IDNLearn.com: Your trusted source for accurate and reliable answers. Join our interactive Q&A community and access a wealth of reliable answers to your most pressing questions.
Sagot :
To find [tex]\( P(C \mid Y) \)[/tex], we need to use the concept of conditional probability. The conditional probability formula is:
[tex]\[ P(C \mid Y) = \frac{P(C \cap Y)}{P(Y)} \][/tex]
From the table, we have the following relevant information:
1. The total number of events where [tex]\( Y \)[/tex] occurs, [tex]\( P(Y) \)[/tex]:
- [tex]\( Y \)[/tex] appears in the columns with totals: [tex]\( 10 + 5 + 15 = 30 \)[/tex].
2. The total number of events where both [tex]\( C \)[/tex] and [tex]\( Y \)[/tex] occur, [tex]\( P(C \cap Y) \)[/tex]:
- There are 15 events where both [tex]\( C \)[/tex] and [tex]\( Y \)[/tex] are true.
Using this information, we can calculate:
[tex]\[ P(C \mid Y) = \frac{P(C \cap Y)}{P(Y)} = \frac{15}{30} = 0.5 \][/tex]
Thus, the value of [tex]\( P(C \mid Y) \)[/tex], rounded to the nearest tenth, is [tex]\( 0.5 \)[/tex].
So the correct answer is:
[tex]\[ \boxed{0.5} \][/tex]
[tex]\[ P(C \mid Y) = \frac{P(C \cap Y)}{P(Y)} \][/tex]
From the table, we have the following relevant information:
1. The total number of events where [tex]\( Y \)[/tex] occurs, [tex]\( P(Y) \)[/tex]:
- [tex]\( Y \)[/tex] appears in the columns with totals: [tex]\( 10 + 5 + 15 = 30 \)[/tex].
2. The total number of events where both [tex]\( C \)[/tex] and [tex]\( Y \)[/tex] occur, [tex]\( P(C \cap Y) \)[/tex]:
- There are 15 events where both [tex]\( C \)[/tex] and [tex]\( Y \)[/tex] are true.
Using this information, we can calculate:
[tex]\[ P(C \mid Y) = \frac{P(C \cap Y)}{P(Y)} = \frac{15}{30} = 0.5 \][/tex]
Thus, the value of [tex]\( P(C \mid Y) \)[/tex], rounded to the nearest tenth, is [tex]\( 0.5 \)[/tex].
So the correct answer is:
[tex]\[ \boxed{0.5} \][/tex]
Your presence in our community is highly appreciated. Keep sharing your insights and solutions. Together, we can build a rich and valuable knowledge resource for everyone. IDNLearn.com provides the best answers to your questions. Thank you for visiting, and come back soon for more helpful information.