Probability Distribution — MHT-CET Maths

Formulas (5)

• Classical Probability — Favourable over Total · Classical probability of an event
  $$P(E) = \dfrac{n(E)}{n(S)} = \dfrac{\text{favourable outcomes}}{\text{total equally-likely outcomes}},\qquad P(E') = 1 - P(E)$$
• Counting Probabilities with Combinations and Arrangements · Combination count and word-arrangement count
  $${}^nC_r = \dfrac{n!}{r!\,(n-r)!},\qquad \text{(word, one letter } k \text{ times)}\ \#\text{arrangements} = \dfrac{n!}{k!}$$
• Mutually Exclusive and Exhaustive Events · Mutually exclusive and exhaustive events sum to 1
  $$A_1, A_2, \ldots, A_n \text{ mut. excl. \& exhaustive} \;\Longrightarrow\; \sum_{i=1}^{n} P(A_i) = 1$$
• The Addition Theorem — P(A∪B), Exactly One, and Complements · Addition theorem and its derived identities
  $$P(A\cup B) = P(A) + P(B) - P(A\cap B),\qquad P(\text{exactly one}) = P(A\cup B) - P(A\cap B)$$
• Odds in Favour and Odds Against a Probability · Odds and probability
  $$\text{odds in favour } m:n \;\Longleftrightarrow\; P(E) = \dfrac{m}{m+n},\qquad \text{odds against} = \dfrac{P(E')}{P(E)} = \dfrac{1-P(E)}{P(E)}$$

Watch out for (15)

• Probability needs EQUALLY-likely outcomes
  → Classical Probability — Favourable over Total
• A probability can never exceed 1 or go below 0
  → Classical Probability — Favourable over Total
• Use combinations when the draw order does NOT matter
  → Counting Probabilities with Combinations and Arrangements
• Multiply combinations for 'one of each category'
  → Counting Probabilities with Combinations and Arrangements
• 'With replacement' means $k^m$ ordered outcomes
  → Counting Probabilities with Combinations and Arrangements
• 'Not together' = 1 − 'together' (glue the alike letters)
  → Counting Probabilities with Combinations and Arrangements
• Mutually exclusive is NOT the same as independent
  → Mutually Exclusive and Exhaustive Events
• Only add all the pieces to 1 when the events are BOTH exclusive AND exhaustive
  → Mutually Exclusive and Exhaustive Events
• ADD the intersection back, don't subtract, to get P(A)+P(B)
  → The Addition Theorem — P(A∪B), Exactly One, and Complements
• 'Exactly one' is the union MINUS the intersection
  → The Addition Theorem — P(A∪B), Exactly One, and Complements
• On a distribution table, an event's probability is a SUM of rows
  → The Addition Theorem — P(A∪B), Exactly One, and Complements
• $P(A')+P(B') = 2 - [P(A)+P(B)]$, not $1-[P(A)+P(B)]$
  → The Addition Theorem — P(A∪B), Exactly One, and Complements
• Odds compare favourable to UNfavourable, not to the total
  → Odds in Favour and Odds Against a Probability
• 'Odds against' puts the unfavourable term first
  → Odds in Favour and Odds Against a Probability
• 'Must and only one can happen' = mutually exclusive and exhaustive
  → Odds in Favour and Odds Against a Probability

Formulas (7)

• Conditional Probability — Restricting the Sample Space · Definition of conditional probability
  $$P(A\mid B) = \dfrac{P(A\cap B)}{P(B)},\qquad P(B) > 0$$
• Multiplication Rule and Sequential Draws Without Replacement · Chain rule for a sequence of dependent draws
  $$P(E_1\cap E_2\cap E_3) = P(E_1)\,P(E_2\mid E_1)\,P(E_3\mid E_1\cap E_2)$$
• Computing P(A|B) by Restriction — Distributions, Counting and Composite Events · Restriction form for composite conditioning
  $$P(A\mid B) = \dfrac{P(A\cap B)}{P(B)} = \dfrac{n(A\cap B)}{n(B)}\ \text{(equally likely)}$$
• Independence and Event Algebra with Unions · Independence and the union it produces
  $$P(A\cap B) = P(A)\,P(B),\qquad P(A\cup B) = P(A)+P(B)-P(A)\,P(B)$$
• At Least One and Exactly One for Independent Trials · At-least-one and exactly-one
  $$P(\text{at least one}) = 1 - \prod_{i}(1-p_i),\qquad P(\text{exactly one of }A,B) = P(A)P(B') + P(A')P(B)$$
• Total Probability Theorem · Total probability theorem
  $$P(E) = \sum_{i=1}^{n} P(H_i)\,P(E\mid H_i)$$
• Bayes' Theorem — Reversing the Conditioning · Bayes' theorem (posterior from priors and likelihoods)
  $$P(H_k\mid E) = \dfrac{P(H_k)\,P(E\mid H_k)}{\sum_{i} P(H_i)\,P(E\mid H_i)}$$

