Question

Interest centers around the life of an electronic component. Suppose it is known that the probability that the component survives for more than 6000 hours is 0.42. Suppose also that the probability that the component survives no longer than 4000 hours is 0.04.(a) What is the probability thatthe life of the component is less than or equal to 6000 hours? (b)What is the probability that the life is greater than 4000 hours?

Answer 1

Answer:

a) The probability that the life of the component is less than or equal to 6000 hours is 58%

b) The probability that the life is greater than 4000 hours is 96%

Step-by-step explanation:

Let's define the variable time as $t$, which is measured in hours. Then, according to the problem:

• The probability that the component survives for more than 6000 hours is 0.42 $\rightarrow P(t> 6000) = 0.42$
• The probability that the component survives no longer than 4000 hours is 0.04 $\rightarrow P(t\leq 4000) =0.04$

a) The probability that the life of the component is less than or equal to 6000 needs to be equal to $1-P(t>6000)$.

To see this, think of a full set, whose elements can be separate in two categories, the ones that survive more than 6000 hours (let's call them A and number of elements of this type Na) and the ones that survive less than or equal to 6000 (let's call them B and number of elements of this type Nb). Thus, the total number of elements can be calculate as $Na+Nb$, which represents the unit.

If you want to calculate the proportion of elements of type A, you need to divide the number of element of type A over the total number of elements, which is $P(A) = Na/(Na+Nb)$. And the same analysis applies to the proportion of elements of type B:  $P(B) = Nb/(Na+Nb)$. Hence, it is easy to see:

$P(A)+P(B) = \frac{Na}{Na+Nb} + \frac{Nb}{Na+Nb} = 1$

of course this only applies when the elements can be only categorized as one type or the other, which is call in probability, mutually exclusive events.

Using this for:

a) $P(t\leq 6000 )=1-P(t>6000) = 1 - 0.42 =0.58\\P(t\leq 6000 )=58\%$

b) $P(t> 4000 )=1-P(t\leq 4000) = 1 - 0.04 =0.96\\P(t> 4000 )=96\%$