SOLUTION: Each packet of a certain cereal contains a small plastic model of one of ve di erent dinosaurs; a given packet is equally likely to contain any one of the ve dinosaurs. Find the

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Question 1207873: Each packet of a certain cereal contains a small plastic model of one of ve di erent
dinosaurs; a given packet is equally likely to contain any one of the ve dinosaurs.
Find the probability that someone buying six packets of the cereal will acquire
models of their three favourite dinosaurs

Answer by ikleyn(52788) About Me  (Show Source):
You can put this solution on YOUR website!
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Each packet of a certain cereal contains a small plastic model of one of ve di erent
dinosaurs; a given packet is equally likely to contain any one of the ve dinosaurs.
Find the probability that someone buying six packets of the cereal will acquire
models of their three favourite dinosaurs.
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See how I edited this post in order for it makes sense.

    Each packet of a certain cereal contains a small plastic model of one of five different
    dinosaurs; a given packet is equally likely to contain any one of the five dinosaurs.
    Find the probability that someone buying six packets of the cereal will acquire
    models of their three favorite dinosaurs.



                 SOLUTION


In this problem, we have random samples of 6 items from unlimited pool of items of 5 different types mixed evenly.


Selection one or another type at each trial is random; each type can be selected with the probability 1/5.


For us, the success at each random trial is to get randomly any item of some of 3 favorite types; 
so, the probability of success is 3/5 at each individual trial.


The number of trials is 6.


So, we have a Binomial distribution experiment with  6 trials, 3 success trials 
and probability of the individual success at each trial of 3/5.


Hence, to answer the question, use the standard formula of partial binomial distribution probability

    P = C%5B6%5D%5E3%2A%283%2F5%29%5E3%2A%281-3%2F5%29%5E%286-3%29 = %28%286%2A5%2A4%29%2F%281%2A2%2A3%29%290.6%5E3%2A0.4%5E3 = 0.27648 (rounded).


ANSWER.  

Solved.