Question

Question 6 One type of Aero-plane engine has a 0.0001 probability of failure while in flying. For a jet that has

four of these engines,
a) Find the probability distribution for number of engines failure?
b) What is the expected number of engines failure?
Ans: n ==
Let x be the random variable denoting # of successes (# of engine failure).
x = 0, 1, 2, 3, 4
p=

Answer 1

a) Let X be the random variable representing the number of engines that fail. Since the probability of failure for each engine is independent of each other, X follows a binomial distribution with n = 4 and p = 0.0001. Therefore, the probability distribution for the number of engine failures is:

P(X = k) = (4 choose k) * (0.0001)^k * (0.9999)^(4-k), for k = 0, 1, 2, 3, 4.

b) The expected number of engine failures is given by:

E(X) = np = 4 * 0.0001 = 0.0004.

Answer 2

To find the probability distribution for the number of engine failures for a jet with four engines, we can use the binomial distribution.

Given:
Probability of engine failure (p) = 0.0001

a) Probability distribution for the number of engine failures:
The probability distribution can be calculated using the formula:

P(x) = C(n, x) * p^x * (1 - p)^(n - x)

Where:
n = Total number of engines (4)
x = Number of engine failures
p = Probability of engine failure (0.0001)

Substituting the values:

P(0) = C(4, 0) * (0.0001)^0 * (1 - 0.0001)^(4 - 0)
P(1) = C(4, 1) * (0.0001)^1 * (1 - 0.0001)^(4 - 1)
P(2) = C(4, 2) * (0.0001)^2 * (1 - 0.0001)^(4 - 2)
P(3) = C(4, 3) * (0.0001)^3 * (1 - 0.0001)^(4 - 3)
P(4) = C(4, 4) * (0.0001)^4 * (1 - 0.0001)^(4 - 4)

The values for P(0), P(1), P(2), P(3), P(4) can be calculated using combinations (C) and the given probabilities.