[tex] \\ \\ \text{if \: \( \tt P(A)=0.5, P(B)=0.6 \) and \: \( \tt P(A \cup B)=0.8 \) \: \: then \: \: find \: \: P(A|B) \: and \: P(B|A)}.[/tex]
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[tex]\large\underline{\sf{Solution-}}[/tex]
Given that,
[tex]\rm \: P(A) = 0.5 \\ [/tex]
[tex]\rm \: P(B) = 0.6 \\ [/tex]
[tex]\rm \: P(A\cup B) = 0.8 \\ [/tex]
We know,
[tex]\rm \: P(A\cup B) = P(A) + P(B) - P(A\cap B) \\ [/tex]
[tex]\rm \: 0.8 = 0.5 + 0.6 - P(A\cap B) \\ [/tex]
[tex]\rm \: P(A\cap B) = 1.1 - 0.8 \\ [/tex]
[tex]\rm\implies \:\boxed{ \rm{ \:P(A\cap B) = 0.3 \: \: }} \\ [/tex]
Now, Consider
[tex]\rm \: P(A|B) \\ [/tex]
[tex]\rm \: = \: \dfrac{P(A\cap B)}{P(B)} \\ [/tex]
[tex]\rm \: = \: \dfrac{0.3}{0.6} \\ [/tex]
[tex]\rm \: = \: 0.5 \\ [/tex]
[tex]\rm\implies \:\boxed{ \rm{ \:P(A|B) = 0.5 \: \: }} \\ [/tex]
Now, Consider
[tex]\rm \: P(B|A) \\ [/tex]
[tex]\rm \: = \: \dfrac{P(A\cap B)}{P(A)} \\ [/tex]
[tex]\rm \: = \: \dfrac{0.3}{0.5} \\ [/tex]
[tex]\rm \: = \: 0.6 \\ [/tex]
[tex]\rm\implies \:\boxed{ \rm{ \:P(B|A) = 0.6 \: \: }} \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information :-
1. If A and B are independent events iff
[tex]\boxed{ \rm{ \:P(A\cap B) = P(A) \times P(B) \: \: }} \\ [/tex]
2. If A and B are independent events then
[tex]\boxed{ \rm{ \:P(A\cap B') = P(A) \times P(B') \: \: }} \\ [/tex]
[tex]\boxed{ \rm{ \:P(A'\cap B') = P(A') \times P(B') \: \: }} \\ [/tex]
Step-by-step explanation:
\large\underline{\sf{Solution-}}
Solution−
Given that,
\begin{gathered}\rm \: P(A) = 0.5 \\ \end{gathered}
P(A)=0.5
\begin{gathered}\rm \: P(B) = 0.6 \\ \end{gathered}
P(B)=0.6
\begin{gathered}\rm \: P(A\cup B) = 0.8 \\ \end{gathered}
P(A∪B)=0.8
We know,
\begin{gathered}\rm \: P(A\cup B) = P(A) + P(B) - P(A\cap B) \\ \end{gathered}
P(A∪B)=P(A)+P(B)−P(A∩B)
\begin{gathered}\rm \: 0.8 = 0.5 + 0.6 - P(A\cap B) \\ \end{gathered}
0.8=0.5+0.6−P(A∩B)
\begin{gathered}\rm \: P(A\cap B) = 1.1 - 0.8 \\ \end{gathered}
P(A∩B)=1.1−0.8
\begin{gathered}\rm\implies \:\boxed{ \rm{ \:P(A\cap B) = 0.3 \: \: }} \\ \end{gathered}
⟹
P(A∩B)=0.3
Now, Consider
\begin{gathered}\rm \: P(A|B) \\ \end{gathered}
P(A∣B)
\begin{gathered}\rm \: = \: \dfrac{P(A\cap B)}{P(B)} \\ \end{gathered}
=
P(B)
P(A∩B)
\begin{gathered}\rm \: = \: \dfrac{0.3}{0.6} \\ \end{gathered}
=
0.6
0.3
\begin{gathered}\rm \: = \: 0.5 \\ \end{gathered}
=0.5
\begin{gathered}\rm\implies \:\boxed{ \rm{ \:P(A|B) = 0.5 \: \: }} \\ \end{gathered}
⟹
P(A∣B)=0.5
Now, Consider
\begin{gathered}\rm \: P(B|A) \\ \end{gathered}
P(B∣A)
\begin{gathered}\rm \: = \: \dfrac{P(A\cap B)}{P(A)} \\ \end{gathered}
=
P(A)
P(A∩B)
\begin{gathered}\rm \: = \: \dfrac{0.3}{0.5} \\ \end{gathered}
=
0.5
0.3
\begin{gathered}\rm \: = \: 0.6 \\ \end{gathered}
=0.6
\begin{gathered}\rm\implies \:\boxed{ \rm{ \:P(B|A) = 0.6 \: \: }} \\ \end{gathered}
⟹
P(B∣A)=0.6
\rule{190pt}{2pt}
Additional Information :-
1. If A and B are independent events iff
\begin{gathered}\boxed{ \rm{ \:P(A\cap B) = P(A) \times P(B) \: \: }} \\ \end{gathered}
P(A∩B)=P(A)×P(B)
2. If A and B are independent events then
\begin{gathered}\boxed{ \rm{ \:P(A\cap B') = P(A) \times P(B') \: \: }} \\ \end{gathered}
P(A∩B
′
)=P(A)×P(B
′
)
\begin{gathered}\boxed{ \rm{ \:P(A'\cap B') = P(A') \times P(B') \: \: }} \\ \end{gathered}
P(A
′
∩B
′
)=P(A
′
)×P(B
′
)
[tex]answer}[/tex]