NIMCET 2027 Probability Previous Year Questions (PYQs) – Page 1 of 9

Probability — No Black Ball is Chosen

Given:

• Yellow balls = 5
• Black balls = 4
• Green balls = 3
• Total balls = 5 + 4 + 3 = 12

We are to find:

Probability that no black ball is selected when 3 balls are drawn at random.

Step 1: Total number of ways to choose any 3 balls from 12:

\[ \text{Total ways} = \binom{12}{3} = 220 \]

Step 2: Ways to choose 3 balls such that no black ball is chosen:

Only yellow and green balls are allowed ⇒ Total = 5 (yellow) + 3 (green) = 8
\[ \text{Favorable ways} = \binom{8}{3} = 56 \]

Step 3: Probability

\[ P(\text{no black ball}) = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{56}{220} = \frac{14}{55} \]

\[ \boxed{\text{Probability} = \frac{14}{55}} \]

Conditional Probability – Bayes' Theorem

Given: One ball is transferred from Bag I to Bag II, and then a ball is drawn from Bag II and is black.

Goal: Find the probability that the transferred ball was red, given that a black ball was drawn.

Using Bayes' theorem: \[ P(R|A) = \frac{P(R \cap A)}{P(A)} = \frac{\frac{3}{10} \cdot \frac{5}{10}}{\frac{3}{10} \cdot \frac{5}{10} + \frac{4}{10} \cdot \frac{6}{10} + \frac{3}{10} \cdot \frac{5}{10}} = \frac{15}{54} = \boxed{\frac{5}{18}} \]

✅ Final Answer: \( \boxed{\frac{5}{18}} \)

Given:

P(dies before 90) \(= \dfrac{1}{3}\)   P(survives till 90) \(= \dfrac{2}{3}\)

Key Idea — Symmetry:

If \(k\) persons die before 90, each is equally likely to be first.

\[P(A_1 \text{ is first to die}) = \sum_{k=1}^{4} P(\text{exactly } k \text{ persons die}) \times \frac{1}{k}\]

Case k = 1:

\[P = \binom{3}{0}\left(\frac{1}{3}\right)^1 \left(\frac{2}{3}\right)^3 \times \frac{1}{1} = \frac{8}{81}\]

Case k = 2:

\[P = \binom{3}{1}\left(\frac{1}{3}\right)^2 \left(\frac{2}{3}\right)^2 \times \frac{1}{2} = \frac{2}{27}\]

Case k = 3:

\[P = \binom{3}{2}\left(\frac{1}{3}\right)^3 \left(\frac{2}{3}\right)^1 \times \frac{1}{3} = \frac{2}{81}\]

Case k = 4:

\[P = \binom{3}{3}\left(\frac{1}{3}\right)^4 \left(\frac{2}{3}\right)^0 \times \frac{1}{4} = \frac{1}{324}\]

Final Answer:

\[P = \frac{32}{324} + \frac{24}{324} + \frac{8}{324} + \frac{1}{324}\]

\[\boxed{P = \frac{65}{324}}\]