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TS EAMCET 10-Sep-2020 Shift 2 Solved Paper
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© examsnet.com
Question : 38
Total: 160
In an examination there are four Yes/No type of questions. The probability that the answer by the student to a question without guess to be correct is
2
3
. The probability that a student guesses a correct answer is
1
2
. A student writes the examination either by without guessing answers to all the 4 questions or by guessing answers to all 4questions. The probability that he attempt the exam by guessing answers to all questions is
3
7
. Given that a student answered at least 3 questions correctly, the probability that he answered all the questions without guessing is
13
15
405
1429
1024
1429
2
15
Validate
Solution:
Consider the events
E
1
=
Answer by student without guessing
E
2
=
Answer by student guessing
A
=
At least three question correctly
P
(
E
1
)
=
3
7
,
p
(
B
2
)
=
4
7
p
(
A
E
1
)
=
4
C
3
(
1
2
)
4
+
4
C
4
(
1
2
)
4
=
(
1
2
)
4
(
4
+
1
)
=
5
16
p
(
A
E
2
)
=
4
C
3
(
2
3
)
3
×
1
3
+
4
C
4
(
2
3
)
4
=
(
2
3
)
3
(
4
3
+
2
3
)
p
(
A
E
2
)
=
16
27
Required probability
=
p
(
E
2
A
)
=
p
(
E
2
)
p
(
A
E
2
)
p
(
E
1
)
p
(
A
E
1
)
+
p
(
E
2
)
p
(
A
E
2
)
=
4
7
×
64
27
3
7
×
5
16
+
4
7
×
64
27
=
1024
1429
© examsnet.com
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