probabilty questions (1 Viewer)

[u-borat]

from the cambridge 3u Y12 textbook:

10G question 28:

how many different combinations are there of three different integers between one and thirty inclusive such that their sum is divisble by three?

question 29:

a)how many doubles tennis games are possible, given a group of four players?

b)in how many ways can two games of doubles tennis be arranged, given a group of eight players?

c)six married couples are to play in three games of doubles tennis. find how many ways the pairings can be arranged if;
i)there are no restrictions.
ii)each game is to be a game of mixed doubles.

also for those with the textbook, i'm also having trouble with question 25 from the same exercise, but its got a diagram so i can't type it up.

 

[lolokay]

28. I think you would take this one by cases.
all three divisible by 3 = 10!/7!3! = 120
all three 1 more than a multiple of 3 " = 120
all three 2 more than a multiple of 3 " = 120
one from each = 10³ = 1000
total = 1360

29. a) 3 I would think (since for any person, there are 3 others to choose from, and the other 2 make the second pair) or 4!/2!2!2! = 3

b) 8!/2!2!2!2!2!2!2! = 315 (order players, divide by 2 for each pair, each side of the court and then to unorder the courts) maybe I divided by 2 too many times though

c)i) 12 people, so 3 courts
12!/2⁶2³3! = 155925 (order players, unorder pairs, unorder pairs of pairs/court sides, unorder courts)

ii) 6 of each gender
6!6!/2³3! = 10800 (order each gender, unorder court sides, unorder courts)

 

[u-borat]

woah you're insane at this.
thanks a lot.

 

[u-borat]

anda another:

the letters of ENTERTAINMENT are arranged ina row. find the probability that two Es are together and one is apart.

 

[YannY]

anda another:

the letters of ENTERTAINMENT are arranged ina row. find the probability that two Es are together and one is apart.

There are 13 letters with 3 Es 3 Ts 3 Ns

So there are 13!/(3!3!3!) arrangements. If all two Es are together then then we shall treat them as one letter so then there are 12!/(2!3!3!) arrangments. Hence probability.

 

[u-borat]

na man that aint right cos if you get ee then get your 12 other letters, there's a possibility that the e in the group of 12 happens to be at the end so thus you get 3e's not 2.

 

[lyounamu]

anda another:

the letters of ENTERTAINMENT are arranged ina row. find the probability that two Es are together and one is apart.

12!/3!3!2! (the entire probability, not counting the probability where 3Es come together) - 12!/3!3!3! (the chance of 3 Es coming together) = 4435200

 

[lyounamu]

na man that aint right cos if you get ee then get your 12 other letters, there's a possibility that the e in the group of 12 happens to be at the end so thus you get 3e's not 2.

Well, you are right on that part. BUT there are possibilities where they are not together which is evident from Yanny's working out. His working out just didn't include the possibility that they are together.

By the way, if you knew how to to it from the beginning, why did you ask?

 

[u-borat]

yeah that's what i got, but the answer in the back of the book says differently so I was a bit confused. Thanks for the help.