Watch out for (20)

• P(A|B) and P(B|A) are not the same number
  → Conditional Probability — Restricting the Sample Space
• Divide by the GIVEN event's probability, not by 1
  → Conditional Probability — Restricting the Sample Space
• Without replacement: shrink BOTH the numerator and the denominator
  → Multiplication Rule and Sequential Draws Without Replacement
• Add over all favourable orderings for a composition
  → Multiplication Rule and Sequential Draws Without Replacement
• 'Alternately O,E,O OR E,O,E' means add both patterns
  → Multiplication Rule and Sequential Draws Without Replacement
• The overlap A∩B is measured inside B, not over the whole space
  → Computing P(A|B) by Restriction — Distributions, Counting and Composite Events
• Compute 'at least one' as the complement
  → Computing P(A|B) by Restriction — Distributions, Counting and Composite Events
• Watch for a 'None of these' answer when your value is not listed
  → Computing P(A|B) by Restriction — Distributions, Counting and Composite Events
• P(A'|B) = P(A') needs INDEPENDENCE
  → Independence and Event Algebra with Unions
• The union formula loses its cross-term only when independent
  → Independence and Event Algebra with Unions
• Convert odds to probability before plugging in
  → Independence and Event Algebra with Unions
• Independent is not the same as mutually exclusive
  → Independence and Event Algebra with Unions
• 'At least one' is 1 − P(none), NOT the sum of individual probabilities
  → At Least One and Exactly One for Independent Trials
• Exactly one ≠ at least one
  → At Least One and Exactly One for Independent Trials
• Complement each event correctly inside a composite pattern
  → At Least One and Exactly One for Independent Trials
• The routes must partition the space — exclusive AND exhaustive
  → Total Probability Theorem
• In draw-then-add problems, update the bag before the conditional
  → Total Probability Theorem
• Numerator is ONE route; denominator is ALL routes
  → Bayes' Theorem — Reversing the Conditioning
• Do not swap priors and likelihoods
  → Bayes' Theorem — Reversing the Conditioning
• Equal priors cancel — reduce to a likelihood ratio
  → Bayes' Theorem — Reversing the Conditioning

Formulas (9)

• Discrete Random Variable and Its Probability Mass Function · The two pmf axioms
  $$0 \le P(X=x_i) \le 1, \qquad \sum_{i} P(X=x_i) = 1$$
• Finding the Constant k from a Linear Probability Table · Linear normalisation
  $$\sum_x P(X=x) = 1 \;\Longrightarrow\; (\text{multiple of } k) = 1 \;\Longrightarrow\; k = \frac{1}{\text{that multiple}}$$
• Reading a Range Probability from the pmf Table · Complement for a tail probability
  $$P(X \ge a) = 1 - P(X < a) = 1 - \sum_{x < a} P(X=x)$$
• Finding k from a Quadratic Probability Table · The two recurring MHT-CET quadratics
  $$6k^2+5k-1=(6k-1)(k+1),\qquad 10k^2+9k-1=(10k-1)(k+1)$$
• Finding k for an Exponential pmf on a Finite Range · Finite geometric normalisation
  $$\sum_{x=0}^{n} k\,r^{x} = k\cdot\frac{r^{\,n+1}-1}{r-1} = 1$$
• Finding k for an Infinite pmf k(x+1)rˣ · AGP normalisation for k(x+1)rˣ
  $$\sum_{x=0}^{\infty}(x+1)r^{x} = \frac{1}{(1-r)^{2}} \;\Longrightarrow\; k = (1-r)^{2}$$
• Constructing a Probability Distribution from an Experiment · Binomial and hypergeometric building blocks
  $$P(X=r)=\binom{n}{r}p^{r}(1-p)^{n-r}, \qquad P(X=r)=\frac{\binom{D}{r}\binom{N-D}{n-r}}{\binom{N}{n}}$$
• Cumulative Distribution Function and pmf ↔ CDF Differencing · CDF definition and differencing
  $$F(x)=P(X\le x)=\sum_{x_i\le x}P(X=x_i), \qquad P(X=x_i)=F(x_i)-F(x_{i-1})$$
• Continuous Random Variables — pdf, Normalisation, CDF and P(a < X < b) · Continuous normalisation, CDF, interval
  $$\int_{-\infty}^{\infty} f(x)\,dx = 1,\quad F(x)=\int_{-\infty}^{x} f(t)\,dt,\quad P(a<X<b)=\int_a^b f(x)\,dx$$

