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Joint Entrance Examination
Mathematics
Probability
Question 197
Question 197 of 417
Medium
MCQ
Five machines; probability any one is out of service on a day is
1
4
\dfrac14
4
1
. If
P
(
at most 2 out of service
)
=
(
3
4
)
3
k
P(\text{at most 2 out of service})=\left(\dfrac34\right)^3 k
P
(
at most 2 out of service
)
=
(
4
3
)
3
k
, find
k
k
k
.
A
17
/
8
17/8
17/8
B
17
/
4
17/4
17/4
C
17
/
2
17/2
17/2
D
4
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Official solution
\[P(\le2) = {}^5C_0\left(\frac34\right)^5+{}^5C_1\left(\frac14\right)\left(\frac34\right)^4+{}^5C_2\left(\frac14\right)^2\left(\frac34\right)^3 = \frac{459}{512} = \left(\frac34\right)^3\times\frac{17}{8}\]
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