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Joint Entrance Examination
Mathematics
Probability
Question 126
Question 126 of 417
Medium
NUMERICAL
25\% of the population are smokers. A smoker has 27 times more chance of developing lung cancer than a non-smoker. A person diagnosed with lung cancer --- the probability that this person is a smoker is
k
10
\dfrac{k}{10}
10
k
. Find
k
k
k
. Answer: 9
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Official solution
P
(
smoker
)
=
1
4
P(\text{smoker})=\dfrac14
P
(
smoker
)
=
4
1
,
P
(
non-smoker
)
=
3
4
P(\text{non-smoker})=\dfrac34
P
(
non-smoker
)
=
4
3
.
P
(
E
∣
smoker
)
=
27
28
P(E\mid\text{smoker})=\dfrac{27}{28}
P
(
E
∣
smoker
)
=
28
27
,
P
(
E
∣
non-smoker
)
=
1
28
P(E\mid\text{non-smoker})=\dfrac{1}{28}
P
(
E
∣
non-smoker
)
=
28
1
. \[P(\text{smoker}\mid E) = \frac{\frac14\cdot\frac{27}{28}}{\frac14\cdot\frac{27}{28}+\frac34\cdot\frac{1}{28}} = \frac{27}{30}=\frac{9}{10} \implies k=9\]
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