1 / 4

# Solutions to the HWPT problems

Solutions to the HWPT problems. How many distinguishable arrangements are there of the letters in the word PROBABILITY? Answer : 11!/(2! 2!)

Télécharger la présentation

## Solutions to the HWPT problems

E N D

### Presentation Transcript

1. Solutions to the HWPT problems

2. How many distinguishable arrangements are there of the letters in the word PROBABILITY?Answer: 11!/(2! 2!) • A and B are special dice. The faces of A are 2,2,5,5,5,5 and the faces of B are 3,3,3,6,6,6. The two dice are rolled. What is the probability that the number showing on die B is greater than the number showing on die A? • Hint: try presenting the event F= “the number showing on die B is greater than the number showing on die A” as a combination of the elementary events using “either or” and “and” operations. • Solution: F =[(A=2) and (Any B) ]or [(A=5) and (B=6)] =(2/6) + (4/6)*(1/2)=2/3.

3. The integers 1,2,3,4 are randomly permuted. What is the probability that 4 is to the left of 2? Hint: Find the total number of permutations. What is the share of the permutations keeping 4 to the left of 2?Solution: There are 4!= 24 permutations. In half of them 4 is to the left of 2.Answer: 12/24=0.5 • 4. A box contains 8 red, 3 white and 9 blue balls. Three balls are to be drawn without replacement. What is the probability that more blues than whites are drawn? (42) • Hint: Present this event as a union of simpler events (use “B” (for blue), “W” and “R” letters. Do not forget that BRR, for example, also satisfies the requirement. • Solution: F= “More blues than whites are drawn”. • F= BBB + BBR + BRR + BBW; P (F)= ( C9,3 + C9,2 8 + 9C8,2 + C9,23)/C20,3=61/95~0.64.

4. 5. A production lot has has 100 units of which 25 are known to be defective. A random sample of 4 units is chosen without replacement. What is the probability that the sample will contain no more than 2 defective units. Hint: (a) Notice that “At most 2 defective” = “0 defective” + “1 defective” + “2 defective”. (b) Notice that the number of outcomes corresponding to “0 defective” is N(0)=C75,4 (meaning that all four units are taken out of 75 non-defective). Using the same approach, find N(1) and N(2). Notice that they are described as intersections (“and”) of two events each. (c) Add up the numbers of outcomes corresponding to all three events in the right-hand side, and divide by the total number of possible selections (4 out of 100). Answer:.

More Related