Binomial Distribution Probability Problem

  • Thread starter mkir
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  • #1
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Homework Statement


The mailing list of an agency that markets scuba-diving trips to the Florida Keys contains 70% males and 30% females. The agency calls 30 people chosen at random from its list.

What is the probability that the first woman is reached on the fourth call? (That is, the first 4 calls give MMMF.)


Homework Equations


P(x=k) = (n choose k)p^k(1-p)^(n-k)


The Attempt at a Solution


since they want MMMF does it make sense to do just go

.7 * .7 * .7 * .3 and then divide by 4C1 =4

.1029/4
=.025725
 
Last edited:

Answers and Replies

  • #2
statdad
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This is not a binomial distribution problem - in a binomial distribution you are interested in the total number of "successes", but not the order in which they are obtained, or the exact trial on which the first success is obtained.

You could use the multiplication rule, as you did at the end of your post (don't divide by anything), but there is another probability distribution that fits this type of problem - do you know what that other distribution is?
 
  • #3
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No, I don't know what the other distribution is.
 
  • #4
HallsofIvy
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Actually, you don't have to worry about either kind of distribution for this particular question. In order that we get "MMMF" the result must be in that particular order.
What is the probability that the first call be to a man? What is the probability that the second call be to a man? What is the probability that the third call be to a man? What is the probability that the fourth call is to a woman? Now multiply those four probabilities together.
 
  • #5
statdad
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As Halls says, and I mentioned, you don't need the other formula, as you've already done the work in your first post.
 
  • #6
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Thanks guys, I get it now.
 

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