Fig 13 1
Fig 13 2
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not to be republished
 538
MATHEMATICS
Let F be the event that ‘there is at least one tail’ and E be the event ‘the die shows
a number greater than 4’ Then
F = {(H,T), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)}
E = {(T,5), (T,6)} and E ∩ F = {(T,5), (T,6)}
Now
P(F) = P({(H,T)}) + P ({(T,1)}) + P ({(T,2)}) + P ({(T,3)})
+ P ({(T,4)}) + P({(T,5)}) + P({(T,6)})
= 1
1
1
1
1
1
1
3
4
12
12
12
12
12
12
4
 
 
 
 
 
 
and
P(E ∩ F) = P ({(T,5)}) + P ({(T,6)}) = 1
1
1
12
12
6
 
 
Hence
P(E|F) =
1
P(E
F)
2
36
P(F)
9
4
∩
=
=
EXERCISE 13