SOLUTION: Please help me to solve this
a factory supervisor finds that 20 percent of the bolts produced by a machine will be defective . If 5 bolts are chosen at random,find the probabili
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a factory supervisor finds that 20 percent of the bolts produced by a machine will be defective . If 5 bolts are chosen at random,find the probabili
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Question 1086784: Please help me to solve this
a factory supervisor finds that 20 percent of the bolts produced by a machine will be defective . If 5 bolts are chosen at random,find the probability that at most 2 bolts will be defective.
thank you :) Answer by jim_thompson5910(35256) (Show Source):
You can put this solution on YOUR website! Binomial Distribution:
Probability of defect = p = 20% = 20/100 = 0.20 (use the decimal form)
sample size = n = 5
We need to compute the probabilities for k = 0, k = 1, k = 2
"At most 2" means "2 is the ceiling" or "2 is the highest we can go"
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For k = 0, we would have this combination
n C k = (n!)/(k!*(n-k)!)
5 C 0 = (5!)/(0!*(5-0)!)
5 C 0 = (5!)/(0!*5!)
5 C 0 = (1)/(0!*1)
5 C 0 = (1)/(1)
5 C 0 = 1
Leading to this binomial probability
P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 0) = (5 C 0)*(0.2)^(0)*(1-0.2)^(5-0)
P(X = 0) = (5 C 0)*(0.2)^(0)*(0.8)^(5)
P(X = 0) = (1)*(0.2)^(0)*(0.8)^5
P(X = 0) = (1)*(1)*(0.32768)
P(X = 0) = 0.32768
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For k = 1, we would have this combination
n C k = (n!)/(k!*(n-k)!)
5 C 1 = (5!)/(1!*(5-1)!)
5 C 1 = (5!)/(1!*4!)
5 C 1 = (5*4!)/(1!*4!)
5 C 1 = (5)/(1!)
5 C 1 = (5)/(1)
5 C 1 = 5/1
5 C 1 = 5
Leading to this binomial probability
P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 1) = (5 C 1)*(0.2)^(1)*(1-0.2)^(5-1)
P(X = 1) = (5 C 1)*(0.2)^(1)*(0.8)^(4)
P(X = 1) = (5)*(0.2)^(1)*(0.8)^4
P(X = 1) = (5)*(0.2)*(0.4096)
P(X = 1) = 0.4096
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For k = 2, we would have this combination
n C k = (n!)/(k!*(n-k)!)
5 C 2 = (5!)/(2!*(5-2)!)
5 C 2 = (5!)/(2!*3!)
5 C 2 = (5*4*3!)/(2!*3!)
5 C 2 = (5*4)/(2!)
5 C 2 = (5*4)/(2*1)
5 C 2 = 20/2
5 C 2 = 10
Leading to this binomial probability
P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 2) = (5 C 2)*(0.2)^(2)*(1-0.2)^(5-2)
P(X = 2) = (5 C 2)*(0.2)^(2)*(0.8)^(3)
P(X = 2) = (10)*(0.2)^(2)*(0.8)^3
P(X = 2) = (10)*(0.04)*(0.512)
P(X = 2) = 0.2048
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If we keep going, we would get this full Binomial distribution
X = k
P(X = k)
0
0.32768
1
0.4096
2
0.2048
3
0.0512
4
0.0064
5
0.00032
Now add up the probabilities for k = 0, k = 1, k = 2.
Answer: 0.94208
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