Answer:
a) the probability that the defective board was produced during the first hour of operation is [tex]\frac{1}{10}[/tex] or 0.1000
b) the probability that the defective board was produced during the last hour of operation is [tex]\frac{1}{10}[/tex] or 0.1000
c) the required probability is 0.2000
Step-by-step explanation:
Given the data in the question;
During a specific ten-hour period, one defective circuit board was found.
Lets X represent the number of defective circuit boards coming out of the machine , following Poisson distribution on a particular 10-hours workday which one defective board was found.
Also let Y represent the event of producing one defective circuit board, Y is uniformly distributed over ( 0, 10 ) intervals.
f(y) = [tex]\left \{ {{\frac{1}{b-a} }\\\ }} \right _0[/tex]; ( a ≤ y ≤ b )[tex]_{elsewhere[/tex]
= [tex]\left \{ {{\frac{1}{10-0} }\\\ }} \right _0[/tex]; ( 0 ≤ y ≤ 10 )[tex]_{elsewhere[/tex]
f(y) = [tex]\left \{ {{\frac{1}{10} }\\\ }} \right _0[/tex]; ( 0 ≤ y ≤ 10 )[tex]_{elsewhere[/tex]
Now,
a) the probability that it was produced during the first hour of operation during that period;
P( Y < 1 ) = [tex]\int\limits^1_0 {f(y)} \, dy[/tex]
we substitute
= [tex]\int\limits^1_0 {\frac{1}{10} } \, dy[/tex]
= [tex]\frac{1}{10} [y]^1_0[/tex]
= [tex]\frac{1}{10} [ 1 - 0 ][/tex]
= [tex]\frac{1}{10}[/tex] or 0.1000
Therefore, the probability that the defective board was produced during the first hour of operation is [tex]\frac{1}{10}[/tex] or 0.1000
b) The probability that it was produced during the last hour of operation during that period.
P( Y > 9 ) = [tex]\int\limits^{10}_9 {f(y)} \, dy[/tex]
we substitute
= [tex]\int\limits^{10}_9 {\frac{1}{10} } \, dy[/tex]
= [tex]\frac{1}{10} [y]^{10}_9[/tex]
= [tex]\frac{1}{10} [ 10 - 9 ][/tex]
= [tex]\frac{1}{10}[/tex] or 0.1000
Therefore, the probability that the defective board was produced during the last hour of operation is [tex]\frac{1}{10}[/tex] or 0.1000
c)
no defective circuit boards were produced during the first five hours of operation.
probability that the defective board was manufactured during the sixth hour will be;
P( 5 < Y < 6 | Y > 5 ) = P[ ( 5 < Y < 6 ) ∩ ( Y > 5 ) ] / P( Y > 5 )
= P( 5 < Y < 6 ) / P( Y > 5 )
we substitute
[tex]= (\int\limits^{6}_5 {\frac{1}{10} } \, dy) / (\int\limits^{10}_5 {\frac{1}{10} } \, dy)[/tex]
[tex]= (\frac{1}{10} [y]^{6}_5) / (\frac{1}{10} [y]^{10}_5)[/tex]
= ( 6-5 ) / ( 10 - 5 )
= 0.2000
Therefore, the required probability is 0.2000
