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Statistics

Notes Solution.

Let A be the event that the unit is assembled on line A, B be the event that it is assembled on line
B and D be the event that it is defective.
Thus, we can write
2 3 a 5 a 10
P A a f = , P B a f = , P D/Af = and P D/Bf =
5 5 100 100
Further, we have
P A b Dg  2 ´ 5  1 and P B  b Dg  3 ´ 10  3
5 100 50 5 100 50
The required probabilities are computed form the following table:

A B Total
1 3 4
D
50 50 50
19 27 46
D
50 50 50
20 30
Total 1
50 50

From the above table, we can write

1 50 1 3 50 3
)
)
P ( /A D  ´  , ( /P B D  ´ 
50 4 4 50 4 4
)
)
P ( /A D  19 ´ 50  19 , ( /P B D  27 ´ 50  27
50 46 46 50 46 46
7.4 Summary of Formulae

1. (a) The number of permutations of n objects taking n at a time are n!

! n
n
(b) The number of permutations of n objects taking r at a time, are P 
r
(n r )!
(c) The number of permutations of n objects in a circular order are (n - 1)!
(d) The number of permutations of n objects out of which n are alike, n are alike, ......
1 2
! n
n are alike, are
k
n !n ! ... !
n
1 2 k
! n
n
(e) The number of combinations of n objects taking r at a time are C 
r
( ! r n r )!
2. (a) The probability of occurrence of at least one of the two events A and B is given by :
P (A  ) B  P A ( ) P (A  B ) 1 P A  (  B ) .
( ) P B+
(b) The probability of occurrence of exactly one of the events A or B is given by :
P (A  B ) ( A+ P  B ) or P (A ) B  P (A ) B

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