2% of the fuses manufactured by a firm are found to be defective. Find the probability that a box containing 200 fuses contain, no defective fuses also 3 or more defective fuses
Answers
Answer:
concept: the concept used here is finding the probability in that to we use poisson's distribution
it is given by
The probability density function of Poisson’s distribution is p ( x ) = e − λ ⋅ λ x /x
Where λ = mean = np
n = number of total outcomes
p = probability of success
Step-by-step explanation:
calculation:
In our given question probability for finding fuse bulb is 2% = p = 0.02
Since p is very small Therefore, we use
Poisson distribution. ⇒ p ( x ) = e − λ ⋅ λ x x ! ⇒p(x)=e−λ⋅λxx! ---(1)
Where λ = np n = 200 (given) p = 0.02 (calculated above) ⇒ λ = 4 ---(2)
For probability that almost 5 bubs may defectives p(x ≤ 5) = p(1) + p(2) + p(3) + p(4) + p(5) + p(0)
So from equation (1) & (2) we can,
p ( x ≤ 5 ) = 4 e − 4 + 16 ⋅ e − 4 2 + 64 6 ⋅ e − 4 + ( 4 ) 4 24 e − 4 + ( 4 ) 5 120 e − 4 + e − 4
p(x≤5)=4e−4+16⋅e−42+646⋅e−4+(4)424e−4+(4)5120e−4+e−4
solve and get
p(x ≤ 5) = 0.7851
we get ≈ 0.79
HENCE THIS IS THE SOLUTION
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Answer:
probability of no fuse and 3 or more defective fuses are = 0.0183 and = 0.726
Step-by-step explanation:
probability of the defected fuse = 2% = 0.02P
the mean number of defective fuse (M) = n × P
mean = 200×0.02 = 4
we know the possion distribution is,
P(x) = e^-m × m^x/ x!
1) probabilty of no defective :
P(x=0) = e^-4 × m^0 / 0!
P(x=0) = e^-4 = 0.0183
2) probability of 3 or more defective
P(x>and equal to3) = 1 - P(x>3)
= 1 - e^-4 (1 + 4 + 8)
= 0.726
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