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Section:
Mathematics
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© examsnet.com
Question : 7 of 92
Marks:
+1
,
-0
In a school there are 3 sections A, B and C. Section A contains 20 girls and 30 boys, section
B
B
B
contains 40 girls and 20 boys and section
C
C
C
contains 10 girls and 30 boys. The probabilities of selecting the section
A
,
B
A, B
A
,
B
and C are
0.2
,
0.3
0.2,0.3
0.2
,
0.3
and 0.5 respectivley. If a student selected at random from the school is a girl, then the probability that she belongs to section A is
[AP EAPCET 24 May 2025 S1]
  
  
121
200
\;\;\frac{121}{200}
200
121
​
  
  
16
121
\;\;\frac{16}{121}
121
16
​
  
  
14
81
\;\;\frac{14}{81}
81
14
​
  
16
81
\;\frac{16}{81}
81
16
​
Validate
Solution:
Let
A
,
B
A, B
A
,
B
and
C
C
C
be the event that a student is from section
A
,
B
A, B
A
,
B
and
C
C
C
.
So,
P
(
A
)
=
0.2
,
P
(
B
)
=
0.3
,
P
(
C
)
=
0.5
P(A)=0.2, P(B)=0.3, P(C)=0.5
P
(
A
)
=
0.2
,
P
(
B
)
=
0.3
,
P
(
C
)
=
0.5
Now, section A : 20 girls, 30 boys, total 50 students
Section B : 40 girls, 20 boys, total 60 students
Section C : 10 girls, 30 boys, total 40 students
P
(
G
/
A
)
  
=
  
20
50
=
0.4
,
P
(
G
/
B
)
=
  
40
60
=
  
2
3
P(G / A)\;=\;\frac{20}{50}=0.4, P(G / B)=\;\frac{40}{60}=\;\frac{2}{3}
P
(
G
/
A
)
=
50
20
​
=
0.4
,
P
(
G
/
B
)
=
60
40
​
=
3
2
​
P
(
G
/
C
)
  
=
  
10
40
=
0.25
P(G / C)\;=\;\frac{10}{40}=0.25
P
(
G
/
C
)
=
40
10
​
=
0.25
P
(
G
)
  
=
P
(
G
/
A
)
â‹…
P
(
A
)
+
P
(
G
/
B
)
â‹…
P
(
B
)
+
P
(
G
/
C
)
â‹…
P
(
C
)
P(G)\;=P(G / A)\cdot P(A)+P(G / B)\cdot P(B)+P(G / C)\cdot P(C)
P
(
G
)
=
P
(
G
/
A
)
â‹…
P
(
A
)
+
P
(
G
/
B
)
â‹…
P
(
B
)
+
P
(
G
/
C
)
â‹…
P
(
C
)
  
=
0.4
×
0.2
+
  
2
3
×
0.3
+
0.25
×
0.5
\;=0.4 \times 0.2+\;\frac{2}{3}\times 0.3+0.25 \times 0.5
=
0.4
×
0.2
+
3
2
​
×
0.3
+
0.25
×
0.5
  
=
0.08
+
0.2
+
0.125
\;=0.08+0.2+0.125
=
0.08
+
0.2
+
0.125
  
=
0.405
\;=0.405
=
0.405
Now,
P
(
G
∩
A
)
=
P
(
G
/
A
)
â‹…
P
(
A
)
P(G\cap A)=P(G / A)\cdot P(A)
P
(
G
∩
A
)
=
P
(
G
/
A
)
â‹…
P
(
A
)
=
0.4
×
0.2
=
0.08
=0.4 \times 0.2=0.08
=
0.4
×
0.2
=
0.08
Using Bayes' theorem,
  
P
(
A
/
G
)
=
  
P
(
A
∩
G
)
P
(
G
)
\;P(A / G)=\;\frac{P(A\cap G)}{P(G)}
P
(
A
/
G
)
=
P
(
G
)
P
(
A
∩
G
)
​
  
=
  
0.08
0.405
=
  
80
405
=
  
16
81
\;=\;\frac{0.08}{0.405}=\;\frac{80}{405}=\;\frac{16}{81}
=
0.405
0.08
​
=
405
80
​
=
81
16
​
So, the probability that a randomly selected girl belongs to section
A
A
A
is
  
16
81
\;\frac{16}{81}
81
16
​
.
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