A defective coin minting machine produces coins whose probability of Heads is a random variable Q with PDF
fQ(q)={5*q^4 if q∈[0,1]
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)=?
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)=?
P(B∣A)=
Well, if we have a defective coin minting machine, it's safe to say that the probability of anything happening may be a bit... off. Let's dive into this and see what we can come up with.
First, let's find the probability of event A, which is the first toss resulting in Heads. We can calculate this by integrating the PDF of Q from 0 to 1:
P(A) = ∫(0 to 1) (5*q^4) dq
Now, let's calculate that integral:
P(A) = [q^5] from 0 to 1
P(A) = (1^5) - (0^5)
P(A) = 1 - 0
So, P(A) = 1. That means that the probability of the first toss resulting in Heads is 1. Perfect! Now, let's move on to the next part.
Given event A, we want to find the conditional PDF of Q. It's denoted as fQ|A(q). In this case, we need to find the PDF of Q, but with the condition that the first toss resulted in Heads.
To find fQ|A(q), we can use Bayes' theorem:
fQ|A(q) = fQ(q) * P(A|q) / P(A)
Since P(A) is equal to 1, we can simplify the equation:
fQ|A(q) = fQ(q) * P(A|q)
This means that the conditional PDF of Q given event A is simply equal to the original PDF of Q, fQ(q), multiplied by the probability of event A given a certain value of Q, P(A|q).
Lastly, let's find P(B|A), which is the probability that the second toss results in Heads given that the first toss resulted in Heads. Since the tosses are assumed to be independent, the result of the first toss doesn't affect the second toss. So:
P(B|A) = P(B)
In other words, the probability of the second toss resulting in Heads is the same as if we didn't have event A.
And that, my friend, is as far as we can go with this problem. Probability can be tricky, but as long as we don't take it too seriously, we'll be fine. Happy coin tossing!
To find P(A), we need to calculate the probability that the first toss of the coin results in Heads. Since the coin's probability of Heads is a random variable Q with PDF fQ(q), we can integrate the PDF over the range of q where the event A occurs.
Given fQ(q) = 5*q^4 for q ∈ [0,1], the probability density function (PDF) of Q, we can calculate P(A) as:
P(A) = ∫[0,1] fQ(q) dq
= ∫[0,1] 5*q^4 dq
Using the power rule to integrate, we have:
= (5/5)*[q^5/5] evaluated from 0 to 1
= q^5/5 evaluated from 0 to 1
= (1^5/5) - (0^5/5)
= 1/5 - 0
= 1/5
Therefore, P(A) = 1/5.
Next, we need to find the conditional PDF of Q given event A, denoted as fQ|A(q).
The conditional PDF is defined as:
fQ|A(q) = P(Q = q | A)
To calculate this, we can use Bayes' theorem:
fQ|A(q) = (P(A | Q = q) * fQ(q)) / P(A)
We already know P(A) = 1/5 from earlier calculations. Now, we need to find P(A | Q = q).
P(A | Q = q) is the probability that the first toss results in Heads when the probability of Heads is q. Since the tosses are assumed to be independent, this probability is simply q.
Therefore, P(A | Q = q) = q.
Now, we can substitute these values into the conditional PDF equation:
fQ|A(q) = (q * fQ(q)) / P(A)
= (q * 5*q^4) / (1/5)
= 25q^5
Therefore, fQ|A(q) = 25q^5 for 0 ≤ q ≤ 1.
Now, to find P(B | A), the probability that the second toss of the coin results in Heads given that the first toss resulted in Heads.
Since the events A and B are assumed to be independent and the first toss has already occurred, the probability of getting Heads on the second toss is simply q, the probability of Heads for that particular coin.
Therefore, P(B | A) = q.
Since the probability of Heads, q, is a random variable Q with a PDF fQ(q) = 5*q^4, we can substitute this PDF into the expression:
P(B | A) = ∫[0,1] q * 5*q^4 dq
= 5 ∫[0,1] q^5 dq
Using the power rule to integrate, we have:
= (5/6) * [q^6/6] evaluated from 0 to 1
= (1/6) - (0/6)
= 1/6
Therefore, P(B | A) = 1/6.
To find the probability of event A, we need to calculate the integral of the probability density function (PDF) over the range where the event A occurs.
Given the PDF of Q, we have:
fQ(q) = 5*q^4, for q ∈ [0,1]
To find P(A), we need to integrate fQ(q) over the range where the first toss results in Heads, which is q ∈ [0,1]:
P(A) = ∫[0,1] fQ(q) dq
Let's calculate this integral:
P(A) = ∫[0,1] 5*q^4 dq
Integrating the function 5*q^4:
P(A) = [5/5 * q^5/5] evaluated from 0 to 1
P(A) = [q^5/5] evaluated from 0 to 1
P(A) = (1^5/5) - (0^5/5)
P(A) = 1/5 - 0/5
P(A) = 1/5
So, P(A) = 1/5.
Now, let's find the conditional PDF of Q given event A, denoted as fQ|A(q).
To calculate fQ|A(q), we need to divide the PDF of Q, fQ(q), by P(A). Since P(A) = 1/5, we have:
fQ|A(q) = (1/(1/5)) * fQ(q)
fQ|A(q) = 5 * fQ(q)
Therefore, for 0 ≤ q ≤ 1, the conditional PDF of Q given event A is:
fQ|A(q) = 5 * 5 * q^4
fQ|A(q) = 25 * q^4
Now, let's calculate P(B|A), the probability of event B given event A.
If the first toss resulted in Heads (event A), the second toss is still an independent event. Therefore, the probability of event B given event A is the same as the probability of getting Heads in a single toss:
P(B|A) = P(Heads)
Since we are not given any information about the bias of the defective coin minting machine affecting the second toss, the probability of getting Heads in a single toss remains the same, regardless of event A.
Thus, P(B|A) = P(Heads) = 1/2.