Complete Question
The complete question is shown on the first uploaded image
Answer:
a
The probability is [tex]P_T= 0.4560[/tex]
b
The probability is [tex]P_F= 0.0013[/tex]
Step-by-step explanation:
From the question we are told that
The mean for the exponential density function of bulbs failure is [tex]\mu = 1600 \ hours[/tex]
Generally the cumulative distribution for exponential distribution is mathematically represented as
[tex]1 - e^{- \lambda x}[/tex]
The objective is to obtain the p=probability of the bulbs failure within 1800 hours
So for the first bulb the probability will be
[tex]P_1(x < 1800)[/tex]
And for the second bulb the probability will be
[tex]P_2 (x< 1800)[/tex]
So from our probability that we are to determine the area to the left of 1800 on the distribution curve
Now the rate parameter [tex]\lambda[/tex] is mathematically represented as
[tex]\lambda[/tex] [tex]= \frac{1}{\mu}[/tex]
[tex]\lambda = \frac{1}{1600}[/tex]
The probability of the first bulb failing with 1800 hours is mathematically evaluated as
[tex]P_1(x < 1800) = 1 - e^{\frac{1}{1600} * 1800 }[/tex]
[tex]= 0.6753[/tex]
Now the probability of both bulbs failing would be
[tex]P_T=P_1(x < 1800) * P_2(x < 1800)[/tex]
[tex]= 0.6375 * 06375[/tex]
[tex]P_T= 0.4560[/tex]
Let assume that one bulb failed at time [tex]T_a[/tex] and the second bulb failed at time [tex]T_b[/tex] then
[tex]T_a + T_b = 1800\ hours[/tex]
The mathematical expression to obtain the probability that the first bulb failed within between zero and [tex]T_a[/tex] and the second bulb failed between [tex]T_a \ and \ 1800[/tex] is represented as
[tex]P_F=\int_{0}^{1800}\int_{0}^{1800-x} \f{\lambda }^{2}e^{-\lambda x}* e^{-\lambda y}dx dy[/tex]
[tex]=\int_{0}^{1800} {\lambda }e^{-\lambda x}\int_{0}^{1800-x} {\lambda } e^{-\lambda y}dx dy[/tex]
[tex]=\int_{0}^{1800} {\frac{1}{1600} }e^{-\lambda x}\int_{0}^{1800-x} \frac{1}{1600 } e^{-\lambda y}dx dy[/tex]
[tex]=\int_{0}^{1800} {\frac{1}{1600} }e^{-\lambda x}[e^{- \lambda y}]\left {1800-x} \atop {0}} \right. dx[/tex]
[tex]=\int_{0}^{1800} {\frac{1}{1600} }e^{-\frac{x}{1600} }[e^{- \frac{1800 -x}{1600} }-1] dx[/tex]
[tex]=[ {\frac{1}{1600} }e^{-\frac{1800}{1600} }-\frac{1}{1600}[e^{- \frac{x}{1600} }] \left {1800} \atop {0}} \right.[/tex]
[tex]=[ {\frac{1}{1600} }e^{-\frac{1800}{1600} }-\frac{1}{1600}[e^{- \frac{1800}{1600} }] -[[ {\frac{1}{1600} }e^{-\frac{1800}{1600} }-\frac{1}{1600}[e^{-0}][/tex]
[tex]=[\frac{1}{1600} e^{-\frac{1800}{1600} } - \frac{1}{1600} e^{-0} ][/tex]
[tex]=0.001925 -0.000625[/tex]
[tex]P_F= 0.0013[/tex]
