GO Classes Test Series 2024 | Mock GATE | Test 14 | Question: 21

Question

Suppose we have events $A, B$ in a sample space. And we know that $\mathrm{P}(\mathrm{A})=0.3, \mathrm{P}\left(\mathrm{B} \mid \mathrm{A}^c\right)=0.25, \mathrm{P}(\mathrm{B} \mid \mathrm{A})=0.45$. 

What is $\mathrm{P}\left(\mathrm{A}^c \mid \mathrm{B}\right) ?$

• A. 0.75
• B. 0.55
• C. 0.2
• D. 0.56

Comments

1. Suppose the $P(A), P(A^c)$ are positive then for any event $B$, $P(B) = P(A)P(B|A) + P(A^C)P(B|A^C)$.
2. Suppose $P(B) > 0$ then for any event $A\implies P(A | B) + P(A^c |B) = 1 \implies P(A^c |B)  = 1 - P(A|B) $
3. For any event $A$,   $\mathbb{P}(A) = 1 -\mathbb{P}(A^c)$

$\begin{align}\text{In the Question we are interested in finding }P(A^c | B )  &=1- P(A | B)\\ &= 1 - \dfrac{P(A)P(B |A)}{P(B)}\end{align}$

From the given information, $P(B) = 0.3\times 0.45 + 0.7\times 0.25 = 0.31$

$\begin{align}\text{Then }P(A^c | B ) &= 1 - \dfrac{P(A)P(B |A)}{P(B)}\\&=1-\dfrac{0.3\times0.45}{0.31}\\&= 1-0.43548\\ &\approx 0.56 \end{align}$

 

Answer 1

I hope this finds helpful...

Answer 2

This question can be solved simply using the concepts of Baye's Theorem in conditional probability.

The following data is given: -

P(A) = 0.3, P(B | A^c) = 0.25 and P(B | A) = 0.45

From the given data, we can find out: -

P(A^c) = 1- P(A) = 1- 0.3 = 0.7

Also, P(B) = [P(A) x P(B | A)] + [P(A^c) x P(B | A^c)] = [0.3 x 0.45] + [0.7 x 0.25] = 0.135 + 0.175 = 0.31

Now, according to Bayes' Theorem: -

P(A^c | B) = [P(B | A^c) x P(A^c)] / P(B) = [0.7 x 0.25] / 0.31 = 0.175 / 0.31 = 0.564 ≈ 0.56

Therefore, the correct answer is Option (D).