DISCRETE PROBABILITY DISTRIBUTIONS 问题

zipp_882000

4)In a game a fair die is toosed .If the number 6 appear a person receives rm 10 if an odd number appears that person receinves rm 5 and if other numbers appear ,he will be fined rm 6 .Determine the expected gain if the person pays rm 3 to play once .

5)A bag contains 4 red coloured beads and 6 green coloured beads.two beads are drawn,one after the order ,without replacement .suppose X is a random variable which represents the number of red beads drawn ,find E(x)

6)     2  /  5  /4  /1  / 3
       5  / 1  /  2 /3  /4
       3  / 4  / 1  /2  /5
       1  / 3  / 2  /5  /4

In a game ,a player is bilndfolded and he is required to place two rm 1 coin at random on the square labelled ,as shown in the above diagram.both coin can be placed on the same square and assume that each square has an equal chance of being filled by the coins, the player's score is the sum ofthe numbers on the square filled by the coins,construct a probability distribution table for all the possible scores and hence find the player 's expected score after playing 10 times /


thx !!!

nikuang04

感觉上题目很熟悉
原来是pelangi参考书的题目
前几天刚刚做完

4。得到10令吉的几率是1/6
5令吉的几率是1/2
被扣除6令吉的几率是1/3
所以expected value= 10(1/6)+5(1/2)+(-6)(1/3)=RM13/6
那个人给3令吉来玩，所以expected gain=RM13/6 -RM3=RM-5/6
或者expected loss=RM-5/6

5。P(X=0)=6/10 X 5/9
x=1= 4/10 X 6/9 X 2
2: 4/10 X 3/9
剩下的就是乘

6。注意1-5的几率都是1/5
所以S=2 : (1/5)^2
3: 1/25 *2 。。。。。。
得expected value 6
expected score after 10 times: 6X10=60

~HeBe~_@

我们看看第二题。。。
Conditions:
1) 2 outcomes
2) without replacement

所以我们用Hypergeometric Distribution.

x=0     P(X=0)= (4C0 .  6C2)/(10C2) = 1/3
x=1     P(X=1)= (4C1 .  6C1)/(10C2) = 8/15
x=2     P(X=2)= (4C2 .  6C0)/(10C2) = 2/15

E(X)= 0(1/3) + 1(8/15) + 2(2/15) = 12/15


如果你看到Conditions是：
1) 2 outcomes
2) with replacement

所以我们用Binomial  Distribution.

[ 本帖最后由 ~HeBe~_@ 于 26-3-2007 04:24 PM 编辑 ]