Question

$\\ \\ \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)}.$

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Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm \: P(A) = 0.5$

$\rm \: P(B) = 0.6$

$\rm \: P(A\cup B) = 0.8$

We know,

$\rm \: P(A\cup B) = P(A) + P(B) - P(A\cap B)$

$\rm \: 0.8 = 0.5 + 0.6 - P(A\cap B)$

$\rm \: P(A\cap B) = 1.1 - 0.8$

$\rm\implies \:\boxed{ \rm{ \:P(A\cap B) = 0.3 \: \: }}$

Now, Consider

$\rm \: P(A|B)$

$\rm \: = \: \dfrac{P(A\cap B)}{P(B)}$

$\rm \: = \: \dfrac{0.3}{0.6}$

$\rm \: = \: 0.5$

$\rm\implies \:\boxed{ \rm{ \:P(A|B) = 0.5 \: \: }}$

Now, Consider

$\rm \: P(B|A)$

$\rm \: = \: \dfrac{P(A\cap B)}{P(A)}$

$\rm \: = \: \dfrac{0.3}{0.5}$

$\rm \: = \: 0.6$

$\rm\implies \:\boxed{ \rm{ \:P(B|A) = 0.6 \: \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

1. If A and B are independent events iff

$\boxed{ \rm{ \:P(A\cap B) = P(A) \times P(B) \: \: }}$

2. If A and B are independent events then

$\boxed{ \rm{ \:P(A\cap B') = P(A) \times P(B') \: \: }}$

$\boxed{ \rm{ \:P(A'\cap B') = P(A') \times P(B') \: \: }}$

Answer 2

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

′

)

$answer}$