Random Variables, Expectation and Variance for CUET Applied Maths

AnswerB · The number of heads in 5 tosses of a coin

1. A discrete random variable takes separate, countable values; a continuous one can take any value in an interval.

2. The number of heads can only be 0, 1, 2, 3, 4 or 5. Height, time and weight are measured, and can take any value in a range.

AnswerD · $0, 1, 2$

1. The outcomes are $HH, HT, TH, TT$, with 2, 1, 1 and 0 heads.

2. So $X$ takes the values 0, 1 and 2.

AnswerA · Only I

1. I is the definition of a random variable. True.

2. II: calls are counted, 0, 1, 2, …, so the variable is discrete. False.

AnswerB · The height of a student picked at random

1. A continuous random variable can take any value in an interval; a discrete one takes separate, countable values.

2. A height can be 162.4 cm or 162.43 cm or anything between; the others are counts.

AnswerC · $11$

1. The sum runs from $1 + 1 = 2$ to $6 + 6 = 12$, and every whole number between occurs.

2. 2, 3, …, 12: 11 values.

AnswerA · Only I

1. I is the definition of a random variable. True.

2. II is false: $HT$ and $TH$ both give 'one head'.

AnswerC · $\frac{3}{8}$

1. Of the 16 equally likely outcomes, $^{4}C_{2} = 6$ have exactly 2 tails.

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

AnswerD · $5$

1. There can be 0, 1, 2, 3 or 4 tails.

2. That is 5 values.

AnswerA · $6$

1. The larger number can be 1 (only from (1, 1)) up to 6.

2. So $X$ takes the 6 values 1, 2, 3, 4, 5, 6.

AnswerB · $-3, -1, 1, 3$

1. With $t$ tails there are $3 - t$ heads, so $X = t - (3 - t) = 2t - 3$.

2. For $t = 0, 1, 2, 3$: $X = -3, -1, 1, 3$.

AnswerD · $0, 1, 2$

1. At most 3 bulbs are picked, but only 2 defective ones exist.

2. So $X$ can be 0, 1 or 2.

AnswerD · $\frac{15}{16}$

1. The only way to get no tail is HHHH, with probability $\frac{1}{16}$.

2. $1 - \frac{1}{16} = \frac{15}{16}$.

AnswerB · $1$

1. $E(X) = 2$, and $E(X^{2}) = \frac{0 + 4 + 24 + 36 + 16}{16} = 5$.

2. $\text{Var}(X) = 5 - 2^{2} = 1$.

AnswerB · $0.1$

1. The probabilities of a distribution add up to 1: $k + 3k + 4k + 2k = 10k = 1$.

2. So $k = 0.1$.

AnswerC · $\frac{1}{4}$

1. There are $2^{4} = 16$ equally likely outcomes.

2. Exactly one head: $HTTT, THTT, TTHT, TTTH$, 4 outcomes, so $\frac{4}{16} = \frac{1}{4}$.

AnswerD · $0.8$

1. $P(X \le 2) = P(0) + P(1) + P(2)$.

2. $0.1 + 0.3 + 0.4 = 0.8$, or $1 - P(3) = 1 - 0.2$.

AnswerB · $\frac{1}{10}$

1. The probabilities must add up to 1: $k(1 + 2 + 3 + 4) = 10k = 1$.

2. $k = \frac{1}{10}$.

AnswerB · $0.1$

1. The probabilities add to 1: $k + 2k + 3k + 2k + 2k = 10k = 1$.

2. $k = 0.1$.

AnswerB · $0.4$

1. $P(X \ge 3) = P(3) + P(4) = 2k + 2k$.

2. $0.2 + 0.2 = 0.4$.

AnswerC · $\frac{3}{8}$

1. Of the 8 equally likely outcomes, $HHT$, $HTH$ and $THH$ have exactly two heads.

2. $P(X = 2) = \frac{3}{8}$.

AnswerD · $\frac{7}{10}$

1. $k(1 + 2 + 3 + 4) = 10k = 1$, so $k = \frac{1}{10}$.

2. $P(X \ge 3) = P(3) + P(4) = \frac{3}{10} + \frac{4}{10} = \frac{7}{10}$.

