A defective coin minting machine produces coins whose probability of Heads is a random variable [mathjaxinline]Q[/mathjaxinline] with PDF
[mathjax]f_{Q}(q) = \left\{ \begin{array}{ll} 5q^4, & \mbox{if $q \in [0,1]$},\\ 0, & \mbox{otherwise}. \end{array}\right.[/mathjax]
A coin produced by this machine is tossed repeatedly, with successive tosses assumed to be independent. Let [mathjaxinline]A[/mathjaxinline] be the event that the first toss of this coin results in Heads, and let [mathjaxinline]B[/mathjaxinline] be the event that the second toss of this coin results in Heads.
[mathjaxinline]\mathbf{P}(A)=\,[/mathjaxinline]
(Your answer should be a number.)
Find the conditional PDF of [mathjaxinline]Q[/mathjaxinline] given event [mathjaxinline]A[/mathjaxinline]. Express your answer in terms of [mathjaxinline]q[/mathjaxinline] using standard notation.
For [mathjaxinline]0 \leq q \leq 1[/mathjaxinline], [mathjaxinline]f_{Q\mid A}(q)=\,[/mathjaxinline]
\(\)
[mathjaxinline]\mathbf{P}(B\mid A)=\,[/mathjaxinline
Since the first toss of the coin resulting in Heads is one of the conditions for event A to occur, we can write:
[mathjax]\mathbf{P}(A) = \int_{0}^{1} \left( 5q^4 \cdot \mathbf{P}(A|Q=q) \right) dq[/mathjax]
Since [mathjax]\mathbf{P}(A|Q=q) = q[/mathjax], we can simplify the expression for [mathjax]\mathbf{P}(A)[/mathjax]:
[mathjax]\mathbf{P}(A) = \int_{0}^{1} 5q^5 dq = \frac{5}{6}[/mathjax]
Next, we can find the conditional PDF of [mathjax]Q[/mathjax] given event A using Bayes' rule:
[mathjax]f_{Q|A}(q) = \frac{f_Q(q) \cdot \mathbf{P}(A|Q=q)}{\mathbf{P}(A)}[/mathjax]
Substituting the given values, we get:
[mathjax]f_{Q|A}(q) = \frac{5q^4 \cdot q}{\frac{5}{6}} = 6q^5[/mathjax]
Finally, we can find [mathjax]\mathbf{P}(B|A)[/mathjax]:
[mathjax]\mathbf{P}(B|A) = \int_{0}^{1} \left( 6q^5 \cdot \mathbf{P}(B|Q=q,A) \right) dq[/mathjax]
Since the second toss of the coin is assumed to be independent of the first toss, [mathjax]\mathbf{P}(B|Q=q,A) = q[/mathjax]. Therefore:
[mathjax]\mathbf{P}(B|A) = \int_{0}^{1} (6q^6) dq = \frac{6}{7}[/mathjax]
So, [mathjax]\mathbf{P}(B|A) = \frac{6}{7}[/mathjax].
To find [mathjaxinline]\mathbf{P}(A)[/mathjaxinline], we need to integrate the PDF [mathjaxinline]f_{Q}(q)[/mathjaxinline] over the range of possible values for [mathjaxinline]Q[/mathjaxinline]. Given that [mathjaxinline]q \in [0,1][/mathjaxinline], the probability of event [mathjaxinline]A[/mathjaxinline] can be calculated as:
[mathjax]\mathbf{P}(A) = \int_{0}^{1} f_{Q}(q) \, dq[/mathjax]
Using the given PDF [mathjaxinline]f_{Q}(q) = 5q^4[/mathjaxinline], we can substitute it into the integral:
[mathjax]\mathbf{P}(A) = \int_{0}^{1} 5q^4 \, dq[/mathjax]
Evaluating the integral gives us:
[mathjax]\mathbf{P}(A) = \left[ \frac{5}{5} \cdot \frac{q^5}{5} \right]_{0}^{1} = \frac{1}{5}[/mathjax]
So, [mathjaxinline]\mathbf{P}(A) = \frac{1}{5}[/mathjaxinline].
To find the conditional PDF of [mathjaxinline]Q[/mathjaxinline] given event [mathjaxinline]A[/mathjaxinline], we can use Bayes' theorem. The conditional PDF, denoted as [mathjaxinline]f_{Q\mid A}(q)[/mathjaxinline], can be calculated as:
[mathjax]\begin{align*}
f_{Q\mid A}(q) &= \frac{f_{AQ}(a, q)}{f_A(a)} \\
&= \frac{\mathbf{P}(AQ)}{\mathbf{P}(A)}
\end{align*}[/mathjax]
Since [mathjaxinline]A[/mathjaxinline] is the event that the first toss of the coin results in Heads, we have [mathjaxinline]\mathbf{P}(AQ) = f_{Q}(q) \cdot \mathbf{P}(A) = 5q^4 \cdot \frac{1}{5} = q^4[/mathjaxinline]. Substituting this into the formula, we get:
[mathjax]\begin{align*}
f_{Q\mid A}(q) &= \frac{q^4}{\frac{1}{5}} \\
&= 5q^4
\end{align*}[/mathjax]
Therefore, for [mathjaxinline]0 \leq q \leq 1[/mathjaxinline], [mathjaxinline]f_{Q\mid A}(q) = 5q^4[/mathjaxinline].
Finally, to find [mathjaxinline]\mathbf{P}(B\mid A)[/mathjaxinline], we can use the fact that the tosses are assumed to be independent, so the probability of event [mathjaxinline]B[/mathjaxinline] given event [mathjaxinline]A[/mathjaxinline] is the same as the probability of a Heads on the second toss, which is 0.5. Therefore, [mathjaxinline]\mathbf{P}(B\mid A) = 0.5[/mathjaxinline].