4 a) A coin is tossed twice. Calculate the probability of each of the following

Question

4      a) A coin is tossed twice. Calculate the probability of each of the following occurring, commenting briefly on the basis of your calculation:

A head on the first toss;

A tail on the second toss, given that the first toss was a head;

Two tails;

A tail on the first toss and a head on the second;

A tail on the first and a head on the second, or a head on the first and a tail on the second;

At least one head on the two tosses.

                                                                                                                             [20%]

Do you think the Poisson distribution, which assumes independent arrivals, is a good estimation of arrival rates in the following queuing systems? Explain your reasoning in each case:

University cafeteria or coffee bar;

Hairdresser's shop;

Hardware store;

Dentist's surgery;

University lecture;

Movie cinema.

                                                                                                                             [50%]

Why is a different queuing model needed if the population of potential customers for a system is limited rather than unlimited? Use examples of real or hypothetical systems to illustrate your answer.                                                                             [30%]

Explanation / Answer

Answer:-

A silver dollar is flipped twice.

Sample space, S = {HH, HT, TH, TT}

Probability = Favorable number of outcomes

                        Total number of possible outcomes

Favorable number of outcomes = 2 {HH, HT}

Total number of possible outcomes = 4 {HH, HT, TH, TT}

            Probability of a head on first flip = 2/4 = 0.5

Let event A be occurring of tail on the second flip

Let event B be that first toss was a head

P (A|B) =         P (AB)

                        P (B)

Consider, P (AB) = P (first toss was a head and second toss was a tail)

Here, Favorable number of outcomes = 1 {HT}

Total number of possible outcomes = 4 {HH, HT, TH, TT}

Thus, P (AB) = ¼

Also consider, P (B) = P (first toss was a head)

Here, Favorable number of outcomes = 2 {HT, HH}

Total number of possible outcomes = 4 {HH, HT, TH, TT}

P (B) = 2/4

Thus, P (A|B) =           (1/4) = 1 = 0.5

                                    (2/4)    2

Probability of a tail on the second flip given that the first toss was a head = 0.5

Favorable number of outcomes = 1 {TT}

Total number of possible outcomes = 4 {HH, HT, TH, TT}

Probability of two tails = 1/4 = 0.25

Favorable number of outcomes = 1 {TH}

Total number of possible outcomes = 4 {HH, HT, TH, TT}

Probability of a tail on the first and a head on the second = 1/4 = 0.25

Let event A denote occurrence of a tail on the first and a head on the second
let event B denote occurrence of or a head on the first and a tail on the second

P (A) or P (B) = P (A) + P (B)

Consider P (A) = P (tail on the first and a head on the second)

Here, Favorable number of outcomes = 1 {TH}

Total number of possible outcomes = 4 {HH, HT, TH, TT}

Thus, P (A) = ¼

Consider P (B) = P (head on the first and a tail on the second)

Here, Favorable number of outcomes = 1 {HT}

Total number of possible outcomes = 4 {HH, HT, TH, TT}

Thus, P (B) = ¼

P (A) + P (B) = ¼ +1/4 = 2/4 = ½ = 0.5

Probability of a tail on the first and a head on the second or a head on the first and a tail on the second = 0.5

Favorable number of outcomes = 3 {HT, TH, HH}

Total number of possible outcomes = 4 {HH, HT, TH, TT}

Probability of at least one head on the two flips = 3/4

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