Question

A fair, “relatively thick” coin, when flipped, can show heads (H) or tails (T) upon coming to rest, with equal probabilities, but it may decide to come to rest on its edge (E) with a probability of 6%. Showing formulas and explanation, answer the following questions:

1. If the coin were flipped twice, what would be the probability that it would show heads the first time and would sit on its edge the second time? Provide a brief explanation to support your computation(s). Show the probability value in 4 decimal places.
2. If the coin is flipped 12 times, what is the probability that the coin would sit on its edge (“getting edge”) at most once? Showing your work, report the probability value in 4 decimal places.
3. Determine the probability of getting tails twice when the coin is flipped 12 times. Showing your work, report the probability value in 4 decimal places.
4. What is the probability of getting heads or tails at least once when the coin is flipped 12 times? Showing your work, report the probability value in 4 decimal places.

Answer 1

P(rest on oits edge) = 0.06

P(H)=P(T) = (1-0.6)/2 = 0.94/2 = 0.47

a)

flipping of coin is independent event

P(first head and then edge) = P(first hea)*P(then edge) = 0.47*0.06=0.0282

b)

it is a binomial probability distribution,

and probability is given by

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

here, n=12

P(sit on edge) =0.06

P(sit on edge atmost once)=P(X=0)+P(X=1) =

C(12,0)*0.06⁰*(1-0.06)^(12-0) + C(12,1)*0.06¹*(1-0.06)^(12-1) = 0.47592+0.3645=0.8405

c)

P(tail)=0.47

P(Tail twice) = C(12,2)*0.47²*(1-0.47)^(12-2) =0.0255

d)

P(head or tail)=0.47+0.47 = 0.94

P( getting heads or tails at least once)= 1 - P(no heads or tails) = 1 - C(12,0)*0.94⁰*(1-0.94)^(12-0)=1