Question 1203299: Consider a box containing 3 faulty and 2 good fuses. Two fuses are chosen at random from the box. What is the probability of choosing:
a) No faulty fuse
b) 1 faulty fuse
c) 2 faulty fuses
Hence, draw the probability distribution table for event X, the number of faulty fuses obtained.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the probability of choosing a faulty fuse on the first try is 3/5.
the probability of choosing a good fuse on the first tey is 2/5.
you choose two fuses at random.
the probability of choosing two good fuses is 2/5 * 1/4 = 2/20 = 1/10.
this is the same as choosing 0 faulty fuses.
the probability of choosing two faulty fuses is 3/5 * 2/4 = 6/20 = 3/10.
the probability of choosing one faulty fuse is (3/5 * 2/4) + (2/5 * 3/4) = 6/20 + 6/20 = 12/20 = 6/10.
the total possible combinations of getting 2 elements out of 5 elements, when 3 are faulty and 2 are good would be:
let the faulty fuses be f1, f2, and f3.
let the good fuses be g1, g2.
you are drawing 2 fuses out of 5.
the possible combinations are:
f1 f2 = 0 good
f1 f3 = 0 good
f1 g1 = 1 faulty 1 good
f1 g2 = 1 faulty 1 good
f2 f3 = 0 good
f2 g1 = 1 faulty 1 good
f2 g2 = 1 faulty 1 good
f3 g1 = 1 faulty 1 good
f3 g2 = 1 faulty 1 good
g1 g2 = 0 faulty
total of 0 good is 3/10
total of 0 faulty is 1/10
total of 1 faulty 1 good is 6/10
you could also annalyze this using combination formulas of c(n,x), where n is the total number of possibilities and x is the desired number of possibilities.
the formulas becomes:
for 0 good = c(3,2) / c(5,2) = 3/10.
for 0 faulty = c(2,2) / c(5,2) = 1/10.
for 1 faulty 1 good = c(3,1) * c(2,1) / c(5,2) = 6/10.
note that c(n,x) is equal to n! / (x! * (n-x)!)
as an example: c(5,2) is equal to:
5! / (2! * 3!) which is equal to:
(5*4*3!) / (2! * 3!) which is equal to:
(5*4) / 2! which is equal to:
(5*4) / (2*1) which is equal to:
20 / 2 which is equal to:
10
your solutions are:
a) No faulty fuse = 1/10
b) 1 faulty fuse = 6/10
c) 2 faulty fuses = 3/10
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