Correct Answer - Mean =
3
4
,variance =
39
80
Three bulbs drawn one by one without replacement is the same as drawing 3 bulbs simultaneously.
Let X = number of defective bulbs in a lot of 3 bulbs drawn.
Then, X=0,1,2 or 3.
P(X=0)=P(one of the bulbs is defective)
=
12C3
16C3
=(
12×11×10
3×2×1
×
3×2×1
16×15×14
)=
11
28
.
P(X=1)=P(1 defective bulb and 2 nondefective bulbs)
4C1×12C2
16C3
=(
4×12×11
2×1
×
3×2×1
16×15×14
)=
33
70
.
P(X=2)=P(2 defective bulbs and 1 nondefective bulb)
=
(4C2×12C1)
16C3
=(
4×3
2×1
×12×
3×2×1
16×15×14
)=
9
70
.
P(X=3)=P(3 defective bulbs)
=
4C3
16C3
=(
4×3×2
3×2×1
×
3×2×1
16×15×14
)=
1
140
.
