NIMCET Probability Previous Year Questions (PYQs) – Page 1 of 9
Study Stuff for MCA Examinations
Solution
Solution
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
\[ P(\text{no black ball}) = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{56}{220} = \frac{14}{55} \]
\[ \boxed{\text{Probability} = \frac{14}{55}} \]
Solution
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: \( \boxed{\frac{5}{18}} \)
Solution
Solution
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:
\[P = \frac{32}{324} + \frac{24}{324} + \frac{8}{324} + \frac{1}{324}\]
\[\boxed{P = \frac{65}{324}}\]
Solution
Solution
Solution
Solution
Solution
Given: \(P(T_1)=0.2,\ P(T_2)=0.8,\ P(D)=0.07,\) and \(P(D\mid T_1)=10\,P(D\mid T_2)\).
Let \(p_2=P(D\mid T_2)\). Then \(P(D\mid T_1)=10p_2\). Using total probability: \[ 0.07=P(D)=0.2(10p_2)+0.8(p_2)=(2+0.8)p_2=2.8p_2 \Rightarrow p_2=\frac{0.07}{2.8}=0.025. \] Hence \(P(D\mid T_1)=0.25\).
We need: \(P(T_2\mid \overline D)=\dfrac{P(T_2)\,P(\overline D\mid T_2)}{P(\overline D)}\), where \(P(\overline D)=1-0.07=0.93\) and \(P(\overline D\mid T_2)=1-0.025=0.975\).
\[ P(T_2\mid \overline D)=\frac{0.8\times 0.975}{0.93} =\frac{0.78}{0.93} =\frac{26}{31}\approx 0.8387. \]
Answer: \(\boxed{\dfrac{26}{31}}\).
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