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].
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
1 answer