An **electronic chess game has a useful life** that is exponential with a mean of 30 months. Determine each of the following:

a. The probability that any given unit will operate for at least (1) 39 months, (2) 48 months, (3) 60 months.

b. The probability that any given unit will fail sooner than (1) 33 months, (2) 15 months, (3) 6 months.

c. The length of service time after which the percentage of failed units will approximately equal (1) 50 percent, (2) 85 percent, (3) 95 percent, (4) 99 percent.

#### Solution

Let $X$ be the life of an electronic chess game. $X\sim \exp(30)$.

The pdf of $X$ is

` $$ \begin{aligned} f(x) & = \frac{1}{30}e^{-x/30}, x > 0 \end{aligned} $$ `

The distribution function of $X$ is

` $$ \begin{aligned} F(x) & = 1-e^{-x/30} \end{aligned} $$ `

a. The probability that any given unit will operate for

(1) at least 39 months is

` $$ \begin{aligned} P(X > 39) &= 1- P(X\leq 39)\\ & = 1- F(39)\\ &= 1 - (1 - e^{-39/30})\\ & = e^{-39/30} \\ &= 0.2725 \end{aligned} $$ `

(2) The probability that any given unit will operate for at least 48 months is

` $$ \begin{aligned} P(X > 48) &= 1- P(X\leq 48)\\ & = 1- F(48)\\ &= 1 - (1 - e^{-48/30})\\ & = e^{-48/30} \\ &= 0.2019 \end{aligned} $$ `

(3) The probability that any given unit will operate for at least 60 months is

` $$ \begin{aligned} P(X > 60) &= 1- P(X\leq 60)\\ & = 1- F(60)\\ &= 1 - (1 - e^{-60/30})\\ & = e^{-60/30} \\ &= 0.1353 \end{aligned} $$ `

b. The probability that any given unit will fail

(1) sooner than 33 months is

` $$ \begin{aligned} P(X < 33) &= F(33)\\ &= e^{-33/30}\\ &= 0.3329 \end{aligned} $$ `

(2) The probability that any given unit will fail sooner than 15 months is

` $$ \begin{aligned} P(X < 15) &= F(15)\\ &= e^{-15/30}\\ &= 0.6065 \end{aligned} $$ `

(3) The probability that any given unit will fail sooner than 6 months is

` $$ \begin{aligned} P(X < 6) &= F(6)\\ &= e^{-6/30}\\ &= 0.8187 \end{aligned} $$ `

(c) We have

` $$ \begin{aligned} F(x) &= 1- e^{-x/30}\\ e^{-x/30}&= 1-F(x)\\ -x/30 &= \ln(1-F(x))\\ x&= -30\times \ln(1-F(x)) \end{aligned} $$ `

(1) The length of service time after which the percentage of failed units will approximately equal 50 percent is

$x=-30\times \ln(1-F(x))= -30\times \ln(1-0.5) = 20.7944$

(2) The length of service time after which the percentage of failed units will approximately equal 85 percent is

$x=-30\times \ln(1-F(x))= -30\times \ln(1-0.85) = 56.9136$

(3) The length of service time after which the percentage of failed units will approximately equal 95 percent is

$x=-30\times \ln(1-F(x))= -30\times \ln(1-0.95) = 89.872$

(4) The length of service time after which the percentage of failed units will approximately equal 99 percent is

$x=-30\times \ln(1-F(x))= -30\times \ln(1-0.99) = 138.1551$

## Conclusion

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