There were 3 sections namely A, B and C
in a test. Out of three sections, 33
students passed in Section A, 34
students passed in Section B and 32
passed in Section (C) 10 students passed
in Section A and Section B, 9 passed in
Section B and Section C, 8 passed in
Section A and Section (C) The number of
students who passed each section alone
was equal and was 21 for each section.
39. How many passed all the three
sections?
(A) 3
(B) 6
(C) 5
(D) 7
1. B
2. D
3. C
4. A
Answers
Answer:
Let the events A,B,C are the sections in which students pass.
⇒P(A)=p,P(B)=q,P(C)=
2
1
The student will be successful if he able to pass the section A and either of section B or Section C=(Probability of passing in A)\times (Probability of selecting either of section B or C \times (Probability of passing B +Probability of passing C))=P(A)×(
2
1
×(P(B)+P(C))=p×(
2
1
×(q+
2
1
))=
2
pq
+
4
p
But Probability that student is successful in the exam =
2
1
⇒
2
1
=
2
pq
+
4
p
⇒2pq+p=2
From the given option only Option 4 i.e.,p=1,q=
2
1
satisfies the equation.
Hence, options 1,2,3 are wrong.
Answered By
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6 students passed all three sections of the test (Option B).
Given,
In a test, out of three sections, 33 students passed in Section A, 34 students passed in Section B and 32 passed in Section C. 10 students passed in Section A and Section B, 9 passed in Section B and Section C, 8 passed in Section A and Section C. The number of students who passed each section alone was equal and was 21 for each section.
To Find,
The number of students passed all three sections.
Solution,
The method of finding the number of students passed in all three sections is as follows -
Let, A, B, and C be the number of students passed in sections A, B, and C respectively.
Let, a, b, and c be the number of students only passed in sections A, B, and C respectively.
Let, d, e, and f be the number of students passed in both the sections A and C, A and B, B and C respectively.
Also, let g be the number of students passed in all three sections.
So, according to the question A = 33, B = 34, C = 32.
Also, e + g = 10, g + f = 9, d + g = 8, and a = b = c = 21.
Putting the known values in A = 33, we get, a + d + g + e = 33
⇒ 21 + 8 + e = 33
⇒ e = 4.
Now we will put the value of e in the equation, e + g = 10
⇒4 + g = 10
⇒ g = 6.
Hence, 6 students passed all three sections of the test.
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