Solution
Let 0 be the event that the batch has 0 defectives, 1 be the event the batch has 1 defective, and 2 be
the event the batch has 2 defectives. Let 0 be the event that neither selected component is defective.
(0
) = 0.5,(1
) = 0.3, (2
) = 0.2
The event 0 can happen in three different ways: (i) Our batch of 10 is perfect, and we get no defectives in
our sample of two; (ii) Our batch of 10 has 1 defective, but our sample of two misses them; (iii) Our batch
has 2 defective, but our sample misses them.
For (i), the probability is (0.5)(1).
For (ii), the probability that our batch has 1 defective is 0.3. Given that it has 1 defective, the probability
that our sample misses it is (
9
2
)
(
10
2
)
, which is 8
10
. So the probability of (ii) is (0.3) (
8
10).
For (iii), the probability our batch has 2 defective is 0.2. Given that it has 2 defective, the probability that
our sample misses both is (
8
2
)
(
10
2
)
, which is 56
90
. So the probability of (iii) is (0.2) (
56
90). We have therefore found
that
(0
) = (0.5)(1) + (0.3) (
8
10) + (0.2) (
56
90).
We use the general conditional probability formula:
(0
|0
) =
(0 ∩ 0
)
(0
)
=
(0.5)(1)
(0.5)(1) + (0.3) (
8
10) + (0.2) (
56
90)
= 0.5784. (1
|0
) =
(1∩0
)
(0
)
=
(0.3)(
8
10)
(0.5)(1)+(0.3)(
8
10)+(0.2)(
56
90)
= 0.2776.
(2
|0
) =
(2∩0
)
(0
)
=
(0.2)(
56
90)
(0.5)(1)+(0.3)(
8
10)+(0.2)(
56
90)
= 0.1440.
Answer: 0.5784; 0.2776; 0.1440.
