A defective coin minting machine produces coins whose probability of Heads is a random variable Q with PDF
fQ(q)={5q4,0,if q∈[0,1],otherwise.
A coin produced by this machine is tossed repeatedly, with successive tosses assumed to be independent. Let A be the event that the first toss of this coin results in Heads, and let B be the event that the second toss of this coin results in Heads.
P(A)= 0.83333
(Your answer should be a number.)
Find the conditional PDF of Q given event A. Express your answer in terms of q using standard notation.
For 0≤q≤1, fQ∣A(q)=6⋅q5
P(B∣A)= 0.8571
(Your answer should be a number.)
Thanks, Friend. (Y)
To find the conditional PDF of Q given event A, we need to use Bayes' rule. Bayes' rule states that:
P(Q|A) = P(A|Q) * P(Q) / P(A)
Since we are given the PDF of Q, fQ(q), we can calculate P(Q) as the integral of fQ(q) over its range:
P(Q) = ∫[0,1] fQ(q) dq
Calculating the integral, we get:
P(Q) = ∫[0,1] 5q^4 dq = [q^5] from 0 to 1 = 1 - 0 = 1
Next, we need to find P(A|Q), which is the probability of event A given a specific value of Q. Since we are given that P(A) = 0.83333, we can set up the following equation:
P(A|Q) = P(A and Q) / P(Q) = P(Q|A) * P(A) / P(Q)
Given that P(A) = 0.83333 and P(Q) = 1, the equation becomes:
P(Q|A) = P(A|Q) * 1 / 0.83333 = P(A|Q) / 0.83333
Therefore, we have:
P(B|A) = 0.8571 = P(A|B) * P(B) / P(A)
Given that P(A) = 0.83333, we can rearrange the equation as:
P(A|B) = P(B|A) * P(A) / P(B)
Substituting the given values, we get:
P(A|B) = 0.8571 * 0.83333 / P(B)
Since P(B) is not given, we cannot calculate P(A|B) directly.
To find the conditional PDF of Q given event A, we need to apply Bayes' theorem. The formula for the conditional PDF of Q given event A is:
f(Q|A) = (f(Q) * P(A|Q)) / P(A)
Let's calculate each component of this equation step by step:
1. P(A|Q): This is the probability of event A occurring given a specific value of Q. Since we are given that P(A) = 0.83333, this implies that P(A|Q) = P(A). Therefore, P(A|Q) = 0.83333.
2. P(A): This is the probability of event A occurring. We are given that P(A) = 0.83333.
3. f(Q): This is the PDF of Q. Given that fQ(q) = 5q^4 for 0≤q≤1 and 0 otherwise, we need to integrate the PDF to find the total probability density. Integrating fQ(q) over the range [0,1], we have:
∫[0,1] 5q^4 dq = [q^5] from 0 to 1 = 1^5 - 0^5 = 1
Therefore, f(Q) = 1.
Now we can substitute these values into the formula for f(Q|A):
f(Q|A) = (f(Q) * P(A|Q)) / P(A)
= (1 * 0.83333) / 0.83333
= 1
So the conditional PDF of Q given event A, f(Q|A), is equal to 1 for 0≤q≤1.
Moving on to find P(B|A):
P(B|A) = P(A∩B) / P(A)
We are given that P(A) = 0.83333. Since the tosses are assumed to be independent, the probability of both events A and B occurring is the product of their individual probabilities:
P(A∩B) = P(A) * P(B|A) = 0.83333 * 0.8571 = 0.7143
So P(B|A) = 0.7143.