Binomial Distribution — MHT-CET Maths

Formulas (3)

• The Binomial Setting — n Fixed Independent Success or Failure Trials · Binomial variable and its parameters
  $$X \sim B(n, p),\qquad q = 1 - p,\qquad X \in \{0, 1, 2, \dots, n\}$$
• The Binomial PMF — Probability of Exactly r Successes · Binomial probability mass function
  $$P(X = r) = {}^{n}C_r\, p^{r}\, q^{\,n-r}$$
• Building the Full Probability Distribution Table · Distribution terms sum to one via the binomial expansion
  $$\sum_{r=0}^{n} P(X = r) = \sum_{r=0}^{n} {}^{n}C_r\, p^r q^{n-r} = (q + p)^{n} = 1$$

Watch out for (10)

• Binomial needs WITH-replacement (or constant p), not without-replacement
  → The Binomial Setting — n Fixed Independent Success or Failure Trials
• q = 1 − p is derived, so a binomial has only TWO parameters
  → The Binomial Setting — n Fixed Independent Success or Failure Trials
• 'Not a swimmer is 1/5' means success p = 4/5, not p = 1/5
  → The Binomial PMF — Probability of Exactly r Successes
• 'None defective' is P(X = 0) = qⁿ, and it needs WITH-replacement
  → The Binomial PMF — Probability of Exactly r Successes
• Don't forget the ⁿCᵣ multiplier
  → The Binomial PMF — Probability of Exactly r Successes
• Match the exponents to r and n − r, in that order
  → The Binomial PMF — Probability of Exactly r Successes
• Order the table by ascending r — P(X = 0) uses qⁿ, P(X = n) uses pⁿ
  → Building the Full Probability Distribution Table
• Fix which colour is 'success' before building the table
  → Building the Full Probability Distribution Table
• The probabilities must sum to 1 — use it as a check
  → Building the Full Probability Distribution Table
• Middle term of B(2, p) carries a factor 2 (not 1)
  → Building the Full Probability Distribution Table

Formulas (6)

• Adding PMF Terms to Get a Whole Answer · PMF term and the total-probability identity
  $$P(X=r) = \binom{n}{r} p^r q^{\,n-r},\qquad \sum_{r=0}^{n} P(X=r) = (p+q)^n = 1$$
• At Least and At Most — Cumulative Probabilities · Two-term tails you meet most often
  $$P(X \le 1) = q^n + n\,p\,q^{\,n-1},\qquad P(X \ge n-1) = n\,p^{\,n-1}q + p^n$$
• At Least One — the 1 minus qⁿ Shortcut · The at-least-one complement
  $$P(X \ge 1) = 1 - q^n,\qquad \text{smallest } n:\; q^n < 1 - t$$
• Ranges and Symmetric Events by Complement · Absolute-value condition and the complement of a range
  $$|X - a| \le b \iff a-b \le X \le a+b,\qquad P(a{-}b \le X \le a{+}b) = 1 - \!\!\sum_{r \,\notin\, [a-b,\,a+b]} \!\!P(r)$$
• Special Counting — Even Successes, Expected Frequency, and Fixed-Trial Events · Even-count identity and expected frequency
  $$\sum_{r\text{ even}}\binom{n}{r} = 2^{\,n-1},\qquad \mathbb{E}[\text{count over } N] = N \cdot P(\text{event})$$
• Finding p First When the Stem Hides It · p by counting, then the at-least-3 binomial
  $$p = \dfrac{\text{favourable outcomes}}{\text{total outcomes}},\qquad P(X \ge 3) = \binom{4}{3}p^3 q + p^4 = p^3(4q + p)$$

Watch out for (15)

• 'At least k' includes k itself, not just above it
  → Adding PMF Terms to Get a Whole Answer
• A compound event is a SUM of terms, not a single term
  → Adding PMF Terms to Get a Whole Answer
• 'At most one defective' has two terms, not one
  → At Least and At Most — Cumulative Probabilities
• Decide which outcome 'success' labels before counting
  → At Least and At Most — Cumulative Probabilities
• Factor the shared power to match the printed option
  → At Least and At Most — Cumulative Probabilities
• At least one = 1 − qⁿ, not p or np
  → At Least One — the 1 minus qⁿ Shortcut
• For 'smallest n', solve the inequality — don't just plug the mean
  → At Least One — the 1 minus qⁿ Shortcut
• Cap the interval at 0 and n before counting
  → Ranges and Symmetric Events by Complement
• Use the complement when the range is most of 0…n
  → Ranges and Symmetric Events by Complement
• Even number of heads on a fair coin is exactly 1/2
  → Special Counting — Even Successes, Expected Frequency, and Fixed-Trial Events
• Expected frequency is N × P, not N × p
  → Special Counting — Even Successes, Expected Frequency, and Fixed-Trial Events
• 'Second success at the third trial' fixes the last trial
  → Special Counting — Even Successes, Expected Frequency, and Fixed-Trial Events
• Count the favourable numbers carefully — this is where marks are lost
  → Finding p First When the Stem Hides It
• Read the sample-space range: 00–99 is 100, 10–99 is 90
  → Finding p First When the Stem Hides It
• After finding p, still add all the terms for 'at least 3'
  → Finding p First When the Stem Hides It

Formulas (4)

