Answer:
the probability that system’s failure is due to the radio = [tex]\dfrac{1}{3}[/tex]
Step-by-step explanation:
From the question given;
Let the mean lifetime of the radio [tex]\dfrac{1}{\lambda_1} = 1000[/tex] and the mean lifetime of the speaker [tex]\dfrac{1}{\lambda_2} = 500[/tex]
we can re-write both expressions as:
[tex]\lambda_1= \dfrac{1}{1000}[/tex] and [tex]\lambda_2= \dfrac{1}{500}[/tex]
Let consider [tex]X_1 \ and \ X_2[/tex] to be the variables which are independent to the exponentially distributed mean of [tex]\dfrac{1}{\lambda _1} \ and \ \dfrac{1}{\lambda _2}[/tex]
∴
[tex]P(X_1< X_2) = \int ^{\infty}_{0} \ P (X_1<X_2|X_1 =x) \lambda_1 e^{-\lambda_1 \ x} \ dx[/tex]
[tex]P(X_1< X_2) = \int ^{\infty}_{0} \ P (X_1<X_2) \lambda_1 e^{-\lambda_1 \ x} \ dx[/tex]
[tex]P(X_1< X_2) = \int ^{\infty}_{0} \ e^{-\lambda_2 \ x} \lambda _1 \ e^{-\lambda_1 \ x} \ dx[/tex]
[tex]P(X_1< X_2) = \int ^{\infty}_{0} \lambda _1 \ e^{(-\lambda_1 +\lambda_2)} \ dx[/tex]
[tex]P(X_1< X_2) = \dfrac{\lambda_1}{\lambda_1 + \lambda_2}[/tex]
replace the values now; we have:
[tex]P(X_1< X_2) = \dfrac{\dfrac{1}{1000} }{\dfrac{1}{1000} + \dfrac{1}{500}}[/tex]
[tex]P(X_1< X_2) = \dfrac{\dfrac{1}{1000} }{\dfrac{1+2}{1000} }[/tex]
[tex]P(X_1< X_2) = \dfrac{\dfrac{1}{1000} }{\dfrac{3}{1000} }[/tex]
[tex]P(X_1< X_2) = {\dfrac{1}{1000} } \times {\dfrac{1000}{3}[/tex]
[tex]P(X_1< X_2) = {\dfrac{1}{3} }[/tex]
Thus, the probability that system’s failure is due to the radio = [tex]\dfrac{1}{3}[/tex]
