RAFFLES INSTITUTION
H2 Mathematics (9740)
2014 Year 6
__________________________________________________________________________________________
Assignment S2 Probability
1
N01/2/11(b)
A man writes 5 letters, one each to A, B, C, D and E. Each letter is placed in a separate
envelope and sealed. He then addresses the envelopes, at random, one each to A, B, C, D
and E.
(i)
Find the probability that the letter to A is in the correct envelope and the letter to B is
in an incorrect envelope.
[1]
(ii)
Find the probability that the letter to A is in the correct envelope, given that the letter
to B is in an incorrect envelope.
[3]
(iii) Find the probability that both of the letters to A and B are in incorrect envelopes. [5]
Solution:
Let X denote the event that letter to X is in the correct envelope.
1 3 3
P( A  B ')   
5 4 20 .
(i)
3
P( A  B ') 20 3
P( A | B ') 


4 16
P( B ')
5
(ii)
.
4
3  13
P( A ' B ')  P( B ')  P( A ' | B ')  1   
5  16  20
(iii)
P( A '  B ')  P( B ')  P( A  B ') 
Or from Venn diagram,
4 3 13


5 20 20 .
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Assignment S2: Probability
Raffles Institution H2 Mathematics
2014 Year 6
__________________________________________________________________________________________
2
9740/2010/02/Q7
For events A and B it is given that P( A)  0.7 , P( B)  0.6 and P( A | B ')  0.8 . Find
(i)
P( A  B ') ,
[2]
(ii)
P( A  B) ,
[2]
(iii)
P( B ' | A) .
[2]
For a third event C, it is given that P(C) = 0.5 and that A and C are independent.
(iv)
Find P( A ' C ) .
[2]
(v)
Hence state an inequality satisfied by P( A ' B  C ) .
[1]
Solution:
(i) P( A | B ') 
P( A  B ')
P( B ')
 P( A  B ')  P( A | B ')  P( B ')  0.8  (1  0.6)  0.32
(ii) P( A  B)  P( A  B ')  P( B)  0.32  0.6  0.92
(iii) P( B ' | A) 
P( B ' A) 0.32

 0.457
P( A)
0.7
(iv) P( A ' C)  P( A ')  P(C) since A and C are independent  A’ and C are independent
 (1  0.7)  0.5  0.15
(v) P( A ' B  C)  0.15
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Assignment S2: Probability
Page 2 of 3
Raffles Institution H2 Mathematics
2014 Year 6
__________________________________________________________________________________________
3.
Four girls, Amanda, Beryl, Clare and Dorothy, and three boys, Edward, Frank and
George, stand in a queue in random order.
(i)
(ii)
Find the probability that the first two in the queue are Amanda and Beryl, in that
order.
[1]
Find the probability that either Frank is first or Edward is last (or both).
[2]
(iii) Find the probability that no two girls stand next to each other.
[2]
(iv) Find the probability that all four girls stand next to each other.
[2]
(v)
Find the probability that all four girls stand next to each other given that at least
two girls stand next to each other.
[2]
Solution:
(i)
P(first two in the queue are Amanda and Beryl, in that order) =
(ii)
P(either Frank is first or Edward is last or both) =
1 1 1  1  11
   
7 7 7  6  42
(iii) P(no two girls stand next to each other) =
4!3! 1

7! 35
(iv) P(all four girls stand next to each other) =
4!4! 4

7! 35
(v)
11 1
 
7  6  42
P(all 4 girls stand next to each other│at least two girls stand next to each other)
P(all 4 girls stand next to each other and at least 2 girls stand next to each other)
P(at least 2 girls stand next to each other)
P(all 4 girls stand next to each other)

P(at least 2 girls stand next to each other)

4
2
= 35 
1 17
1
35
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Assignment S2: Probability
Page 3 of 3