• The Mean of a Binomial Variable is np · Mean of a binomial variable
  $$\mu = E(X) = np$$
• Variance is npq and Standard Deviation is the Square Root of npq · Variance and standard deviation of a binomial variable
  $$\sigma^2 = npq,\qquad \sigma = \sqrt{npq},\qquad npq < np$$
• Recovering n and p from the Mean and Variance · Recover q, then p and n
  $$q = \dfrac{\sigma^2}{\mu} = \dfrac{npq}{np},\qquad p = 1 - q,\qquad n = \dfrac{\mu}{p}$$
• Solving When the Mean and Variance are Combined into One Equation · Sum of mean and variance
  $$np + npq = np(1 + q) = np(2 - p)$$

Watch out for (11)

• The mean is np, never p or p^n
  → The Mean of a Binomial Variable is np
• 'With replacement' is what makes the trials binomial
  → The Mean of a Binomial Variable is np
• Variance is npq, not np or npq^2
  → Variance is npq and Standard Deviation is the Square Root of npq
• Variance is always smaller than the mean for a binomial variable
  → Variance is npq and Standard Deviation is the Square Root of npq
• SD is the square root of the variance, not of npq-then-forgotten
  → Variance is npq and Standard Deviation is the Square Root of npq
• Divide variance by mean to get q — not p
  → Recovering n and p from the Mean and Variance
• P(X = 0) is q^n, and P(X ≥ 1) = 1 − q^n
  → Recovering n and p from the Mean and Variance
• For a lower tail sum the terms up to r, then divide by 2^n only if p = 1/2
  → Recovering n and p from the Mean and Variance
• Substitute q = 1 − p to reduce the sum to a single-variable equation
  → Solving When the Mean and Variance are Combined into One Equation
• Reject the root outside [0, 1]
  → Solving When the Mean and Variance are Combined into One Equation
• Read whether p, q, or the variance is being asked
  → Solving When the Mean and Variance are Combined into One Equation

Formulas (6)

• The Binomial PMF, Mean and Variance (Recall) · PMF, mean and variance of B(n, p)
  $$P(X=r) = {}^{n}C_{r}\,p^{r}q^{\,n-r},\qquad \text{mean} = np,\qquad \text{var} = npq,\qquad \text{SD} = \sqrt{npq}$$
• The Successive-Term Ratio of a Binomial Distribution · Ratio of consecutive binomial probabilities
  $$\frac{P(X=k)}{P(X=k-1)} = \frac{n-k+1}{k}\cdot\frac{p}{q}$$
• Finding p from a Condition a·P(X=i) = b·P(X=j) · Cancelling a condition to a linear relation
  $$a\,{}^{n}C_{i}\,p^{i}q^{\,n-i} = b\,{}^{n}C_{j}\,p^{j}q^{\,n-j}\;\Longrightarrow\;\text{linear in }p,q,\quad q = 1-p$$
• Finding p from Given Numerical Probabilities · Divide two given probabilities to expose p/q
  $$\frac{P(X=i)}{P(X=j)} = \frac{{}^{n}C_{i}}{{}^{n}C_{j}}\left(\frac{p}{q}\right)^{\,i-j}$$
• Combination Identities: ⁿCₐ = ⁿC_b and PMF Normalisation · The two n-pinning identities
  $${}^{n}C_{a} = {}^{n}C_{b}\;(a\ne b)\;\Rightarrow\; a+b = n,\qquad \sum_{r=0}^{n}{}^{n}C_{r} = 2^{n}$$
• The Most Probable Value (Mode) of a Binomial Distribution · Most probable value for a fair coin B(n, ½)
  $$n\text{ even: } r = \tfrac{n}{2};\qquad n\text{ odd: } r = \tfrac{n-1}{2}\ \text{and}\ \tfrac{n+1}{2}$$

Watch out for (15)

• Variance is npq, not np or np·q with q = p
  → The Binomial PMF, Mean and Variance (Recall)
• The exponent of q is n − r, not r
  → The Binomial PMF, Mean and Variance (Recall)
• The coefficient ratio is (n−k+1)/k, not (n−k)/k or (n−k+1)/(k+1)
  → The Successive-Term Ratio of a Binomial Distribution
• Do not invert the ratio: it is p/q, not q/p
  → The Successive-Term Ratio of a Binomial Distribution
• Cancel powers of BOTH p and q before solving
  → Finding p from a Condition a·P(X=i) = b·P(X=j)
• Always substitute q = 1 − p at the end, not p = 1 − q inconsistently
  → Finding p from a Condition a·P(X=i) = b·P(X=j)
• Read what the question finally asks — p, or the variance/probability that follows
  → Finding p from a Condition a·P(X=i) = b·P(X=j)
• Dividing the two given probabilities is faster than substituting numbers
  → Finding p from Given Numerical Probabilities
• Recover p, then evaluate the REQUESTED probability — not the ones given
  → Finding p from Given Numerical Probabilities
• Read a single given P(X=r) as a product of powers to spot p and q
  → Finding p from Given Numerical Probabilities
• ⁿCₐ = ⁿC_b gives a + b = n (or a = b), not a − b = n
  → Combination Identities: ⁿCₐ = ⁿC_b and PMF Normalisation
• The coefficients cancel only for a FAIR coin
  → Combination Identities: ⁿCₐ = ⁿC_b and PMF Normalisation
• Simplify the final probability into the option's power of 2
  → Combination Identities: ⁿCₐ = ⁿC_b and PMF Normalisation
• For odd n there are TWO modes, both central
  → The Most Probable Value (Mode) of a Binomial Distribution
• The mode is the middle of the range, not the mean np unless p = ½
  → The Most Probable Value (Mode) of a Binomial Distribution