Watch out for (22)

• A pmf must sum to exactly 1, over ALL values
  → Discrete Random Variable and Its Probability Mass Function
• Probabilities can never exceed 1 or go negative
  → Discrete Random Variable and Its Probability Mass Function
• Include every row — even a fixed number — in ΣP = 1
  → Finding the Constant k from a Linear Probability Table
• Match the range operator exactly: strict vs inclusive
  → Finding the Constant k from a Linear Probability Table
• $P(X > a)$ and $P(X \ge a)$ are not the same
  → Reading a Range Probability from the pmf Table
• Use the complement only when it has fewer terms
  → Reading a Range Probability from the pmf Table
• Reject the negative root of the k-quadratic
  → Finding k from a Quadratic Probability Table
• Don't drop the $k^2$ terms when evaluating a range
  → Finding k from a Quadratic Probability Table
• $P(X\ge 2)=1-P(X=1)$, not $1-P(X\le 2)$
  → Finding k from a Quadratic Probability Table
• Sum a FINITE range fully — don't stop early
  → Finding k for an Exponential pmf on a Finite Range
• A finite exponential pmf is NOT the infinite geometric sum
  → Finding k for an Exponential pmf on a Finite Range
• $\sum(x+1)r^x = \dfrac{1}{(1-r)^2}$, NOT $\dfrac{1}{1-r}$
  → Finding k for an Infinite pmf k(x+1)rˣ
• The range is INFINITE here — use $|r|<1$
  → Finding k for an Infinite pmf k(x+1)rˣ
• With replacement is binomial; without replacement is hypergeometric
  → Constructing a Probability Distribution from an Experiment
• $P(X=1)$ counts BOTH orders — include the factor of 2
  → Constructing a Probability Distribution from an Experiment
• In a bounded 'until' experiment, the last cell POOLS two branches
  → Constructing a Probability Distribution from an Experiment
• Read $P(X=x_i)$ as a CDF DIFFERENCE, not the CDF value
  → Cumulative Distribution Function and pmf ↔ CDF Differencing
• $P(X\le 0)=F(0)$; $P(X>0)=1-F(0)$
  → Cumulative Distribution Function and pmf ↔ CDF Differencing
• Integrate over the SUPPORT only
  → Continuous Random Variables — pdf, Normalisation, CDF and P(a < X < b)
• A two-unknown pdf needs TWO equations
  → Continuous Random Variables — pdf, Normalisation, CDF and P(a < X < b)
• The CDF is the running integral, and $F'(x)=f(x)$
  → Continuous Random Variables — pdf, Normalisation, CDF and P(a < X < b)
• $P(|X|<2)$ is a symmetric integral $\int_{-2}^{2}$
  → Continuous Random Variables — pdf, Normalisation, CDF and P(a < X < b)

Formulas (7)

