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A lamp has two bulbs, each of a type with an average lifetime of 9 hours. The probability density function for the lifetime of a bulb is f(t)=19ᵉ−ᵗ/9,t≥0. What is the probability that both of the bulbs will fail within 5 hours?

User Mina Samir
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The probability that both of the bulbs will fail within 5 hours is 0.287.

How to find probability?

To find the probability that both bulbs will fail within 5 hours, find the joint probability density function and then integrate it over the given range.

The probability density function (pdf) for the lifetime of a single bulb is given by:


\[ f(t) = (1)/(9) e^(-t/9), \quad t \geq 0 \]

To find the joint probability density function for both bulbs failing within a certain time, multiply the individual pdfs. Let T₁ be the lifetime of the first bulb, and T₂ be the lifetime of the second bulb. The joint pdf is given by:


\[ f_(T_1, T_2)(t_1, t_2) = f(t_1) \cdot f(t_2) \]

Now, substitute the given pdf into this expression:


\[ f_(T_1, T_2)(t_1, t_2) = (1)/(81) e^(-t_1/9) e^(-t_2/9) \]

The integral for the joint probability is:


\[ P(T_1 \leq 5, T_2 \leq 5) = \int_(0)^(5) \int_(0)^(5) (1)/(81) e^(-t_1/9) e^(-t_2/9) \, dt_1 \, dt_2 \]

Calculate it:


\[ P(T_1 \leq 5, T_2 \leq 5) = \int_(0)^(5) \int_(0)^(5) (1)/(81) e^(-t_1/9) e^(-t_2/9) \, dt_1 \, dt_2 \]


\[ = (1)/(81) \int_(0)^(5) \int_(0)^(5) e^(-t_1/9) e^(-t_2/9) \, dt_1 \, dt_2 \]


\[ = (1)/(81) \int_(0)^(5) \left( \int_(0)^(5) e^(-t_1/9) e^(-t_2/9) \, dt_2 \right) dt_1 \]


\[ = (1)/(81) \int_(0)^(5) \left( \int_(0)^(5) e^(-(t_1 + t_2)/9) \, dt_2 \right) dt_1 \]


\[ = (1)/(81) \int_(0)^(5) \left( -9e^(-(t_1 + t_2)/9) \Big|_0^5 \right) dt_1 \]


\[ = (1)/(81) \int_(0)^(5) \left( -9e^(-5/9) + 9 \right) dt_1 \]


\[ = (1)/(81) \left( -9e^(-5/9) + 9 \right) \int_(0)^(5) dt_1 \]


\[ = (1)/(81) \left( -9e^(-5/9) + 9 \right) t_1 \Big|_0^5 \]


\[ = (1)/(81) \left( -9e^(-5/9) + 9 \right) \cdot 5 \]


\[ = (1)/(81) \left( -45e^(-5/9) + 45 \right) \]

The calculated value for
\( P(T_1 \leq 5, T_2 \leq 5) \) is approximately 0.287.

User Csf
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