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NDA 2013
Exam
NDA
Year
2013
Subject
Mathematics
Chapter
Probability
NDA 2013 · Mathematics · Probability
If A and B are independent events such that P(A)=(1)/(5), P(A...
If A and B are independent events such that
P
(
A
)
=
1
5
,
P
(
A
∪
B
)
=
7
10
\mathrm{P}(\mathrm{A})=\frac{1}{5}, \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{7}{10}
P
(
A
)
=
5
1
,
P
(
A
∪
B
)
=
10
7
, then what is
P
(
B
‾
)
\mathrm{P}(\overline{\mathrm{B}})
P
(
B
)
equal to?
Options
A
2
7
\frac{2}{7}
7
2
B
3
7
\frac{3}{7}
7
3
C
3
8
\frac{3}{8}
8
3
D
7
9
\frac{7}{9}
9
7
Correct answer
C.
3
8
\frac{3}{8}
8
3
Step-by-step solution
As A and B are independent events, so
P
(
A
∩
B
)
=
P
(
A
)
.
P
(
B
)
\mathrm{P}(\mathrm{ A} \cap \mathrm{ B})=\mathrm{P}(\mathrm{ A}) . \mathrm{P}(\mathrm{ B})
P
(
A
∩
B
)
=
P
(
A
)
.
P
(
B
)
Now, from
P
(
A
∪
B
)
=
P
(
A
)
+
P
(
B
)
−
P
(
A
∩
B
)
7
10
=
1
5
+
P
(
B
)
−
1
5
⋅
P
(
B
)
\begin{aligned} \mathrm{P}(\mathrm{ A} \cup \mathrm{ B}) & =\mathrm{P}(\mathrm{ A})+\mathrm{P}(\mathrm{ B})-\mathrm{P}(\mathrm{ A} \cap \mathrm{ B}) \\ \frac{7}{10} & =\frac{1}{5}+\mathrm{P}(\mathrm{ B})-\frac{1}{5} \cdot \mathrm{P}(\mathrm{ B}) \end{aligned}
P
(
A
∪
B
)
10
7
=
P
(
A
)
+
P
(
B
)
−
P
(
A
∩
B
)
=
5
1
+
P
(
B
)
−
5
1
⋅
P
(
B
)
or,
1
2
=
4
5
P
(
B
)
⇒
P
(
B
)
=
5
8
P
(
B
‾
)
=
1
−
P
(
B
)
=
3
8
\frac{1}{2}=\frac{4}{5} \mathrm{P}(\mathrm{B})\Rightarrow \mathrm{P}(\mathrm{B})=\frac{5}{8}\mathrm{P}(\overline{\mathrm{B}})=1-\mathrm{P}(\mathrm{B})=\frac{3}{8}
2
1
=
5
4
P
(
B
)
⇒
P
(
B
)
=
8
5
P
(
B
)
=
1
−
P
(
B
)
=
8
3
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