Probability | CBSE - Wyatt's Notes

sources:

• text: Standard textbook reference

Probability quantifies the likelihood of events. This topic covers conditional probability, Bayes’ theorem, random variables, and probability distributions.

• Conditional probability: $P(A|B) = \frac{P(A \cap B)}{P(B)}$
• Multiplication rule: $P(A \cap B) = P(A|B) \cdot P(B)$
• Bayes’ theorem: $P(A_i|B) = \frac{P(B|A_i) \cdot P(A_i)}{\sum_j P(B|A_j) \cdot P(A_j)}$
• Total probability: $P(B) = \sum_i P(B|A_i) \cdot P(A_i)$
• Random variable: a function assigning numerical values to outcomes
• Mean (expected value): $E(X) = \sum x_i P(X = x_i)$
• Variance: $\text{Var}(X) = E(X^2) - [E(X)]^2$
• Binomial distribution: $P(X = r) = \binom{n}{r} p^r q^{n-r}$, mean $= np$, variance $= npq$

Problem: A bag contains 5 red and 3 blue balls. Two balls are drawn without replacement. Find the probability that both are red.

Solution:

$P(\text{both red}) = P(\text{1st red}) \times P(\text{2nd red}|\text{1st red})$

$= \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$

Common mistake: Treating the draws as independent when they are without replacement. The second draw depends on the first.

Problem: A factory has three machines. Machine A produces 30% of items with 5% defect rate, Machine B produces 45% with 3% defect rate, and Machine C produces 25% with 4% defect rate. An item is selected at random and found to be defective. Find the probability it came from Machine A.

Solution:

Let $D$ = defective, $A$, $B$, $C$ = machines.

$P(D) = P(D|A)P(A) + P(D|B)P(B) + P(D|C)P(C)$

$= 0.05 \times 0.30 + 0.03 \times 0.45 + 0.04 \times 0.25$

$= 0.015 + 0.0135 + 0.01 = 0.0385$

By Bayes’ theorem: $P(A|D) = \frac{P(D|A)P(A)}{P(D)} = \frac{0.015}{0.0385} = \frac{15}{38.5} \approx 0.3896$

Common mistake: Forgetting to use the total probability $P(D)$ in the denominator. Using only $P(D|A)P(A)$ gives the wrong answer.

Problem: A coin is tossed 5 times. Find the probability of getting exactly 3 heads.

Solution:

This is a binomial distribution with $n = 5$, $r = 3$, $p = 0.5$, $q = 0.5$:

$P(X = 3) = \binom{5}{3} (0.5)^3 (0.5)^2 = 10 \times 0.125 \times 0.25 = 0.3125$

Mean: $\mu = np = 5 \times 0.5 = 2.5$

Variance: $\sigma^2 = npq = 5 \times 0.5 \times 0.5 = 1.25$

Common mistake: Using $\binom{n}{r} = \frac{n!}{r!(n-r)!}$ incorrectly. $\binom{5}{3} = \frac{5!}{3!2!} = 10$.

1. Two cards are drawn from a deck. Find the probability that both are aces given that at least one is an ace.
2. A die is rolled 4 times. Find the probability of getting exactly 2 sixes.
3. 60% of patients recover from a disease. If 5 patients are selected, find the probability that at least 4 recover.

Probability is fundamental to statistics, data science, risk assessment, and decision-making. Bayes’ theorem underpins modern machine learning, medical diagnosis, and legal reasoning. Understanding probability distributions is essential for any quantitative field.

Quantifying uncertainty with numbers: Probability assigns a number between 0 (impossible) and 1 (certain) to how likely an event is. Conditional probability is like narrowing your focus — “given that it’s raining, what’s the chance my picnic is ruined?” uses different information than “what’s the chance it rains?” Bayes’ theorem is the reverse-engineering tool: it takes an observed effect (like a positive test result) and works backward to find the most likely cause (like whether you actually have the disease). The binomial distribution counts successes in a fixed number of independent trials — like flipping a coin 10 times and counting heads.

Why it matters: Probability is the language of uncertainty — it underpins statistics, risk assessment, medical diagnosis, weather forecasting, insurance, gambling odds, and machine learning. Bayes’ theorem is the foundation of modern AI and medical testing. Without probability, we couldn’t make rational decisions under uncertainty.

The key insight: Conditional probability P(A|B) ≠ P(B|A) — “probability of rain given clouds” is very different from “probability of clouds given rain.” This asymmetry is why Bayes’ theorem is so powerful and so commonly misunderstood.

