Binomial and Poisson Distributions for CUET Applied Maths

AnswerB · Tossing a coin 10 times and noting head or tail each time

1. Bernoulli trials are a fixed number of independent trials, each with two outcomes and the same chance of success.

2. The coin fits all of these. Drawing without replacement changes the chance each time; 'until a six' has no fixed number; a height is not a two-way outcome.

AnswerA · Only I

1. I is one of the conditions for Bernoulli trials. True.

2. II: without replacement, each draw changes what is left in the bag, so the chance of success changes. False.

AnswerC · $n = 8,\ p = \frac{1}{5}$

1. Each of the 8 questions is one trial: $n = 8$.

2. A guess is right with probability $\frac{1}{5}$: $p = \frac{1}{5}$.

AnswerB · $n = 10,\ p = \frac{1}{6}$

1. Each throw of the pair is one trial: $n = 10$.

2. A doublet comes in 6 of 36 outcomes: $p = \frac{1}{6}$.

AnswerD · $n = 6,\ p = \frac{1}{3}$

1. $n$ is the number of trials: 6 throws.

2. Success is 5 or 6, two faces of six: $p = \frac{2}{6} = \frac{1}{3}$.

AnswerD · Drawing 4 balls one by one without replacement from a bag of 6 red and 4 blue, and noting red or not

1. Bernoulli trials need the same probability of success on every trial, independently.

2. Without replacement, the chance of red changes after each draw ($\frac{6}{10}$, then $\frac{5}{9}$ or $\frac{6}{9}$), so those draws are not Bernoulli trials.

AnswerD · I, II and IV only

1. Bernoulli trials are finite in number, independent, and each has exactly two outcomes.

2. III is the opposite of a condition: the probability of success must stay the same, $p$, on every trial.

AnswerC · $\frac{3}{8}$

1. $P(X = 2) = {}^{4}C_{2} \left(\frac{1}{2}\right)^{2} \left(\frac{1}{2}\right)^{2} = 6 \times \frac{1}{16}$.

2. $= \frac{6}{16} = \frac{3}{8}$.

AnswerC · $\frac{5}{16}$

1. $P(X = 3) = {}^{6}C_3\left(\frac{1}{2}\right)^{6}$.

2. $= \frac{20}{64} = \frac{5}{16}$.

AnswerA · $\frac{81}{256}$

1. $P(X = 4) = \left(\frac{3}{4}\right)^{4}$.

2. $= \frac{81}{256}$.

AnswerB · $\frac{243}{1024}$

1. Each answer is wrong with probability $\frac{3}{4}$, independently.

2. $\left(\frac{3}{4}\right)^{5} = \frac{243}{1{,}024}$.

AnswerB · $\frac{91}{216}$

1. 'At least one' is the complement of 'none'. The chance of no six is $\left(\frac{5}{6}\right)^{3} = \frac{125}{216}$.

2. So the chance of at least one six is $1 - \frac{125}{216} = \frac{91}{216}$.

AnswerC · $0.59049$

1. Each bulb is good with probability 0.9, independently.

2. The chance that none is faulty is $(0.9)^{5} = 0.59049$.

AnswerA · ${}^{7}C_{3}\,(0.4)^{3}(0.6)^{4}$

1. $P(X = r) = {}^{n}C_{r}\,p^{r}q^{n - r}$, with $q = 1 - p = 0.6$.

2. For $r = 3$: ${}^{7}C_{3}\,(0.4)^{3}(0.6)^{4}$.

AnswerA · $\frac{1}{1024}$

1. Each guess is right with probability $\frac{1}{4}$, independently.

2. $P(X = 5) = \left(\frac{1}{4}\right)^{5} = \frac{1}{1{,}024}$.

AnswerC · $\frac{25}{216}$

1. $P(X = 2) = {}^{4}C_2\left(\frac{1}{6}\right)^{2}\left(\frac{5}{6}\right)^{2}$.

2. $= 6 \times \frac{25}{1{,}296} = \frac{25}{216}$.

AnswerA · $1 - q^{n}$

1. The only way to have no success is to fail every time: $q^{n}$.

2. So at least one success has probability $1 - q^{n}$.

AnswerC · $\frac{64}{125}$

1. Each seed sprouts with probability 0.8, independently.

2. $0.8^{3} = 0.512$, which is $\frac{64}{125}$.

