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Hey there, Lets solve this problem together.
The First step is to line up the numbers.
<span>We calculate </span>
<span>the result of which is </span>
<span>
</span>
<span>We calculate </span>
<span> the result of which is </span>
<span>.
</span>
Since we get a negative number in the next column, we must take 1 from the next column and carry it over to this column. Now the number will be changed to 10.
We calculate
, and the result is 
.
<span>We calculate </span>
<span> the result of which is</span>
<span>.
</span>
Therefore,
The First step is to line up the numbers.
<span>We calculate </span>
</span>
<span>We calculate </span>
</span>
Since we get a negative number in the next column, we must take 1 from the next column and carry it over to this column. Now the number will be changed to 10.
We calculate
<span>We calculate </span>
</span>
Therefore,
Answer:
See steps below
Step-by-step explanation:
Let X be the random variable that measures the lifespan of a bulb.
If the random variable X is exponentially distributed and X has an average value of 24 month, then its probability density function is
and its cumulative distribution function (CDF) is
• What is probability that the red bulb will need to be replaced at the first inspection?
The probability that the bulb fails the first year is
• If the bulb is in good condition at the end of 18 months, what is the probability that the bulb will be in good condition at the end of 24 months?
Let A and B be the events,
A = “The bulb will last at least 24 months”
B = “The bulb will last at least 18 months”
We want to find P(A | B).
By definition P(A | B) = P(A∩B)P(B)
but B⊂A, so A∩B = B and
We have
hence,
• If the signal has six red bulbs, what is the probability that at least one of them needs replacement at the first inspection? Assume distribution of lifetime of each bulb is independent
If the distribution of lifetime of each bulb is independent, then we have here a binomial distribution of six trials with probability of “success” (one bulb needs replacement at the first inspection) p = 0.39347
Now the probability that exactly k bulbs need replacement is
<em>Probability that at least one of them needs replacement at the first inspection = 1- probability that none of them needs replacement at the first inspection. </em>
This means that,
<em>Probability that at least one of them needs replacement at the first inspection = </em>