Answer to Question #245839 in Statistics and Probability for Happy

Question

A company has 5 products in it’s Basics range, 5 products in it’s Standard range and 4 products in it’s Premium range. The company wishes to advertise in a local newspaper, but due to space constraints, it can only advertise 6 products.

1. How many possible 6-product groupings/selections are possible?
2. What is the probability that an advert of 6 randomly selected products contains an equal number of products from each of the ranges? (Round off to 1 decimal place.)
3. What is the probability that an advert of 6 randomly selected products contains the whole Basics range and one item from the Standardrange? (Round off to 3 decimal places.)
4. What is the probability that an advert of 6 randomly selected products contains the whole Basics range? (Round off to 3 decimal places.)

Answer 1

Solution.

1.

$$C_{14}^6=\frac{14!}{6!8!}=\frac{9•10•11•12•13•14}{1•2•3•4•5•6}=11•3•13•7=3003$$

Answer. 3003

2.

$$\frac{C_5^2C_5^2C_4^2}{C_{14}^6}=\frac{\frac{5!}{2!3!}\frac{5!}{2!3!}\frac{4!}{2!2!}}{3003}=\frac{10•10•6}{3003}=0.1998...=0.2$$

Answer. 0.2

3.

$$\frac{C_5^5C_5^1}{C_{14}^6}=\frac{\frac{5!}{5!0!}\frac{5!}{1!4!}}{3003}=\frac{5}{3003}=0.001665...=0.002$$

Answer. 0.002

4.

$$\frac{C_5^5C_5^1}{C_{14}^6}+\frac{C_5^5C_4^1}{C_{14}^6}=\frac{\frac{5!}{5!0!}\frac{5!}{1!4!}}{3003}+\frac{\frac{5!}{5!0!}\frac{4!}{1!3!}}{3003}=\frac{5}{3003}+\frac{4}{3003}=\frac{9}{3003}=0.002997...=0.003$$

Answer. 0.003

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