f.5 probability

2008-11-21 6:10 am
1) there are five envelopes, 3 of which contain a coupon each and the rest are
empty. Another couponis put into one of the envelopes at random. After that, an envelope is randomly selected. Find the probability that the selected
envelope is empty

2)there are 56 marbles in a tank. The probability of randomly drawing a red
marble from the tank is 5/8
The rest of the marbles are either blue or yellow. when two marbles are
randomly drawn ,the probability for one to be blue and one to be yellow is
7/110. Find the number of yellow marbles in the tank.
更新1:

ai_kawah :but why the marbles are without replacement

回答 (4)

2008-11-21 6:29 am
✔ 最佳答案
Explanation as follows:

圖片參考:http://i238.photobucket.com/albums/ff245/chocolate328154/Maths552.jpg?t=1227191319


2008-11-20 23:10:08 補充:
Sorry that I have skipped some steps.
Plx press reload for it.

I can answer yr follow up Q:
The marbles are surely without replacement since you draw them one followed by one...
That is the meaning of [ 2 marbles are randonly drawn ]

2008-11-20 23:10:12 補充:
And if replacement is allowed, it must be specified in the Q i.e. { ... with replacement }
So without these words, you may assume it is without replacement.

2008-11-20 23:26:18 補充:
I have already answered your Q, plx feel free to ask if you hv problems again =]
參考: My Maths Knowledge
2008-11-21 6:52 am
1)
p(coupon put in an envelop already have coupon) = 0.6
p(coupon put in an empty envelop) = 0.4

therefore,
p(3 envelop have coupon) = 0.6
p(4 envelop have coupon) = 0.4

so,
p(randomly select an empty envelop)
= 0.4 x p(3 envelop have coupon) 0.2 x p(4 envelop have coupon)
= 0.4 x 0.6 0.2 x 0.4
= 0.32//

2)
#red = 56 x 5/8 = 35
#blue #yellow = 56 - #red = 21
#blue = 21 - # yellow

7/110 = p(1 blue & 1 yellow)
7/110 = p(1st blue, 2nd yellow) p(1st yellow, 2nd blue)
7/110 = #blue/56 x #yellow/55 #yellow/56 X #blue/55
7/110 = 2 X ( 21 - #yellow)/56 x #yellow/55
#yellow x #yellow - 21 x # yellow 98 = 0
#yellow = 7 or 14//
參考: any ce text book
2008-11-21 6:36 am
Two scenarios:
a. If the first draw is empty envelop and put the coupon in, then there would be only one empty envelop left. The chance for the selected one is empty is
Probability is 2/5 * 1/5 = 2/25
b. If the first draw is not an empty envelop and put the coupon in, then there would be two empty envelop left. The chance for the selected one is empty is
Probability is 3/5* 2/5 = 6/25
The total probability is 2/25 + 6/25 = 8/25

2. r = number of red, b = number of blue, y = number of yellow
r/56 = 5/8
r =35
Two marbles are random drawn so either first one is blue, second one is yellow or the other way round so
b/56 * y/55 + y/56 * b/55 = 7/110
2by=7*3080/110
by = 98
Total marbles is 56 so
b = 56-r-y
b = 56-35-y
b = 21-y
Put into equation by = 98
(21-y)y = 98
y^2 -21y +98 = 0
y = 14 or 7
參考: CLOUDWAH - 獻醜
2008-11-21 6:35 am
1)
2,1,1,0,0 1,2,1,0,0 1,1,2,0,0 1,1,1,1,0 1,1,1,0,1

P(the selected envelope is empty)
3/5 x 2/5 + 2/5 x 1/5 = 8/25

2) the number of red is about 35 the remaining marble is 56-35 = 21
Let x be the number of yellow marble
21-x be the number of blue marble

the probability for one to be blue and one to be yellow is
(21-x/56 x x/55)x2 =7/110
x=7 or x =14


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