Answer on Question #46161 – Math – Statistics and Probability
Question. A box containing 8 light bulbs of which 3 are defective. A bulb is selected from the
box and tested. If it is defective, another bulb is selected and tested until a nondefective
bulb is chosen. Find the expected number and variance of bulbs chosen.
Solution. Let 𝜉 be the number of bulbs chosen. Then
𝐶1
5! 7!
5
𝑃(𝜉 = 1) = 𝐶51 = 4! ∙ 8! = 8;
8
𝑃(𝜉 = 2) = 𝑃(𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑖𝑠 𝑑𝑒𝑓𝑒𝑐𝑡𝑖𝑣𝑒, 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑖𝑠 𝑛𝑜𝑛𝑒𝑑𝑒𝑓𝑒𝑐𝑡𝑖𝑣𝑒) =
3! 7! 5! 6!
3 5
𝐶31 𝐶51
∙
=
𝐶81 𝐶71
15
= 2! ∙ 8! ∙ 4! ∙ 7! = 8 ∙ 7 = 56;
𝑃(𝜉 = 3) = 𝑃(𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑎𝑟𝑒 𝑑𝑒𝑓𝑒𝑐𝑡𝑖𝑣𝑒, 𝑡ℎ𝑒 𝑡ℎ𝑖𝑟𝑑 𝑖𝑠 𝑛𝑜𝑛𝑑𝑒𝑓𝑒𝑐𝑡𝑖𝑣𝑒) =
𝐶1
𝐶1
𝐶1
8
7
6
3 2 5
5
= 𝐶31 ∙ 𝐶21 ∙ 𝐶51 = 8 ∙ 7 ∙ 6 = 56;
𝐶31 𝐶21 𝐶11
𝑃(𝜉 = 4) = 𝑃(𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡, 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑡ℎ𝑖𝑟𝑑 𝑎𝑟𝑒 𝑑𝑒𝑓𝑒𝑐𝑡𝑖𝑣𝑒) = 1 ∙ 1 ∙ 1 =
𝐶8 𝐶7 𝐶6
3 2 1
1
= 8 ∙ 7 ∙ 6 = 56.
Therefore 𝜉 has the next distribution:
Value k of
𝜉
1
2
3
4
Probability
5
8
15
56
5
56
1
56
of 𝜉 = 𝑘
5
15
5
1
3
The expected number of bulbs chosen is 𝐸𝜉 = 1 ∙ 8 + 2 ∙ 56 + 3 ∙ 56 + 4 ∙ 56 = 2.
Calculate
5
15
5
1
39
𝐸𝜉 2 = 1 ∙ 8 + 4 ∙ 56 + 9 ∙ 56 + 16 ∙ 56 = 14;
The variance of bulbs chosen is
39
9
15
𝑉𝑎𝑟𝜉 = 𝐸𝜉 2 − (𝐸𝜉)2 = 14 − 4 = 28.
3
15
Answer. 𝐸𝜉 = 2 , 𝑉𝑎𝑟𝜉 = 28.
www.AssignmentExpert.com