Manhole explosions (usually caused by gas leaks and sparks) are on the rise in your city. On any given day, the manhole cover near your house explodes with some unknown probability, which is the same across all days. We model this unknown probability of explosion as a random variable Q, which is uniformly distributed between 0 and 0.1. Let Xi be a Bernoulli random variable that indicates whether the manhole cover near your house explodes on day i (where today is day 1).
Give numerical answers for parts (1) and (2).
E[Xi]= ?
var(Xi)= ?
Let A be the event that the manhole cover did not explode yesterday (i.e., X0=0). Find the conditional PDF of Q given A. Express your answer in terms of q using standard notation.
For 0≤q≤0.1, fQ∣A(q)= ?
E[Xi]= 0.05
var(Xi)= 0.0475
please tell us the answer
please tell us the answer!!!!!!!!!!!!!!
answer of the third part... please!!
Answer to the third is:
(1-q)/0.095
enjoy the 6.041 from MIT, guys ^_^
How did you get the Var(Xi)?
To find the answers, we can utilize the properties of Bernoulli random variables and conditional probability. Let's go through each part separately.
(1) Expected Value (E[Xi]):
The expected value of a Bernoulli random variable is the probability of a success, which is denoted by p. In this case, p is the unknown probability of the manhole cover exploding on any given day. However, since Q is uniformly distributed between 0 and 0.1, we can consider p as the average of this range.
Therefore, E[Xi] = p = (0 + 0.1) / 2 = 0.05.
Hence, the expected value of Xi is 0.05.
(2) Variance (var(Xi)):
The variance of a Bernoulli random variable is calculated by multiplying the probability of success, p, by the probability of failure, 1-p.
In this case, since p = 0.05, the probability of failure is 1 - p = 1 - 0.05 = 0.95.
var(Xi) = p * (1 - p) = 0.05 * 0.95 = 0.0475.
Therefore, the variance of Xi is 0.0475.
(3) Conditional Probability (fQ|A(q)):
To find the conditional PDF of Q given the event A, we need to determine the probability distribution of Q when we know that X0 = 0 (the manhole cover did not explode yesterday).
Given X0 = 0, the probability that Q falls in the interval [q, q + dq] is the same as the probability that Q is smaller than q + dq, but greater than or equal to 0.
Since Q is uniformly distributed between 0 and 0.1, the probability that Q is smaller than q + dq is (q + dq) / 0.1.
On the other hand, the probability that Q is smaller than 0 (which should be impossible since Q is always positive) is 0. Hence, the probability distribution of Q given A is 0 for q < 0.
Therefore, the conditional PDF of Q given A, fQ|A(q), is given as:
fQ|A(q) = [(q + dq) / 0.1] * 1(q ≥ 0),
where 1(q ≥ 0) is the indicator function that is 1 when q ≥ 0, and 0 otherwise.
Note: It is important to remember that this is a simplified model assuming uniform distribution for Q and independence between the manhole explosions on different days. In reality, other factors may influence the probability of manhole cover explosions.