Question

Is it Exponential distribution? or Geometric distribution?

The life times, Y in years of low-grade lightbulbs follow an exponential distribution with a mean of 0.7 years. A tester makes random observations of the life times of this lightbulbs and records them one by one as either a success if the life time exceeds 1 year, or as a failure otherwise.

Part a)Find the probability that the first success occurs in the fifth observation.

Part b) Find the probability of the second success occurring in the 8th observation given that the first success occurred in the 3rd observation.

Part c) Find the probability to 2 decimal places that the first success occurs in an odd-numbered observation. That is, the first success occurs in the 1st or 3rd or 5th or 7th (and so on) observation.

Answer 1

The life times of low-grade lightbulbs follow an exponential distribution. To find the probability that the first success occurs in the fifth observation, we can use the cumulative distribution function (CDF) of the exponential distribution. The probability is approximately 0.089.

Step-by-step explanation:

The life times of low-grade lightbulbs follow an exponential distribution. In this case, the life times follow an exponential distribution with a mean of 0.7 years, which is given in the question. Since the life times are exponential, we can use the exponential distribution to solve the given problems.

Part a) To find the probability that the first success occurs in the fifth observation, we can use the cumulative distribution function (CDF) of the exponential distribution. The CDF gives us the probability that the variable is less than or equal to a certain value.

The CDF of the exponential distribution is given by:

P(Y <= y) = 1 - e^(-λy)

where Y is the random variable, y is the observed value, and λ is the rate parameter of the exponential distribution.

In this case, the random variable Y represents the life times of the lightbulbs, and the observed value y represents the number of observations.

Since we want the first success to occur in the fifth observation, the probability of success in the first four observations is:

P(Y <= 1)^4 = (1 - e^(-0.7))^4 ≈ 0.180

The probability that the first success occurs in the fifth observation is the probability of success in the first four observations multiplied by the probability of success in the fifth observation:

P(first success occurs in the fifth observation) ≈ 0.180 * (1 - e^(-0.7)) ≈ 0.180 * 0.497 ≈ 0.089

Therefore, the probability that the first success occurs in the fifth observation is approximately 0.089.

Answer 2

I believe you use the exponential distribution to find the probability of success (say "p"). Then, you use the Geometric(p) distribution using the "p" you calculated from the exponential as your success probability. Also, in b), you use a negative binomial (which is really just a generalized geometric distribution).

Part c) Find the probability to 2 decimal places that the first success occurs in an odd-numbered observation. That is, the first success occurs in the 1st or 3rd or 5th or 7th (and so on) observation.