(a) Let X be the number of purchase boxes that do not have the desired prize.
The random variable X follows negative binomial distribution with parameters p=0.1,r=4
nb(x;4,0.1)
(b) In 6 boxes we obtain 4 successes (prizes) and 2 failures
nb(2;4,0.1)=(r−1x+r−1)pr(1−p)x
=(4−12+4−1)(0.1)4(1−0.1)2
=0.00081≈0.0008 (c)
nb(0;4,0.1)+nb(1;4,0.1)+nb(2;4,0.1)
=(4−10+4−1)(0.1)4(1−0.1)0
+(4−11+4−1)(0.1)4(1−0.1)1
+(4−12+4−1)(0.1)4(1−0.1)2
=0.0001+0.00036+0.00081=0.00127
≈0.0013
(d) The mean (the expected value) is
μ=pr(1−p)=0.14(1−0.1)=36This means that we expect to purchase 36 boxes without the prize (failure) until we have obtain the four prizes (success)
In total we expected to need 36+4=40 boxes.
Boxes without the desired prize: 36. do you expect to purchase?
Boxes to purchase: 40.
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