• Bayes’ theorem problems in standard practice involve multiple stages or sources
• Conditional probability problems often involve “given that” statements
• Binomial distribution applies when there are independent trials with two outcomes
• Practice with tree diagrams for multi-stage probability problems
• Always check if events are independent before applying multiplication rule

• Conditional probability: $P(A|B) = \frac{P(A \cap B)}{P(B)}$, $P(B) > 0$
• Addition rule: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$
• Bayes’ theorem: $P(A_i|B) = \frac{P(B|A_i) P(A_i)}{\sum_j P(B|A_j) P(A_j)}$
• Binomial: $P(X = r) = \binom{n}{r} p^r (1-p)^{n-r}$
• Mean of binomial: $\mu = np$, Variance: $\sigma^2 = np(1-p)$
• Variance formula: $\text{Var}(X) = E(X^2) - [E(X)]^2$

Problem: Two cards are drawn from a standard deck of 52 cards without replacement. Given that the first card is an ace, find the probability that the second card is also an ace.

Solution:

Given the first card is an ace, there are 3 aces left out of 51 remaining cards.

$P(\text{2nd ace} | \text{1st ace}) = \frac{3}{51} = \frac{1}{17}$

This is a direct application of conditional probability. The sample space reduces from 52 to 51 cards after the first draw.

Common mistake: Calculating $\frac{4}{52} \times \frac{3}{51}$ instead of just $\frac{3}{51}$. Since we are given the first card is an ace, we don’t multiply by its probability.

Problem: A random variable $X$ has the probability distribution:

Find $E(X)$, $E(X^2)$, and $\text{Var}(X)$.

Solution:

$E(X) = 0(0.1) + 1(0.2) + 2(0.3) + 3(0.25) + 4(0.15) = 0 + 0.2 + 0.6 + 0.75 + 0.6 = 2.15$

$E(X^2) = 0^2(0.1) + 1^2(0.2) + 2^2(0.3) + 3^2(0.25) + 4^2(0.15) = 0 + 0.2 + 1.2 + 2.25 + 2.4 = 6.05$

$\text{Var}(X) = E(X^2) - [E(X)]^2 = 6.05 - (2.15)^2 = 6.05 - 4.6225 = 1.4275$

Common mistake: Forgetting to square $E(X)$ in the variance formula. $\text{Var}(X) = E(X^2) - [E(X)]^2$, not $E(X^2) - E(X)$.

Problem: A survey shows that 60% of students prefer online learning. If 8 students are selected at random, find the probability that exactly 5 prefer online learning. Also find the mean and standard deviation.

Solution:

This is a binomial distribution with $n = 8$, $p = 0.6$, $q = 0.4$.

$P(X = 5) = \binom{8}{5} (0.6)^5 (0.4)^3 = 56 \times 0.07776 \times 0.064 = 0.2787$

Mean: $\mu = np = 8 \times 0.6 = 4.8$

Variance: $\sigma^2 = npq = 8 \times 0.6 \times 0.4 = 1.92$

Standard deviation: $\sigma = \sqrt{1.92} \approx 1.386$

Common mistake: Confusing standard deviation with variance. The standard deviation is the square root of the variance.

1. For Bayes’ theorem problems, always draw a tree diagram to visualize the stages
2. Conditional probability problems: identify what is “given” and reduce the sample space accordingly
3. Binomial distribution requires: fixed number of trials, two outcomes, constant probability, independent trials
4. When calculating variance, always use $\text{Var}(X) = E(X^2) - [E(X)]^2$ (not $E(X^2) - E(X)$)
5. Practice problems involving “at least one” — use the complement: $P(\text{at least one}) = 1 - P(\text{none})$

Confusing conditional probability P(A|B) with joint probability P(A and B). P(A|B) = P(A and B) / P(B). Students often use P(A) * P(B) instead, which only works for independent events. Conditional probability requires dividing by the condition’s probability.

Forgetting that Bayes’ theorem requires the total probability in the denominator. The formula is P(Ai|B) = P(B|Ai) P(Ai) / sum(P(B|Aj) P(Aj)). Students sometimes omit the summation, using only one term in the denominator. The denominator must account for all possible ways B can occur.

Confusing the mean and variance formulas for binomial distribution. For binomial(n, p), the mean is np and the variance is npq where q = 1 - p. Students often write variance as np instead of npq, forgetting to multiply by the probability of failure. The variance is always less than the mean for binomial distributions.

• Statistics — Probability distributions provide the theoretical foundation for statistical inference and hypothesis testing.
• Matrices — Transition matrices in Markov chains use matrix multiplication to compute state probabilities.
• Calculus — Continuous probability distributions require integration to find probabilities over intervals.