CAT 2025 Lesson : Probability - Conditional Probability

A bag contains $4$ red balls and $6$ black balls. John lies $30 \%$ of the times. John picked a ball from the bag and said its colour is red. What is the probability that the colour of the ball is red?

Solution

Let the event of the ball being red be R and the event of John calling it red be JR. We need to find P(R / JR)

Probability of a drawn ball being red is $0.4$ and black is $0.6$.
Probability of John lying is $0.3$ and speaking the truth is $0.7$.

Probability that John would call it red when
- the drawn ball is red and he speaks the truth $= 0.4 \times 0.7 = 0.28$ -----(1)
- the drawn ball is black and he lies $= 0.6 \times 0.3 = 0.18$

P(JR) $= 0.28 + 0.18 = 0.46$

From ($1$), P(R $\cap$ JR) $=$ Probability of drawn ball is red and John Calls it red $= 0.28$

P(R / JR) $= \dfrac{ \text{P} ( \text{R} \cap \text{JR})}{ \text{P}( \text{JR})} = \dfrac{0.28}{0.46} = \dfrac{14}{23}$

Answer: $\dfrac{14}{23}$

In a group of $20$ students, $8$ are girls. $5$ girls and $8$ boys study biology. If $2$ students are randomly selected and both are girls, what is the probability of them studying biology as well?

Solution

Let A denote selecting two students who study biology and B denote selecting two students who are girls.

As there are a total of $8$ girls, $n$(B) $=$ $^{8}C_{2} = 28$
As there are $5$ girls who study biology, $n$(A $\cap$ B) $=$ $^{5}C_{2} = 10$

P(A / B) $= \dfrac{n(\text{A} \cap \text{B})}{n(\text{B})} = \dfrac{10}{28} = \dfrac{5}{14}$

Answer: $\dfrac{5}{14}$

John is a quality inspector. He decides to select one of $4$ consignments – A, B, C and D – and conduct a test on one item in the consignment. The defective items in bags A, B, C and D are $4 \%, 5 \%, 8 \%$ and $10 \%$ respectively. The chances of John selecting each of the consignment A, B, C and D are $40 \%, 30 \%, 20 \%$ and $10 \%$ respectively. John selects a consignment, tests an item and finds it to be defective. What is the probability that the defective item was from consignment A?

Solution

We need to find the probability that the first consignment was selected given that the item tested is defective.

Let the F be the event where the item tested is defective. Let A be the event where the first bag is chosen.

P(D) = P(Defective from $1^{\mathrm{st}}$) + P(Defective from 2$^{\mathrm{nd}}$) + P(Defective from $3^{\mathrm{rd}}$) +
P(Defective from $4^{\mathrm{th}}$)
P(D) $= (0.4 \times 0.04) + (0.3 \times 0.05) + (0.2 \times 0.08) + (0.1 \times 0.1)$
$= 0.016 + 0.015 + 0.016 + 0.010 = \bm{0.057}$

P(A $\cap$ F) $=$ Probability of the defective item being chosen from Bag A
$= 0.4 \times 0.04 = \bm{0.16}$

P(A / F) $= \dfrac{\text{P}(\text{A} \cap \text{F})}{\text{P}(\text{F})} = \dfrac{0.16}{0.57} = \dfrac{16}{57}$

Answer: $\dfrac{16}{57}$