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Section:
Mathematics
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© examsnet.com
Question : 89 of 92
Marks:
+1
,
-0
A student writes an examination which contains eight true or false questions. If he answers six or more questions correctly, he passes the examination. If the student answers all the questions, then the probability that he fails in the examination is
[AP EAPCET 18 May 2024 Shift 1]
37
256
\;\; \frac{37}{256}
256
37
19
256
\;\; \frac{19}{256}
256
19
119
256
\;\; \frac{119}{256}
256
119
219
256
\;\; \frac{219}{256}
256
219
Validate
Solution:
Here,
x
=
8
,
P
=
1
2
x=8, P=\;\frac{1}{2}
x
=
8
,
P
=
2
1
and
q
=
1
2
q=\;\frac{1}{2}
q
=
2
1
frobabllity that he will pass
=
F
X
=
θ
+
P
(
X
=
7
)
+
P
(
X
=
8
\; =F X=\theta+P(X=7)+P(X=8
=
FX
=
θ
+
P
(
X
=
7
)
+
P
(
X
=
8
=
8
C
4
(
1
2
)
6
(
1
2
)
2
+
8
C
7
(
1
2
)
3
(
1
2
)
+
8
C
3
(
1
2
)
2
\; ={}^8 C_4 \left(\;\frac{1}{2}\right)^6 \left(\;\frac{1}{2}\right)^2+{}^8 C_7 \left(\;\frac{1}{2}\right)^3 \left(\;\frac{1}{2}\right)+{}^8 C_3 \left(\;\frac{1}{2}\right)^2
=
8
C
4
(
2
1
)
6
(
2
1
)
2
+
8
C
7
(
2
1
)
3
(
2
1
)
+
8
C
3
(
2
1
)
2
=
(
1
2
)
2
P
3
C
2
+
3
C
1
+
8
C
4
1
=
37
256
\; = \left(\;\frac{1}{2}\right)^2 P^3 C_2+{}^3 C_1+{}^8 C_4 1=\;\frac{37}{256}
=
(
2
1
)
2
P
3
C
2
+
3
C
1
+
8
C
4
1
=
256
37
Probability that he will fail is the cxamination
=
1
−
37
256
=
219
256
=1-\;\frac{37}{256}=\;\frac{219}{256}
=
1
−
256
37
=
256
219
© examsnet.com
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