Question

Probablity http://prntscr.com/5iioqq

Answer 1

http://prntscr.com/5iiox7

Answer 2

@ganeshie8

Answer 3

so at least one used stamp been missing
all stamps are 24 , used are 10
so 1 or 2 or 3 stamps out of 24
1/24+2/24+3/24=6/24=1/4

Answer 4

ermm not sure

Answer 5

I'll be back. This is so mind boggling

Answer 6

sorry I understand now the question

Answer 7

P(C) = the probability of at least one used stamps are missied

Answer 8

this is just hard to calculated let's calculate the Complementary event

Answer 9

1 - P(C^c) = P(C)

Answer 10

what Complementary of P(C) , no stamps used are missing ;

Answer 11

num of the not used = 14 stamps; num of all stamps = 24;
\[\frac{ 14 }{ 24 }\frac{13}{ 23 } \frac{12 }{ 22}\]

Answer 12

probability that all the missing is (not used); ,but we need the Complementary of this
1- (result above) = P(C)

Answer 13

\[P(C\A) = P(C inresect A)/P(A)\]

Answer 14

P(C intersection A) = A AND C ;
mean that Probablity at least one is missing is used stamps AND there are not green;
this mean is simple :-
at least one used and not green;

Answer 15

nice :) you may use `\cap` for \(\cap\)

Answer 16

does this work

\(P(A~\cap~C) = 1 - P((A\cap C)^c) = 1 - [P(A^c) \cup P(C^c)] = \cdots \)

Answer 17

P(C) = 415/506

Answer 18

this the first part;
the second is to solve the P(C/A)

Answer 19

P(C\A) = P(C AND A)\P(A)

Answer 20

P(C AND A) = at least one of the missing is used AND not green;

Answer 21

\[(\frac{ 10 }{ 24 } \frac{ 10 }{ 23} \frac{ 9 }{ 22 }) *3 +(\frac{ 10 }{ 24 } \frac{ 9 }{ 23} \frac{ 10 }{ 22 }) *3 +(\frac{ 10 }{ 24 } \frac{ 9 }{ 23} \frac{8 }{ 22 }) \]\]

Answer 22

P(A) = all the missing is not green, numof not green stamps is 20;
\[(\frac{ 20}{ 24 } \frac{ 19 }{ 23} \frac{ 18 }{ 2 }) \]

Answer 23

P(C AND A) = 255/506
P(A) = 285/506

P(C AND A)/ P(A) = 17/19 // this is the answer

Answer 24

part 3 to check if the events are independent or not :
P(A AND C) = P(C) *P(A) // if this occur then they are independet

Answer 25

P(A AND C) = 255/506;
P(C) = 415/506;
P(A) = 285/506;

Answer 26

\[\frac{ 285 }{ 506 } * \frac {415}{506} \neq \frac {255}{506}\]
this mean that C and A not independent;

Answer 27

@ganeshie8 i put the answer ; if you want to see

Answer 28

@ganeshie8 I think that you also right .

Answer 29

Makes perfect sense thanks @borak :D

Answer 30

@borak hey thanks for making those explanations. The only thing that I don't understand is how you calculated
\[P(A\cap C)\]but I got some 'inspirations' somehow
because I'm fond of using combinations
so it says at least one used is missing so there are 3 cases ie
-1 used and not green and the other 2 not used and not green\[\frac{ 10C1 \times 10C2 }{ 24C3 }=225/1012\]OR
-2 used and not green and the other 1 not used and not green\[\frac{ 10C2\times 10C1 }{ 24C3 }=225/1012\]OR
-3used and not green\[\frac{ 10C3 }{ 24C3 }=15/253\]adding all the independent cases up
\[P(A\cap C)=\frac{225}{506}\]

Answer 31

yea this how we got P( A AND C)

Answer 32

my bad\[P(A\cap C)=\frac{ 255 }{ 506 }\]