Two lamps are to be chosen from pack of 12 lamps where four are defective and the rest are non defective. what is the probablity that
a.both are defective?
b.one is defective?
c.at most one is defective?
A. 4 are defective and 8 are non-defective
P(2 are defective) =
C(4,2)/C(12, 2)
= 1/11
B. One lamp is defective means the other one is non- defective
P(1- defective) =
(8, 1)(4, 1)/(12, 2)= 16/33
C.
P( at most onedefective)=
P(no defective) +
P(1- defective)
=(8, 2)(4, 2)+(8, 1)(4, 1)/(12, 2)
= 10/11
A. 4 are defective and 8 are non-defective
P(2 are defective) =
C(4,2)/C(12, 2)
= 1/11
Two lamps are to be chosen from a pack of 12 lamps where four are defective and the rest are non defective. What is the probability that a both are defective? b One is defective? c at most one is defective?
a. The probability of selecting the first defective lamp is 4/12. Since we don't replace the first lamp, the probability of selecting a second defective lamp from the remaining 3 defective lamps out of the remaining 11 lamps is 3/11. Therefore, the probability that both lamps are defective is (4/12) * (3/11) = 1/11.
b. To calculate the probability of selecting one defective lamp, we need to consider two scenarios: selecting a defective lamp first and then a non-defective lamp, or selecting a non-defective lamp first and then a defective lamp.
Scenario 1: The probability of selecting a defective lamp first is 4/12. The probability of selecting a non-defective lamp second is 8/11 (since there are 8 non-defective lamps left out of the remaining 11 lamps). Therefore, the probability of this scenario is (4/12) * (8/11).
Scenario 2: The probability of selecting a non-defective lamp first is 8/12. The probability of selecting a defective lamp second is 4/11 (since there are still 4 defective lamps left out of the remaining 11 lamps). Therefore, the probability of this scenario is (8/12) * (4/11).
Adding the probabilities of the two scenarios together, we get [(4/12) * (8/11)] + [(8/12) * (4/11)] = 32/132 + 32/132 = 64/132, which simplifies to 8/33.
c. To calculate the probability that at most one lamp is defective, we need to consider three scenarios: selecting both lamps as non-defective, selecting one defective and one non-defective lamp, or selecting both lamps as defective.
The probability of both lamps being non-defective is calculated as (8/12) * (7/11) = 56/132.
The probability of selecting one defective and one non-defective lamp is calculated as in part b: (4/12) * (8/11) + (8/12) * (4/11) = 64/132.
The probability of both lamps being defective is calculated as (4/12) * (3/11) = 12/132.
Adding these three probabilities together, we get (56/132) + (64/132) + (12/132) = 132/132, which simplifies to 1.
Therefore, the probability that at most one lamp is defective is 1.
To find the probability in each case, we need to determine the total number of lamp combinations and the number of desired outcomes.
a. Probability that both lamps are defective:
To compute this probability, we need to consider that we are choosing two lamps from a pack of 12 lamps, out of which 4 are defective.
Total number of combinations = Choose(12, 2) = 12! / (2! * (12-2)!) = 66 (using the combination formula)
Number of desired outcomes = Choose(4, 2) = 4! / (2! * (4-2)!) = 6 (choosing 2 defective lamps out of 4)
Probability = Number of desired outcomes / Total number of combinations = 6 / 66 = 1 / 11
Therefore, the probability that both lamps are defective is 1/11.
b. Probability that one lamp is defective:
For this probability, we need to determine the number of ways to choose one defective lamp out of four, and one non-defective lamp out of the remaining eight.
Total number of combinations = Choose(12, 2) = 66 (same as above)
Number of desired outcomes = Choose(4, 1) * Choose(8, 1) = 4 * 8 = 32 (choosing 1 defective lamp out of 4 and 1 non-defective lamp out of 8)
Probability = Number of desired outcomes / Total number of combinations = 32 / 66 = 16 / 33
Therefore, the probability that one lamp is defective is 16/33.
c. Probability that at most one lamp is defective:
For this probability, we need to determine all the possible outcomes where at most one lamp is defective - this includes the cases of zero defective lamps and one defective lamp.
Total number of combinations = Choose(12, 2) = 66 (same as above)
Number of desired outcomes = Choose(4, 0) * Choose(8, 2) + Choose(4, 1) * Choose(8, 1) = 1 * 28 + 4 * 8 = 60
Probability = Number of desired outcomes / Total number of combinations = 60 / 66 = 10 / 11
Therefore, the probability that at most one lamp is defective is 10/11.
prob(defective) = 4/12 = 1/3
prob(not defective) = 2/3
a) prob(both defective)
= C(2,2) (1/3)^2
= 1/9
b) one defective:
= C(2,1)(1/3)(2/3)
= 2(2/9) = 4/9
c) at most 1 is defective
----> rule out both defective, which we did in a)
so Pro(of at most 1 defective) = 1 - 1/9 = 8/9
analysis:
we have the following cases
d = defective, g = good
gg = (2/3)(2/3) = 4/9
gd = (2/3)(1/3) = 2/9
dg = (1/3)(2/3) = 2/9
dd - (1/3)(1/3) = 1/9
total of prob's = 9/9 = 1