STATS ASSIGNMENT

Stats Assignment



Stats Assignment

Following are the questions that are answered accordingly to the mentioned description these are the last four questions:

Question 4: A coin is tossed until for the first time the same result appears twice in succession.

(a) Describe the sample space of this experiment.

Solution: The sample space of this experiment cannot be created due to the fact that the outcomes are not certain. The reason behind the uncertainty of the creation of sample space lies in the concept that the desired outcome may be achieved in only 2 tosses, or may not be achieved in a 1000 tosses if the 'Head' and 'Tail' came alternatively. This is why a proper sample space for this experiment cannot be created.

(b) What is the probability of a single possible outcome requiring n tosses, where n > 2? Find the probability of the event that the experiment is finished before the sixth toss.

Solution: The achievement of the desired outcome, which is consecutive heads of consecutive tails, can either be attained in exactly 2 tosses, or more than 2 tosses. In order to determine the probability of achieving the desired outcome in more than 2 tosses we can subtract the probability of achieving the outcome in exactly 2 tosses from 1. As the total probability is always equal to 1, the probability of achieving the desired outcome in more than 2 tosses will be 1 - P (exactly 2 tosses). This is calculated and explained below:

Probability of getting one Head = 0.5

Probability of getting one Tail = 0.5

Probability of getting second Head = 0.5

Probability of getting second Tail = 0.5

However, the probability of getting a second Head after getting a Head already would mean getting 'Head and Head', which is calculated by multiplying the probabilities.

The probability of getting both heads = 0.25

The probability of getting both tails is also = 0.25

Thus getting both heads or both tails = 0.25 + 0.25 = 0.5 (because in case of 'or' the probabilities are added)

Thus, the probability of getting the outcome in more than two tosses would be 1 - P (exactly 2 tosses).

So, it is equal to 1 - 0.5 = 0.5

(c) Find the probability that an even number of tosses is required.

Solution: The probability of achieving the outcome in even tosses is exactly the same as getting the desired outcome in odd tosses, thus, the probability will be 0.5.

Question 5: in a town of ...