Answer: 0.3042
Explanation:
Let A and B are the events to that job done by Printer I and Printer II respectively.
Given : P(A)=0.40 P(B)=0.60
Printing time of Printer I is Exponential with the mean of 2 minutes.
i.e. average number of job done in one minute:[tex]\lambda=\dfrac{1}{2}[/tex]
The cumulative distribution function (CDF) for exponential distribution:-
[tex]F(x)=1-e^{-\lambda x}[/tex], where [tex]\lambda[/tex] is the mean.
Then, the cumulative distribution function (CDF) for Printer I:-
[tex]P(X|A)=1-e^{-\dfrac{1}{2} x}[/tex]
i.e. [tex]P(X<1|A)=1-e^{-\dfrac{1}{2}}[/tex]
Printing time of Printer II is Uniform between 0 minutes and 5 minutes.
The cumulative distribution function (CDF) for uniform distribution in interval (a,b) :-
[tex]F(x)=\dfrac{x-a}{b-a}[/tex]
Then, [tex]P(X|B)=\dfrac{x-0}{5-0}=\dfrac{x}{5}[/tex]
i.e. [tex]P(X<1|B)=\dfrac{1}{5}[/tex]
Now, the required probability :-
[tex]\text{P(A}|X<1)=\dfrac{P(A) P(X<1|A)}{P(A) P(X<1|A)+P(B) P(X<1|B)}\\\\=\dfrac{(0.4)(1-e^{-\frac{1}{2}})}{(0.4)(1-e^{-\frac{1}{2}})+(0.6)(\dfrac{1}{5})}}\\\\=\dfrac{(0.4)(0.3935)}{(0.4)(0.3935)+(0.6)(0.6)}\\\\=0.304213374565\approx0.3042[/tex]
Hence, the required probability = 0.3042
