Let \(A\) be the event that the player makes the first throw, and \(B\) be the event that the player makes the second throw.
Given:
\(P( A) = 0.6\)
\(P( B| A) = 0.7\)
\(P( B| A^ c) = 0.45\)
We need to find \(P( A^ c| B) \).
Using Bayes' theorem:
\[P( A^ c| B) = \frac { P( B| A^ c) \cdot P( A^ c) } { P( B) } \]
First, find \(P( B) \):
\[P( B) = P( B| A) \cdot P( A) + P( B| A^ c) \cdot P( A^ c) \]
\[P( B) = ( 0.7 \cdot 0.6) + ( 0.45 \cdot 0.4) \]
\[P( B) = 0.42 + 0.18\]
\[P( B) = 0.6\]
Now, apply Bayes' theorem:
\[P( A^ c| B) = \frac { P( B| A^ c) \cdot P( A^ c) } { P( B) } \]
\[P( A^ c| B) = \frac { 0.45 \cdot 0.4} { 0.6} \]
\[P( A^ c| B) = \frac { 0.18} { 0.6} \]
\[P( A^ c| B) = 0.3\]
To express \(0.3\) as a fraction:
\[0.3 = \frac { 3} { 10} \]
Thus, the probability that he missed his first throw given that he made his second throw is \(\frac { 3} { 10} \).

Supplemental Knowledge

Conditional probability is the likelihood of an event occurring given that another event has already occurred. It is a fundamental concept in probability theory and statistics, often used to update the probability of an event based on new information.
The formula for conditional probability is:
\[P( A| B) = \frac { P( A \cap B) } { P( B) } \]
where \(P( A| B) \) is the probability of event A occurring given that event B has occurred, \(P( A \cap B) \) is the joint probability of both events A and B occurring, and \(P( B) \) is the probability of event B.

From Concepts to Reality

Consider, for instance, a medical test to identify disease. When patients test positive, doctors use conditional probability analysis to calculate the probability that their patient actually suffers from that particular ailment based on factors like accuracy of testing and disease prevalence rates - providing more informed decisions for further testing or treatments.

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