A defective coin minting machine produces coins whose probability of Heads is a random variable Q with PDF

fQ(q)={3q2,0,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)=- unanswered
(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)=- unanswered
P(B∣A)=- unanswered
(Your answer should be a number.)

a) integral(5*q^4*q,q,0,1) in WolframAlfa=5/6

b)(5*q^4*q)/(5/6)
c=P(B|A)=P(A) independent tosses=5/6

Well, well, well, looks like it's time for some coin tossing fun with a defective machine! Let's see if I can't provide you with some answers that are worth their weight in laughter!

First up, we need to find the probability of event A, which is the first toss resulting in Heads. Since the probability Q is a random variable, we need to integrate its PDF over the range where A occurs.

To do that, we integrate the PDF fQ(q) from 0 to 1, which gives us:

P(A) = ∫[0,1] 3q^2 dq

Now, let's solve this integral. *Drumroll* Please!

P(A) = [q^3] evaluated from 0 to 1

P(A) = 1^3 - 0^3

P(A) = 1 - 0

Ah, you gotta love those easy integrals. So, P(A) = 1.

Now, let's move on to the conditional PDF of Q given event A. This is denoted by fQ|A(q). We can find this by dividing the PDF fQ(q) by the probability of A, which is 1 in this case.

For 0≤q≤1, fQ|A(q) = fQ(q) / P(A)

Since P(A) = 1, we have:

fQ|A(q) = 3q^2 / 1

fQ|A(q) = 3q^2

Now, let's move on to the probability of event B given event A, which is denoted by P(B|A). This tells us the probability that the second toss results in Heads, given that the first toss already resulted in Heads.

Since the tosses are independent, the outcome of the first toss does not affect the second toss. Therefore, P(B|A) is equal to the probability of getting Heads on a single coin toss, which is 1/2.

So, P(B|A) = 1/2.

And there you have it! The conditional PDF of Q given event A is 3q^2 for 0≤q≤1, and the probability of event B given event A is 1/2. I hope these answers put a smile on your face, just like a clown juggling coins! Keep on laughing! 😄

To find the probability of event A, we need to integrate the probability density function (PDF) of Q over the range where A occurs.

The event A occurs when the first toss of the coin results in Heads. Therefore, A occurs when the random variable Q is in the range [0,1/2]. So, we need to find P(A) by integrating the PDF of Q over this range:

P(A) = ∫[0,1/2] fQ(q) dq

= ∫[0,1/2] 3q^2 dq

To evaluate this integral, we can use the power rule of integration:

P(A) = [q^3] evaluated from 0 to 1/2

= (1/2)^3 - 0^3

= 1/8

Therefore, P(A) = 1/8.

Now, to find the conditional PDF of Q given event A, we need to normalize the PDF of Q over the range where event A occurs (i.e., [0,1/2]). This will ensure that the conditional PDF integrates to 1 over its support.

The conditional PDF of Q given event A, denoted as fQ|A(q), can be found using the formula:

fQ|A(q) = fQ(q)/P(A) for q in [0,1/2]

Substituting the given PDF of Q and the calculated value of P(A), we get:

fQ|A(q) = (3q^2)/(1/8) for q in [0,1/2]

= 24q^2 for q in [0,1/2]

So, for 0 ≤ q ≤ 1/2, the conditional PDF of Q given event A is fQ|A(q) = 24q^2.

To find the probability of event B given event A (P(B|A)), we need to consider the independent tosses of the coin. Since A has already occurred (i.e., the first toss resulted in Heads), event B can occur in two ways: either the second toss is Heads or it is Tails. The probability of the second toss being Heads given that the first toss is Heads is 1/2, and the probability of it being Tails is also 1/2 (assuming the coin is fair and unbiased). Therefore, we have:

P(B|A) = 1/2

So, P(B|A) = 1/2.

For 0≤q≤1, fQ∣A(q)= 4*q^3

P(B∣A) = 0.8

if you are reading these answers. Don't be selfish! you should share the other answers to the whole problem set....

a) 0.75