AnswerC · $P(X = x) = \frac{x}{6}$

1. A distribution needs every probability from 0 to 1 and a total of 1.

2. $\frac{1}{6} + \frac{2}{6} + \frac{3}{6} = 1$. The others total 2, have $-\frac{1}{2}$ at $x = 1$, or total $\frac{11}{6}$.

AnswerB · $0.55$

1. $1 < X \le 3$ means $X = 2$ or $X = 3$.

2. $0.3 + 0.25 = 0.55$.

AnswerC · $\frac{7}{15}$

1. Both bulbs must come from the 7 good ones: ${}^{7}C_2 = 21$ of ${}^{10}C_2 = 45$ ways.

2. $P(X = 0) = \frac{21}{45} = \frac{7}{15}$.

AnswerD · $2.2$

1. $E(X) = 0(0.1) + 1(0.2) + 2(0.3) + 3(0.2) + 4(0.2)$.

2. $0 + 0.2 + 0.6 + 0.6 + 0.8 = 2.2$.

AnswerD · $\frac{1}{2}$

1. $P(X = 2) = \frac{3}{8}$ (GGB, GBG, BGG) and $P(X = 3) = \frac{1}{8}$.

2. $P(X \ge 2) = \frac{3}{8} + \frac{1}{8} = \frac{1}{2}$.

AnswerD · $\frac{32}{221}$

1. One ace can come first or second: $\frac{4}{52} \times \frac{48}{51}$ for ace then not, and $\frac{48}{52} \times \frac{4}{51}$ for not then ace.

2. $P(X = 1) = 2 \times \frac{4 \times 48}{52 \times 51} = \frac{32}{221}$.

AnswerA · $1.56$

1. $E(X^{2}) = 0 + 0.2 + 1.2 + 1.8 + 3.2 = 6.4$.

2. $\text{Var}(X) = 6.4 - 2.2^{2} = 6.4 - 4.84 = 1.56$.

AnswerB · $\frac{1}{10}$

1. The probabilities add up to 1: $10k^{2} + 9k = 1$, that is $(10k - 1)(k + 1) = 0$.

2. A probability cannot be negative, so $k = \frac{1}{10}$, not $-1$.

AnswerD · $\frac{80}{221}$

1. The pack has 12 face cards. Face card then not: $\frac{12}{52} \times \frac{40}{51}$; not then face card: $\frac{40}{52} \times \frac{12}{51}$.

2. Together: $2 \times \frac{12 \times 40}{52 \times 51} = \frac{80}{221}$.

AnswerB · $0.9$

1. $E(X) = \sum x\,P(X = x) = 0 \times 0.3 + 1 \times 0.5 + 2 \times 0.2$.

2. $= 0.5 + 0.4 = 0.9$.

AnswerC · $3.5$

1. Each face has probability $\frac{1}{6}$: $E(X) = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = \frac{21}{6}$.

2. $= 3.5$, a value the die never shows: an expectation is a long-run average.

AnswerB · $11$

1. $E(aX + b) = aE(X) + b$.

2. $3 \times 5 - 4 = 11$.

AnswerA · $2$

1. Each throw gives a six with probability $\frac{1}{6}$.

2. Over 12 throws: $12 \times \frac{1}{6} = 2$.

AnswerC · $3$

1. $E(X) = \sum x\,P(X = x)$.

2. $0.1 + 0.4 + 0.9 + 1.6 = 3$.

AnswerD · $\frac{2}{3}$

1. The player loses on 1, 2, 3 or 4: four of the six faces.

2. $\frac{4}{6} = \frac{2}{3}$.

AnswerA · ₹500

1. Expected payout $= 0.005 \times 1{,}00{,}000$.

2. ₹500.

AnswerA · ₹0.40

1. $E = 10 \times 0.2 + (-2) \times 0.8 = 2 - 1.6$.

2. $= 0.4$ rupees a game: a small expected gain.

AnswerB · ₹500

1. The payout is ₹50,000 with probability 0.01 and ₹0 otherwise.

2. $E = 50{,}000 \times 0.01 = 500$ rupees; a premium above this covers the risk.