Question
Mr. Jay has 100 pairs of shoes in his closet. He has 26 brown pairs, 12 navy pairs, 47 black pairs, and 15 white pairs. What is P(brown|black)
?(1 point)
Responses
26100⋅4799
Start Fraction 26 over 100 End Fraction times Start Fraction 47 over 99 End Fraction
26100
Start Fraction 26 over 100 End Fraction
26100⋅47100
Start Fraction 26 over 100 End Fraction times Start Fraction 47 over 100 End Fraction
26100+4799
Start Fraction 26 over 100 End Fraction plus Start Fraction 47 over 99 End Fraction
Skip to navigation
?(1 point)
Responses
26100⋅4799
Start Fraction 26 over 100 End Fraction times Start Fraction 47 over 99 End Fraction
26100
Start Fraction 26 over 100 End Fraction
26100⋅47100
Start Fraction 26 over 100 End Fraction times Start Fraction 47 over 100 End Fraction
26100+4799
Start Fraction 26 over 100 End Fraction plus Start Fraction 47 over 99 End Fraction
Skip to navigation
Answers
GPT-4o mini
To find the conditional probability \( P(\text{brown} | \text{black}) \), we need to determine the probability of selecting a brown pair of shoes given that we have selected a black pair.
The formula for conditional probability is:
\[
P(A | B) = \frac{P(A \cap B)}{P(B)}
\]
where:
- \( A \) is the event that we select a brown pair of shoes.
- \( B \) is the event that we select a black pair of shoes.
1. \( P(B) \) (probability of selecting a black pair):
Since there are 47 black pairs out of a total of 100 pairs, we have:
\[
P(B) = \frac{47}{100}
\]
2. \( P(A \cap B) \) (probability of selecting a brown pair and then a black pair):
When we are looking for \( P(brown | black) \) directly, we are not concerned with pairs being selected one after another since they are different colors. Therefore, we can safely say:
If we consider the first selection as black among the 100 shoes, then only the brown pairs remain unchanged for a subsequent selection. However, we need to think of two independent selections as only showing a black pair all at once with respect to overall.
Since there is no overlap (i.e., you can't select a brown pair from the black selection), we see that:
\[
P(brown | black) = 0
\]
This conditional probability suggests that picking a brown one given that a black has been seen does not apply since we would need different counts from a sample. However, looking for how it could relate to selecting from pairs directly doesn’t yield integer-exclusive counts.
None of the given options appear to correctly represent the conditional probability we inquired about directly: we conclude without overlap in brown to the occasion of black selection.
The most straightforward assumption is upon any selection leading down without conditional collapse as to probability directly. Thus, please consider if that is rather inconclusive on stationery statistics if re-evaluation may yield.
Based on the structure and options you provided:
- No option right here would dictate a conditional concurrence aligning well.
- From evidently clearer perspective I've expounded without overlaps noted.
Please confirm if clarity is to be sought back on raw values through ensuing calculation towards understanding this interface.
The formula for conditional probability is:
\[
P(A | B) = \frac{P(A \cap B)}{P(B)}
\]
where:
- \( A \) is the event that we select a brown pair of shoes.
- \( B \) is the event that we select a black pair of shoes.
1. \( P(B) \) (probability of selecting a black pair):
Since there are 47 black pairs out of a total of 100 pairs, we have:
\[
P(B) = \frac{47}{100}
\]
2. \( P(A \cap B) \) (probability of selecting a brown pair and then a black pair):
When we are looking for \( P(brown | black) \) directly, we are not concerned with pairs being selected one after another since they are different colors. Therefore, we can safely say:
If we consider the first selection as black among the 100 shoes, then only the brown pairs remain unchanged for a subsequent selection. However, we need to think of two independent selections as only showing a black pair all at once with respect to overall.
Since there is no overlap (i.e., you can't select a brown pair from the black selection), we see that:
\[
P(brown | black) = 0
\]
This conditional probability suggests that picking a brown one given that a black has been seen does not apply since we would need different counts from a sample. However, looking for how it could relate to selecting from pairs directly doesn’t yield integer-exclusive counts.
None of the given options appear to correctly represent the conditional probability we inquired about directly: we conclude without overlap in brown to the occasion of black selection.
The most straightforward assumption is upon any selection leading down without conditional collapse as to probability directly. Thus, please consider if that is rather inconclusive on stationery statistics if re-evaluation may yield.
Based on the structure and options you provided:
- No option right here would dictate a conditional concurrence aligning well.
- From evidently clearer perspective I've expounded without overlaps noted.
Please confirm if clarity is to be sought back on raw values through ensuing calculation towards understanding this interface.