permutation/probability

2010-04-24 7:33 am

(1)Find the number of integers between1000and4000 which can be formed by using the digits1,2,3,4:(a)if each digits may be used only once,(b)if each digit may be used more than once.
ans: 18,192

(2)In how many different ways can the word''MATHEMATICS'' be arranges?find the number of arrangements in which all the vowels come together.
ans: 120960

(3)Suppose there are 4 white balls and 2 red balls in a bag and the balls have equal chance to be selected. If a ball is drawn out from the bag each time without replacement, what is the probabilityof getting a thrid white ball in the fourth draw?
ans: 2/5

(4)There are two bags, one contain 4 white balls and 2 red balls, the other contains 3 white balls and 3 red balls. A bag is randomly selected from these two bags and a ball is drawn out from the selected bag.The ball is then put into the other bag and a ball is then drawn from this bag.Find the probability that at least one ball drawn out is white.
ans:67/84

(5)Eight cards are marked 1,1,1,2,2,3,4,5respecitvely. Two cards are selected at random, one after the other without being replaced. Find the probbailities that:(a)the numbers on the two cards are the same(b)the numbers on the two cards have a sum of 9(c)the numbers on the two cards have a sum of 4.
ans: 1/7,1/28,1/7

(6)A bag contains five cards numbered 3,4,4,5,6.Two cards are drawn from the bag one by one without replacement.Find the probability that the sum is 9.
ans:3/10

更新1:

(7)A 3-digit number consists of three digits 3, 6 and x. Given that x is one of the digits from 1 to 9. Find the probability that the 3-digit number is divisible by 6.

更新2:

ans: 4/27

回答 (1)

用戶9112

2010-04-24 8:14 am

✔ 最佳答案

(1) (a) 4 must not happen in the first digit, hence total number of integers
=3 x 3 x 2 x 1=18
(b) if repeat is allowed, total number = 3 x 4 x 4 x 4=192
(2) There are 2 M, 2 A, 2 T, 1H, 1E, 1I, 1C and 1S total 11 letters
Total number of words possible = 11!/(2!2!2!)=4989600
If the vowels come together there are (AAIE),M,M,T,T,H,C,S
There are 8! of these 8 items and 4! Permutations of the vowels
Total numbers of words = 8!4!/(2!2!2!)=120960
(3) P(third white ball in 4th draw)
=[P(RWW)+P(WRW)+P(WWR)]*P(3rd white in fourth draw)
=[(2/6)(4/5)(3/4)+(4/6)(2/5)(3/4)+(4/6)(3/5)(2/4)]*(2/3)
=(3/5)(2/3)=2/5
(4) Consider both balls are red
Bag A: 4W + 2R
Bag B: 3W + 3R
Probability to get bag A=1/2
P(First ball is red)=2/6
P(Second ball is red)=(3+1)/(6+1)=4/7
Combined probability =(1/2)(2/6)(4/7)=8/84
Probability to get bag B=1/2
P(First ball is red)=3/6
P(Second ball is red)=(2+1)/(6+1)=3/7
Combined probability =(1/2)(3/6)(3/7)=9/84
Probability of both balls are red = 8/84+9/84=17/84
Probability of at least one white ball = 1-17/84=67/84
(5)(a) Total number of ways to draw 2 cards from 8 is C(8,2)=28
For same numbers it can be both 1 having C(3,2)=3 cases
or both 2 having only one way.
Probability = (3+1)/28=1/7
(b) For sum = 9 there is only one way viz 4+5
Probability =1/28
(c) for sum = 4, it can be any one of the three 1’s plus 3 or the two 2’s
Total 4 ways probability = 4/28 = 1/7
(6) Total number of ways to choose 2 cards from 5 = C(5,2)=10
For sum =9, it can be 3+6 or any of the two 4’s + 5.
Total 3 ways and probability = 3/10

2010-04-24 13:12:34 補充：
There are 9 possibilities for the digit x.
For each possibility there are 3! ways to arrange, so there are totally 9*6=54ways
For the number to be divisible by 6, it must be divisible by 3 and even.
For divisible by 3, x must be 3 , 6 or 9

2010-04-24 13:12:54 補充：
For 3, out of the 6 arrangements, 336 and 336 are even (note 3 is doubled)
For 6, out of the 6 arrangements, 636, 636, 366, and 366 are even (6 is doubled)
For 9, out of the 6 arrangements, 936 and 396 are even
So there are in total 8 ways out of 54, probability =8/54=4/27

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