• Expectation as the Long-Run Average · Expected value of a discrete random variable
  $$E(X) = \mu = \sum_{i} x_i\,P(X=x_i)$$
• Computing the Mean E(X) from a Probability Distribution · Mean of a listed distribution
  $$E(X) = \sum_{i} x_i\,P(X=x_i),\qquad E(X+Y) = E(X)+E(Y)$$
• Variance and Standard Deviation: Var(X) = E(X²) − [E(X)]² · Variance and standard deviation
  $$\mathrm{Var}(X) = E(X^2) - [E(X)]^2,\qquad E(X^2) = \sum x^2 P(x),\qquad \sigma = \sqrt{\mathrm{Var}(X)}$$
• Expected Winnings of a Game: E(g(X)) = Σ g(x)·P(x) · Expected value of a payoff (function of X)
  $$E\big(g(X)\big) = \sum g(x)\,P(x)$$
• Uniform Distribution on 1 to n: E(X) = (n+1)/2, Var(X) = (n²−1)/12 · Discrete uniform on 1..n
  $$E(X) = \dfrac{n+1}{2},\qquad \mathrm{Var}(X) = \dfrac{n^2-1}{12},\qquad \dfrac{\mathrm{Var}(X)}{E(X)} = \dfrac{n-1}{6}$$
• Finding Unknown Probabilities from the Mean and ΣP = 1 · The determining system
  $$\sum P(x) = 1 \quad\text{and}\quad E(X) = \sum x\,P(x) = \mu_{\text{given}}$$
• Expectation of Standard Distributions: Geometric and Hypergeometric · Means of named distributions
  $$\text{Geometric: } E(X) = \dfrac{1}{p},\qquad \text{Hypergeometric: } E(X) = \dfrac{nK}{N}$$

Watch out for (20)

• The mean is a weighted average, not a plain average of the values
  → Expectation as the Long-Run Average
• Always verify $\sum p_i = 1$ before computing anything
  → Expectation as the Long-Run Average
• Use linearity for the sum on dice, don't build all 36 outcomes
  → Computing the Mean E(X) from a Probability Distribution
• Solve for the unknown probability before taking the mean
  → Computing the Mean E(X) from a Probability Distribution
• E(X²) is NOT [E(X)]²
  → Variance and Standard Deviation: Var(X) = E(X²) − [E(X)]²
• Convert a CDF to a pmf before computing an expectation
  → Variance and Standard Deviation: Var(X) = E(X²) − [E(X)]²
• Standard deviation vs variance — don't hand back the wrong one
  → Variance and Standard Deviation: Var(X) = E(X²) − [E(X)]²
• Variance is never negative
  → Variance and Standard Deviation: Var(X) = E(X²) − [E(X)]²
• A loss is a negative payoff — carry the minus sign
  → Expected Winnings of a Game: E(g(X)) = Σ g(x)·P(x)
• Get the all-heads/all-tails probability right
  → Expected Winnings of a Game: E(g(X)) = Σ g(x)·P(x)
• Variance of a winning amount is still E(X²) − [E(X)]²
  → Expected Winnings of a Game: E(g(X)) = Σ g(x)·P(x)
• Memorise both uniform formulas — mean (n+1)/2 AND variance (n²−1)/12
  → Uniform Distribution on 1 to n: E(X) = (n+1)/2, Var(X) = (n²−1)/12
• Cancel the (n+1) factor for 'find n' questions
  → Uniform Distribution on 1 to n: E(X) = (n+1)/2, Var(X) = (n²−1)/12
• P(x) = 2x/[n(n+1)] is NOT the uniform distribution
  → Uniform Distribution on 1 to n: E(X) = (n+1)/2, Var(X) = (n²−1)/12
• Watch the sign in the E(X) equation
  → Finding Unknown Probabilities from the Mean and ΣP = 1
• Use the extra stated relation as your second equation
  → Finding Unknown Probabilities from the Mean and ΣP = 1
• For range problems, apply non-negativity to EVERY row
  → Finding Unknown Probabilities from the Mean and ΣP = 1
• 'Until success' means geometric, mean = 1/p
  → Expectation of Standard Distributions: Geometric and Hypergeometric
• Hypergeometric mean is nK/N — no replacement needed for the mean
  → Expectation of Standard Distributions: Geometric and Hypergeometric
• For E(X²) build the small combination pmf first
  → Expectation of Standard Distributions: Geometric and Hypergeometric