.
50/58=.862069 is the probablity of selecting one doughnut that does not contain jelly
862069^6=.410442
The probability that she randomly select 6 doughnut and they are not jelly is .410

ANDS! · #2 $Old$ October 31st, 2009, 06:34 AM

Are they asking you to treat this as a binomial experiment? Because thats how you have it set up.

david langstaff · #3 $Old$ October 31st, 2009, 06:43 AM

Part of the problem is that I don't know what there asking for. The question doesn't specify with or withour replication. What is confusing me is the 6 random sample because in class and in the text we always pratice ways of finding a single random sample of a population. I have spent waaay to much time on this one problem. So what ever input you can offer that would be great. In the chapter we are also covering the complement method, addition, mutiplication, counting method.

Guess992 · #4 $Old$ October 31st, 2009, 10:25 PM

I'm unsure of the proper notation for this, but to my knowledge, it would be:

$P(x) = \frac{\frac{50!}{(50-6)!}}{\frac{58!}{(58-6)!}}$

Simplified:

$P(x) = \frac{50}{58}*\frac{49}{57}*\frac{48}{56}*\frac{47}{55}*\frac{46}{54}*\frac{45}{53} = .3926$

I'm assuming she would not be replacing the already selected doughnuts. (That would logically make no sense.)
Were she to replace them, then your answer would be correct. $\bigg(\frac{50}{58}\bigg)^6$

matheagle · #5 $Old$ October 31st, 2009, 11:49 PM

that's hypergeometric (wor) vs binomial (wr)