Answer :
Let's solve the given probabilities step-by-step.
### 1. Probability that the student was male given that they got a 'B':
We need to find [tex]\( P(\text{male} \mid B) \)[/tex].
From the given table:
- Number of males who got a 'B': 7
- Total number of students who got a 'B': 15
The formula to calculate [tex]\( P(\text{male} \mid B) \)[/tex] is:
[tex]\[ P(\text{male} \mid B) = \frac{P(\text{male and B})}{P(B)} \][/tex]
Here:
- [tex]\( P(\text{male and B}) \)[/tex] is the probability that a student is both male and got a 'B'. This can be found by the ratio of the number of males who got a 'B' to the total number of students, which is [tex]\( \frac{7}{72} \)[/tex].
- [tex]\( P(B) \)[/tex] is the probability that a student got a 'B'. This can be found by the ratio of the total number of students who got a 'B' to the total number of students, which is [tex]\( \frac{15}{72} \)[/tex].
Now, the probability:
[tex]\[ P(\text{male} \mid B) = \frac{\frac{7}{72}}{\frac{15}{72}} = \frac{7}{15} \][/tex]
### 2. Probability that the student got a 'B' given that they are male:
We need to find [tex]\( P(B \mid \text{male}) \)[/tex].
From the given table:
- Number of students who got a 'B' who are male: 7
- Total number of males: 32
The formula to calculate [tex]\( P(B \mid \text{male}) \)[/tex] is:
[tex]\[ P(B \mid \text{male}) = \frac{P(\text{male and B})}{P(\text{male})} \][/tex]
Here:
- [tex]\( P(\text{male and B}) \)[/tex] is again the probability that a student is both male and got a 'B', which is [tex]\( \frac{7}{72} \)[/tex].
- [tex]\( P(\text{male}) \)[/tex] is the probability that a student is male. This can be found by the ratio of the total number of males to the total number of students, which is [tex]\( \frac{32}{72} \)[/tex].
Now, the probability:
[tex]\[ P(B \mid \text{male}) = \frac{\frac{7}{72}}{\frac{32}{72}} = \frac{7}{32} \][/tex]
Thus, the required probabilities in their reduced fraction forms are:
[tex]\[ P(\text{male} \mid B) = \frac{7}{15} \][/tex]
[tex]\[ P(B \mid \text{male}) = \frac{7}{32} \][/tex]
### 1. Probability that the student was male given that they got a 'B':
We need to find [tex]\( P(\text{male} \mid B) \)[/tex].
From the given table:
- Number of males who got a 'B': 7
- Total number of students who got a 'B': 15
The formula to calculate [tex]\( P(\text{male} \mid B) \)[/tex] is:
[tex]\[ P(\text{male} \mid B) = \frac{P(\text{male and B})}{P(B)} \][/tex]
Here:
- [tex]\( P(\text{male and B}) \)[/tex] is the probability that a student is both male and got a 'B'. This can be found by the ratio of the number of males who got a 'B' to the total number of students, which is [tex]\( \frac{7}{72} \)[/tex].
- [tex]\( P(B) \)[/tex] is the probability that a student got a 'B'. This can be found by the ratio of the total number of students who got a 'B' to the total number of students, which is [tex]\( \frac{15}{72} \)[/tex].
Now, the probability:
[tex]\[ P(\text{male} \mid B) = \frac{\frac{7}{72}}{\frac{15}{72}} = \frac{7}{15} \][/tex]
### 2. Probability that the student got a 'B' given that they are male:
We need to find [tex]\( P(B \mid \text{male}) \)[/tex].
From the given table:
- Number of students who got a 'B' who are male: 7
- Total number of males: 32
The formula to calculate [tex]\( P(B \mid \text{male}) \)[/tex] is:
[tex]\[ P(B \mid \text{male}) = \frac{P(\text{male and B})}{P(\text{male})} \][/tex]
Here:
- [tex]\( P(\text{male and B}) \)[/tex] is again the probability that a student is both male and got a 'B', which is [tex]\( \frac{7}{72} \)[/tex].
- [tex]\( P(\text{male}) \)[/tex] is the probability that a student is male. This can be found by the ratio of the total number of males to the total number of students, which is [tex]\( \frac{32}{72} \)[/tex].
Now, the probability:
[tex]\[ P(B \mid \text{male}) = \frac{\frac{7}{72}}{\frac{32}{72}} = \frac{7}{32} \][/tex]
Thus, the required probabilities in their reduced fraction forms are:
[tex]\[ P(\text{male} \mid B) = \frac{7}{15} \][/tex]
[tex]\[ P(B \mid \text{male}) = \frac{7}{32} \][/tex]
