Adefective coin minting machineproducescoinswhoseprobability ofheadsisarandomvariable P with PDF
1+sin(2πp), if p ∈ [0,1],fP (p)= 0, otherwise.
Inessence,aspecificcoinproducedby thismachinewillhavea fixedprobability P = p of giving heads, but you do not know initially what that probability is. A coin produced by this machine is selected and tossed repeatedly, with successive tosses assumed independent.
(a)
Find the probability that the first coin toss results in heads.
(b)
Given that the first coin toss resulted in heads, find the conditional PDF of P.
(c)
Given that the first coin toss resulted in heads, find the conditional probability of heads on the second toss.
(a) To find the probability that the first coin toss results in heads, we need to integrate the PDF over the range of possible values for P. Since the range of P is from 0 to 1, we can calculate the probability as follows:
P(heads on first toss) = ∫[0,1] (1 + sin(2πp)) dp
However, since the integral involves a trigonometric function, it might be easier to split it into two parts:
P(heads on first toss) = ∫[0,1] 1 dp + ∫[0,1] sin(2πp) dp
The first integral is simply the integral of a constant, which evaluates to the length of the interval [0,1]:
P(heads on first toss) = [p] evaluated from 0 to 1 + ∫[0,1] sin(2πp) dp
Simplifying the first part, we get:
P(heads on first toss) = 1 + ∫[0,1] sin(2πp) dp
To integrate sin(2πp), we can use the following property:
∫ sin(ax) dx = -[cos(ax)]/a
In this case, a = 2π, so the integral becomes:
P(heads on first toss) = 1 - [cos(2πp)]/[2π] evaluated from 0 to 1
Evaluating at the bounds, we get:
P(heads on first toss) = 1 - (cos(2π) - cos(0))/(2π)
= 1 - (1 - 1)/(2π)
= 1 - 0
= 1
Therefore, the probability that the first coin toss results in heads is 1.
(b) Given that the first coin toss resulted in heads, we want to find the conditional PDF of P. This can be calculated using Bayes' theorem:
fP|H(p) = fH|P(H|p) * fP(p) / fH(H)
where fP|H(p) is the conditional PDF of P given that H (heads) occurred on the first toss, fH|P(H|p) is the conditional probability of H given P, fP(p) is the PDF of P, and fH(H) is the probability of H (which we found to be 1 in part (a)).
Since H is heads and we know the probability P, the conditional probability fH|P(H|p) is simply P:
fH|P(H|p) = P
Plugging in the known values, we get:
fP|H(p) = P * fP(p) / fH(H)
Since fH(H) is 1, we can simplify the equation further:
fP|H(p) = P * fP(p)
Therefore, the conditional PDF of P given that the first toss resulted in heads is P multiplied by the original PDF, (1 + sin(2πp)).
(c) Given that the first coin toss resulted in heads, we want to find the conditional probability of heads on the second toss.
Let's denote the second coin toss as H2. Using the law of total probability, we can calculate this conditional probability as follows:
P(H2|H1) = P(H2 and H1) / P(H1)
P(H2|H1) = P(H1 and H2) / P(H1)
Since successive coin tosses are assumed to be independent, P(H1 and H2) is simply the product of the probabilities of heads on the first toss and heads on the second toss:
P(H2|H1) = P(H1) * P(H2)
Given that the first coin toss resulted in heads (which we found to have a probability of 1), we have:
P(H2|H1) = 1 * P(H2)
Therefore, the conditional probability of heads on the second toss given that the first toss resulted in heads is equal to the probability of heads on the second toss.