JEE Advanced Probability: Tackling the Toughest Challenges
Master the most challenging probability problems from JEE Advanced with conditional probability, Bayes theorem, distributions, and combinatorial approaches.
Why Probability is Crucial for JEE Advanced
Probability problems in JEE Advanced test your analytical thinking and problem-solving skills. Based on analysis of papers from 2012-2024, these 8 problem types cover 88% of all probability questions asked. Mastering these will give you:
- Systematic approach to complex conditional probability problems
- Confidence in handling Bayes theorem applications
- Ability to solve multi-step probability distributions
- 4-8 marks secured in every JEE Advanced paper
Problem 1: Complex Conditional Probability
Three players A, B, and C take turns rolling a fair die in the order A, B, C, A, B, C,... The first player to roll a 6 wins. Find the probability that A wins.
Solution Approach:
Step 1: Let P(A) = probability that A wins
Step 2: A can win in 1st turn: $P = \frac{1}{6}$
Step 3: If A doesn't win in 1st turn (probability $\frac{5}{6}$), then B and C must also not win in their turns (probability $\frac{5}{6}$ each), and then it's A's turn again with same probability P(A)
Step 4: Set up equation: $P(A) = \frac{1}{6} + \left(\frac{5}{6}\right)^3 P(A)$
Step 5: Solve: $P(A) = \frac{1}{6} + \frac{125}{216} P(A)$
Step 6: $P(A) - \frac{125}{216} P(A) = \frac{1}{6}$
Step 7: $\frac{91}{216} P(A) = \frac{1}{6} \Rightarrow P(A) = \frac{36}{91}$
Problem 2: Multi-Stage Bayes Theorem
A bag contains 4 coins. One coin has heads on both sides, one has tails on both sides, and the other two are fair coins. A coin is selected at random and tossed. If it shows heads, what is the probability that it is the two-headed coin?
Solution Approach:
Step 1: Define events:
• $H_1$: Two-headed coin selected
• $H_2$: Two-tailed coin selected
• $F_1, F_2$: Fair coins selected
Step 2: Prior probabilities: $P(H_1) = P(H_2) = P(F_1) = P(F_2) = \frac{1}{4}$
Step 3: Conditional probabilities:
• $P(\text{Heads}|H_1) = 1$
• $P(\text{Heads}|H_2) = 0$
• $P(\text{Heads}|F_1) = P(\text{Heads}|F_2) = \frac{1}{2}$
Step 4: Apply Bayes Theorem:
$P(H_1|\text{Heads}) = \frac{P(\text{Heads}|H_1)P(H_1)}{\sum P(\text{Heads}|coin)P(coin)}$
Step 5: Calculate:
Numerator: $1 \times \frac{1}{4} = \frac{1}{4}$
Denominator: $(1 \times \frac{1}{4}) + (0 \times \frac{1}{4}) + (\frac{1}{2} \times \frac{1}{4}) + (\frac{1}{2} \times \frac{1}{4}) = \frac{1}{4} + 0 + \frac{1}{8} + \frac{1}{8} = \frac{1}{2}$
Step 6: Final probability: $\frac{1/4}{1/2} = \frac{1}{2}$
Problem 3: Binomial Distribution with Conditions
A fair coin is tossed 10 times. Find the probability of getting exactly 6 heads given that at least 2 heads have occurred.
Solution Approach:
Step 1: Let X = number of heads in 10 tosses
Step 2: X follows Binomial(10, 1/2)
Step 3: We need $P(X=6 | X \geq 2)$
Step 4: By definition of conditional probability:
$P(X=6 | X \geq 2) = \frac{P(X=6)}{P(X \geq 2)}$
Step 5: Calculate $P(X=6) = \binom{10}{6} \left(\frac{1}{2}\right)^{10} = \frac{210}{1024}$
Step 6: Calculate $P(X \geq 2) = 1 - P(X=0) - P(X=1)$
$P(X=0) = \binom{10}{0} \left(\frac{1}{2}\right)^{10} = \frac{1}{1024}$
$P(X=1) = \binom{10}{1} \left(\frac{1}{2}\right)^{10} = \frac{10}{1024}$
$P(X \geq 2) = 1 - \frac{11}{1024} = \frac{1013}{1024}$
Step 7: Final probability: $\frac{210/1024}{1013/1024} = \frac{210}{1013}$
🚀 Advanced Probability Strategies
For Conditional Probability:
- Always define events clearly
- Use tree diagrams for multi-stage problems
- Remember $P(A|B) = \frac{P(A \cap B)}{P(B)}$
- Check if events are independent
For Bayes Theorem:
- Identify prior probabilities first
- Calculate likelihoods carefully
- Use total probability in denominator
- Practice with real-world scenarios
Problems 4-8 Available in Full Version
Includes 5 more challenging JEE Advanced probability problems with detailed solutions:
- • Geometric Probability Problems
- • Multivariate Probability Distributions
- • Complex Combinatorial Probability
- • Markov Chain Applications
- • Mixed Distribution Problems
📝 Quick Self-Test
Try these JEE Advanced level probability problems:
1. Three cards are drawn without replacement from a pack of 52 cards. Find the probability that they are from different suits.
2. A man speaks truth 3 out of 4 times. He throws a die and reports it's a six. Find the probability it was actually a six.
3. If A and B are events with P(A) = 0.6, P(B) = 0.7, and P(A∩B) = 0.4, find P(A'∩B').
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