prob.distribution math help please?!?

A burger restaurant has a 3% chance of running out of meat every day. 

a)What is the probability the burger restaurant will run out of meat one day in the next two weeks? 

b)What is the probability the burger restaurant will run out of meat at least two days in the next two weeks? 

c)What is the expected number of times the burger restaurant will run out of meat? 

2 Answers

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  • Alan
    Lv 7
    10 months ago

    Use the binomial probability formula .

    P( exactly k events in n trials) = (n k) p^(k) (1-p)^(n-k)

    where (n k ) = nCk = n! / ( (n-k)! k!)

    n = number of trials

    k = exact number of successes

    p = probability of success in one trial

    a.  

    exactly one day of the 14  

    k = 1 , n = 14 , p = 0.03 , just plug that into the

    p(1) =   (14 1) (0.03)^1(0.97)^13   

    (14 1) = 14!/ (13!)*(1!) =   14 

    p(1) = 14*0.03^1*(0.97)^13  = 0.282671378

    b. (at least 2 days)  

    P( at least 2 days) = 1 -P(0 days) - P(1 days) 

    P(0 days)   = (14 0) (0.03)^0 (0.97)^14  = (0.97)^14 

    P( 0 days) =    0.652836277

    p(1) = 14*0.03^1*(0.97)^13 = 0.282671378  

    P(at least 2 days) = 1 -  0.652836277 - 0.282671378 =  

    P(at least 2 days) = 0.064492345

    (c) 

    Expected value of binomial probability distribution = np 

    E(k) =  14*0.03 =  0.42

    so it is expected that it will occur either 0 or 1 times in two weeks, 

    but from (b) you know 6 percent of the time , it will occur 2 or 

    more times. 

  • 10 months ago

    a) the probability of NOT running out one day in the next 2 weeks (14 days) is

    0.97 ^ 14 so see what that is and subtract it from 1.

    b) Now it will be 1 minus (the probability of no days plus exactly one day)

    P(no days) is from part a.

    P(exactly one day) = 14 • (0.97)^13 • (0.03)

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