AnswerD · $\frac{3}{16}$

1. At most one head: $P(0) + P(1) = \frac{1}{32} + \frac{5}{32}$.

2. $= \frac{6}{32} = \frac{3}{16}$.

AnswerB · $\frac{11}{16}$

1. Complement: $1 - P(0) - P(1) = 1 - \frac{1}{16} - \frac{4}{16}$.

2. $= \frac{11}{16}$.

AnswerD · $\frac{405}{1024}$

1. $P(X = 1) = {}^{5}C_{1}\left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^{4}$.

2. $5 \times \frac{1}{4} \times \frac{81}{256} = \frac{405}{1{,}024}$.

AnswerC · $\frac{781}{1024}$

1. $P(X \ge 1) = 1 - P(X = 0)$.

2. $1 - \frac{243}{1{,}024} = \frac{781}{1{,}024}$.

AnswerC · $\frac{15}{16}$

1. $n = 5$, $p = \frac{1}{4}$, $q = \frac{3}{4}$.

2. Variance $= npq = 5 \times \frac{1}{4} \times \frac{3}{4} = \frac{15}{16}$.

AnswerB · $\frac{671}{1296}$

1. 'At least one six' is the complement of 'no six': $1 - \left(\frac{5}{6}\right)^{4}$.

2. $= 1 - \frac{625}{1{,}296} = \frac{671}{1{,}296}$.

AnswerD · $\frac{441}{1000}$

1. $P(X = 2) = {}^{3}C_2(0.7)^{2}(0.3)$.

2. $= 3 \times 0.49 \times 0.3 = 0.441$.

AnswerD · $\frac{3}{16}$

1. At least 4 means 4 or 5: ${}^{5}C_{4} + {}^{5}C_{5} = 5 + 1 = 6$ of the $2^{5} = 32$ outcomes.

2. $P = \frac{6}{32} = \frac{3}{16}$.

AnswerC · $6$

1. The mean of a binomial distribution is $np$.

2. $20 \times 0.3 = 6$.

AnswerB · $20$

1. Mean $= np$.

2. $50 \times 0.4 = 20$.

AnswerC · $5$

1. The number of sixes is binomial with $n = 30$, $p = \frac{1}{6}$.

2. Mean $= np = 5$.

AnswerD · $2.4$

1. Mean $= np$.

2. $4 \times 0.6 = 2.4$.

AnswerA · $0.96$

1. Variance $= npq = 4 \times 0.6 \times 0.4$.

2. $0.96$.

AnswerC · $0.1296$

1. $P(X = 4) = 0.6^{4}$.

2. $0.1296$.

AnswerB · $n = 20,\ p = 0.4$

1. $\frac{npq}{np} = q = \frac{4.8}{8} = 0.6$, so $p = 0.4$.

2. $n = \frac{8}{0.4} = 20$.

AnswerB · $4$

1. Variance $= npq = 100 \times 0.2 \times 0.8 = 16$.

2. Standard deviation $= \sqrt{16} = 4$.

AnswerA · Both A and R are true, and R is the correct explanation of A.

1. Variance 6 with mean 5 would need $q = \frac{6}{5} > 1$, which no probability can be. A is true.

2. R is exactly that reason, so R explains A.

AnswerD · A-II, B-IV, C-I, D-III

1. Binomial: mean $np = 5$, variance $npq = 10 \times 0.5 \times 0.5 = 2.5$.

2. Poisson: the variance equals $\lambda = 4$; with $\lambda = 9$ the standard deviation is $\sqrt{9} = 3$.

AnswerA · $\frac{40}{3}$

1. Variance $= npq = 60 \times \frac{1}{3} \times \frac{2}{3}$.

2. $= \frac{40}{3}$.

AnswerB · $4$

1. $npq = 16p(1 - p)$, and $p(1 - p)$ is largest at $p = \frac{1}{2}$, where it is $\frac{1}{4}$.

2. Largest variance $= 16 \times \frac{1}{4} = 4$.

AnswerC · $\frac{2}{3}$

1. $np = 3npq$, so $q = \frac{1}{3}$.

2. $p = 1 - \frac{1}{3} = \frac{2}{3}$.