Exam 1: Probability and Stat stics art sum is greater than Jutity your .,ws\'. t

Question

Exam 1: Probability and Stat stics art sum is greater than Jutity your .,ws'. that the probability of e Justify your nower inteNooly hrn two co are iped, Otherwise, coins are tossed. What is the probabiiny that two brads turs up Jusaihy yon than 3t 6. Two Saif dice are soned, what the zwotwhilty that that tho Tratnler thwVngonon will be twice the sumber appearing on the ot her Justily your an 7 In how many ways can the letters of"newjyeysouth" be arranged so that r s preceded by an s and followed by an e? Justify your EW 8. Five fair dice are relled. What is the probability that the faces showing coust iate 9. A computer has grierated seven random mrambers over the interal 0 to . mre 10. Holly puls two socks out of a drawer at random. The drawer coutains 3 red, s black mber and the other two laces show . a "full house": three of the faces show one different sumber Justify your answe likely that (a) exactly three will be in the isterval 13, 11 or (b) fres than thaee will be greater than Explain and 6 blac socks, all randomly arranged. What is the probability that the two taken are a match? Extra Credit. How many ways can the letters in the word stumgullion be arranged so that the three L's preoede all the other consonants

Explanation / Answer

4)
Probability that Holly knows the answer, p = 0.95
Number of questions for which Holly knows the answer is 0.95*25 = 23.75

5)
Let A be an event that number turns up is less than 3,
Hence probability of getting two heads, P(A) = 2/6 * (0.5)^2 = 0.0833

In the above expression, 2/6 is the probability of getting number less than 3 and (0.5)^2 is the probability of getting both heads when two coins are tossed.

Let B be an event that number turns up is not less than 3, hence the probability of getting two heads when 3 coins are tossed, P(B) = 4/6 * 3C2 * (0.5)^2 * 0.5 = 4/6 * 3 * (0.5)^2 * 0.5 = 0.25

Hence probability of getting two heads, P = P(A) + P(B) = 0.0833 + 0.25 = 0.3333

6)
Possible outcomes are
(1,2), (2,1), (2,4), (4,2), (3,6), (6,3)
Hence required probability = 6/36 = 1/6 = 0.1667

7)
newjerseysouth
If we consider "rse" as one letter, there will be 12 letters that needs to be arranged. This can be done in 12! ways. Among other letters there are 2 e's.

Hence total possible arrangements are 12!/2! = 239500800

  

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