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Every Friday night I order a delivery takeaway? Everything usually goes perfectly, but sometimes things go wrong :(. The food is cold with probability 0.1. Some part of the order is forgotten with probability 0.2. The delivery driver gets lost and calls for directions with probability 0.3. Let Y denote the number of Fridays needed until everything goes per- fectly with my order. Assuming that all the events are independent:

(i) find the probability that everything goes perfectly on the first Friday night, i.e. that Y = 1.
(ii) determine the distribution that Y follows.
(iii) calculate the probability that Y is a multiple of three.

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Final answer:

The probability that everything goes perfectly on the first Friday is 0.504. The number of Fridays needed until a perfect order is a geometric distribution with success probability 0.504. The probability that Y is a multiple of three can be found using a series sum formula for geometric sequences.

Step-by-step explanation:

To solve this mathematical problem, we can use the concepts of probability and geometric distributions. I'll calculate each part of the question step by step.



Part (i): Probability that everything goes perfectly on the first Friday night (Y = 1)

Since each event is independent, the probability that nothing goes wrong (i.e., the food is not cold, the order is complete, and the delivery driver doesn’t get lost) is given by (1 - probability of each event). Therefore, the probability of a perfect service is:

(1 - 0.1) * (1 - 0.2) * (1 - 0.3) = 0.9 * 0.8 * 0.7 = 0.504.



Part (ii): Distribution that Y follows

The number of Fridays needed until everything goes perfectly with an order follows a geometric distribution, with the success probability p calculated in part (i).



Part (iii): Probability that Y is a multiple of three

To find this probability, we use the formula for the geometric distribution:

P(Y is a multiple of 3) = p * (1-p)^(3k-1) for k = 1, 2, 3, ...

This is an infinite series and can be summed up using the formula for the sum of a geometric series:

∑ (1-p)^(3k-1) from k=1 to infinity = (1-p)^2 / (1 - (1-p)^3)

Plugging in the value of p = 0.504, we get:

P(Y is a multiple of 3) = 0.504 * 0.496^2 / (1 - 0.496^3)

Calculating this gives us the required probability.

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