3 Probability Qs

17(a) Consider P(2 black chocolates) = x/(16+x) * (x-1)/(15+x)So,x/(16+x) * (x-1)/(15+x) < 1/12By trial and errorx = 2,3,4,5,6,7(b) (7C2)(16C1)/23C3 + 7C3/23C3 = 0.20918(a)(i) P(not raining on that day) = 1 - 0.2 = 0.8
(ii) P(not raining on two consecutive days) = 0.8 * 0.8 = 0.64
(iii) (3C1)(0.2)(0.8)^2 = 0.384(b)(i) P(not raining on Saturday and Sunday)
= (0.4)(0.8)
= 0.32(ii) P(raining on Sunday)
= (0.8)(0.2) + (0.2)(0.6)
= 0.16 + 0.12
= 0.28(iii) Let the probability that raining on the just past Friday is p
Raining on Sunday = p * 0.6 * 0.4 + p * 0.4 * 0.8 + (1 - p) * 0.2 * 0.4 + (1 - p) * 0.8 * 0.8
= 0.72 - 0.16pP(raining on the just past Friday| Raining on Sunday) = 0.56p/(0.72 - 0.16p

19(a) 8! = 40320
(b)(i) If Ken sits the first row, then May should be sit the second row
Combinations = 4 * 4 * 6 * 5 * 4 * 3 * 2 * 1 = 11520
Similarly, if Ken sits the second row, then May should be sit the first row
Required probability = 11520 * 2/40320 = 0.571(ii) Consider Ken and Mary as a group
Required probability = 2 * 7!/8! = 0.25

2012-08-05 16:04:24 補充：
18b(iii) 有問題﹐因為沒有提供星期五本身下雨的概率

2012-08-05 21:55:19 補充：
17(a) Consider P(2 black chocolates) = x/16 * (x-1)/15

So,x/16 * (x-1)/15 < 1/12

x(x - 1) < 20

So, -4 < x < 5. So, possible values of x is 2,3,4

(b) (4C2)(12C1)/16C3 + 4C3/16C3 = 0.13571

2012-08-05 22:03:13 補充：
18b(iii) The answer given by the book is assumed that p = 0.5.

19b(ii) Consider Ken and Mary as a group and sit at the first row. Combinations = 2 * 3 * 6! (As Ken and Mary can change their seat)

Similarly for the second row

So, Required probability = 2 * 2 * 3 * 6!/8! = 3/14

1. P(dark chocolate is taken out) =x/16
P( 2 dark chocolates are taken out ) = x/16 * (x-1)/15

x(x-1)/(16*15) < 1/12
x(x-1)< 16*15/12
x(x-1)<20
x^2 -x -20 <0
(x-5)(x+4)<0
-4 <x <5
(a) all possible values of x =2, 3, 4 (2 <=x <= 16 is given)
(b) P(at least two chocolates taken out)
x at its maximum value is 4
P(two chocolates taken out)+ P(three chocolates taken out)
3C2 *(4/16)(3/15)(12/14) + (4/16)(3/15)(2/14)

3C2, 因為 2 dark +others 的排法有 3C2種 e.g. ddo, dod, odd

=3*144/3360 + 24/3360
=3*12/280 + 2/280
=3*6/140 +1/140
=19/140 //

18(ai) P(not raining on that day) =1-0.2 =0.8//
(aii)P(not raining on two consecutive days) =0.8*0.8=0.64//
(aiii)P(only one day raining within 3 days) =3C1* 0.8*0.8*0.2
=3*0.128
=0.384//

b(i) P(not raining on both Saturday & Sunday ) =0.4 * 0.8=0.32 //
b(ii)P(raining on Sunday; given that not raining on Friday)
P(raining on Saturday and raining on Sunday) +P(not raining on Saturday and raining on Sunday)
=0.2* 0.6 + 0.8*0.2
=0.12+0.16
=0.28//
b(iii) P(it is raining on Friday; given that it is not raining on Sunday)

P(it is not raining on Sunday)
=P(it is raining on both Friday & Saturday and not raining on Sunday)
+ P(it is raining on Friday & not raining on both Saturday and Sunday)
+P(it is not raining on both Friday & Sunday and raining on Saturday)
+P(it is not raining on Friday , Saturday and Sunday)
=0.6*0.4+0.4*0.8 + 0.2*0.4 + 0.8*0.8
=0.24+0.32+0.08+0.64
=1.28

P(it is raining on Friday; given that it is not raining on Sunday)
(0.24+0.32)/1.28
=0.56/1.28
=0.4375 //

19(a)8! =40320

(bi)Ken & May sit in different rows
6C3* 4! (no. of the ways that the row Ken sit)
4! (no .of ways that the row May sit)
=6C3 * 4!*4! * 2 (two rows can be exchanged)
=20*24*24*2=23040//

(bii) Ken sit next to May at the same row
6C2 * 3! * 4! * 2 * 2 (Ken and May can exchange and the two rows can exchange too)
=15*6*24*2*2
=8640 //



2012-08-05 16:53:10 補充：
19(bi)=23040/40320
=4/7

2012-08-05 16:54:32 補充：
19(bii) 8640/40320
=3/14//

Just do as much as you can will be good enough =]

18) b) iii) is quite weird...

2012-08-05 16:16:41 補充：
I have a method to find the P(sat. rain) for 18) b) iii), I not sure if it's ok.
First find the P(rain for any given day). Let it = p
p = 0.6p + 0.2(1 - p), p = 1/3
Assume that we don' t know Sun.rained.
P(rain on Sat.& Sun.) = 0.6p = 1/5; P(rain on Sun.but not Sat.) = 0.2(1 - p) = 2/15

2012-08-05 16:17:21 補充：
P(rain on Sat.) = 1/5/(1/5 + 2/15) = 1/3

Is this ok?

2012-08-05 20:48:37 補充：
To freda:

Can you further explain how you get this 0.6*0.4 + 0.4*0.8 + 0.2*0.4 + 0.8*0.8 ?