A certain store has kept records over many years and has found that for people e

Question

A certain store has kept records over many years and has found that for people entering the store:

28% enter and leave without purchase or sample merchandise

42% enter and leave with a purchase

22% enter and leave with sample merchandise

6% enter to return a product and leave without purchase

2% enter to return a product and leave with a purchase

If 25 people enter a store on a given day, what is the probability that at least 7 enter and leave without pruchase or sample merchandise, at least 10 enter and leave with a purchase, at least 5 enter and leave with sample merchandise, and at least 2 enter to return a prodct and leave without a purchase? (there are five cases to consider)

Explanation / Answer

1. There are n=25 independent trials, with probability of success, p=0.28, being constant throughout the trials. The specific number of success in n trials is denoted by r. Use Binomial distribution, P(X,r)=nCr(p)^r(1-p)^n-r to derive the required probability.

P(X>=7)=1-P(X<7)=1-[P(X=0)+P(X=1)+P(X=2)+...+P(X=6)]

=1-[25C0(0.28)^0(1-0.28)^25+25C1(0.28)^1(1-0.28)^24+25C2(0.28)^2(1-0.28)^23+...+25C6(0.28)^6(1-0.28)^19]

=1-0.4247

=0.5753

2. P(X>=10)=1-P(X<10)=1-[P(X=0)+P(X=1)+...+P(X=9)]

=1-[25C0(0.42)^0(1-0.42)^25+25C1(0.42)^1(1-0.42)^24+25C2(0.42)^2(1-0.42)^23+...+25C9(0.42)^9(1-0.42)^16]

=1-0.3465

=0.6535

3. P(X>=5)=1-P(X<5)=1-[25C0(0.22)^0(1-0.22)^25+25C1(0.22)^1(1-0.22)^24+...+25C4(0.22)^4(1-0.22)^21]

=1-0.3282

=0.6718

4. P(X>=2)=1-P(X<2)=1-[P(X=0)+P(X=1)]=1-[25C0(0.06)^0(1-0.06)^25+25C1(0.06)^1(1-0.06)^24]

=1-0.5527

=0.4473

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