Consider a game where the player tosses a six-sided fair die. If the face that comes up is $6$, the player wins ₹36; otherwise he loses ₹$k^2$, where $k$ is the face that comes up, $k\in\{1,2,3,4,5\}$. The expected amount to win at this game in ₹ is (1) $\dfrac{19}6$ (2) $-\dfrac{19}6$ (3) $\dfrac32$ (4) $-\dfrac32$.

Consider a game where the player tosses a six-sided fair die. If the face that comes up is 6, the player wins

Exercise 11.6 · Q3

6

(1)

\dfrac{19}6

(2)

-\dfrac{19}6

(3)

\dfrac32

(4)

-\dfrac32

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Concept understanding — Mathematical Expectation and Variance

Mean (Definition 11.8): for a random variable $X$ with pmf/pdf $f(x)$ ,

E(X)=\sum_x x f(x)\ \ (\text{discrete})\qquad\text{or}\qquad E(X)=\int_{-\infty}^{\infty} x f(x)\,dx\ \ (\text{continuous}).

$E(X)$ generalises the plain numerical average, weighting each value by its true probability rather than by $\tfrac1n$ ; it need not be a value $X$ can actually take, and is best read as the long-run average over many repetitions. Theorem 11.3 extends this to any function $g(X)$ : $E(g(X))=\sum_x g(x)f(x)$ or $\int g(x)f(x)\,dx$ ; taking $g(X)=X^k$ gives the $k$ -th moment $E(X^k)$ .

Variance (Definition 11.9): $V(X)=E\big((X-E(X))^2\big)$ , with the far more usable computing form

V(X)=E(X^2)-\big(E(X)\big)^2.

Standard deviation is $\sigma=\sqrt{V(X)}$ ; both are always $\ge0$ . A smaller $\sigma^2$ means values cluster tightly around the mean; a larger $\sigma^2$ means they scatter more widely — even distributions sharing the same mean can differ sharply here.

Three linearity laws (for constants $a,b$ ): $E(aX+b)=aE(X)+b$ (so $E(aX)=aE(X)$ and $E(b)=b$ ); $V(X)=E(X^2)-(E(X))^2$ (restated); and $V(aX+b)=a^2V(X)$ (so $V(aX)=a^2V(X)$ and $V(b)=0$ ). These make quick work of a shifted/scaled random variable — e.g. a net "winning amount" that is a linear function of a raw count — without recomputing the distribution from scratch.

Worked technique. For a discrete $X$ : tabulate $x$ , $f(x)$ , $xf(x)$ , $x^2f(x)$ ; sum the last two columns to get $E(X)$ and $E(X^2)$ directly, then apply $V(X)=E(X^2)-(E(X))^2$ . For a continuous $X$ : compute $E(X)=\int xf(x)\,dx$ and $E(X^2)=\int x^2f(x)\,dx$ over the support, then the same variance formula.

Each face is equally likely with probability $\tfrac16$ ; sum face-value times payoff across all six faces.

Step 1. List the payoff for each face. Face $6$ : win $+36$ . Faces $1,2,3,4,5$ : lose $k^2$ , i.e. payoff $-1,-4,-9,-16,-25$ respectively.

Step 2. Compute $E(X)$ as the average of the six payoffs (each with probability $\tfrac16$ ).

E(X)=\dfrac16\Big[36-(1+4+9+16+25)\Big]=\dfrac16\big[36-55\big]=\dfrac{-19}6.

Step 3. State the result. $E(X)=-\dfrac{19}6$ (rupees) — a negative expected value, i.e. the player expects to lose on average.

✓Final answer

E(X)=-\dfrac{19}6

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 markMCQ
$2l$
(a) $l,\ \dfrac{l^2}{12}$
(b) $\dfrac{l}{2},\ \dfrac{l^2}{3}$
(c) $\dfrac{l}{2},\ \dfrac{l^2}{12}$
(d) $\dfrac{l}{2},\ \dfrac{l^2}{6}$
›Reveal solutionSolution
Computes mean and variance of the given uniform-type density on $(0,l)$ using $E[X]$ and $E[X^2]-\left(E[X]\right)^2$ .

$\text{Mean}=E[X]=\displaystyle\int_0^lx\,f(x)\,dx=\int_0^lx\cdot\dfrac1l\,dx=\dfrac1l\left[\dfrac{x^2}2\right]_0^l=\dfrac1l\cdot\dfrac{l^2}2=\dfrac l2$ .

$E[X^2]=\displaystyle\int_0^lx^2\,f(x)\,dx=\int_0^lx^2\cdot\dfrac1l\,dx=\dfrac1l\left[\dfrac{x^3}3\right]_0^l=\dfrac1l\cdot\dfrac{l^3}3=\dfrac{l^2}3$ .

$\text{Variance}=E[X^2]-\left(E[X]\right)^2=\dfrac{l^2}3-\left(\dfrac l2\right)^2=\dfrac{l^2}3-\dfrac{l^2}4$ .

Common denominator $12$ : $\dfrac{4l^2}{12}-\dfrac{3l^2}{12}=\dfrac{l^2}{12}$ .

So Mean $=\dfrac l2$ , Variance $=\dfrac{l^2}{12}$ .

✓Final answer
$\dfrac{l}{2},\ \dfrac{l^2}{12}$
CBSE 2024Set ANNUAL1 markMCQ
$\text{Var}(3)$
(a) $0$
(b) $3$
(c) $\text{Var}(3)$
(d) $9$
›Reveal solutionSolution
A constant has no spread around its mean, so its variance is identically zero, regardless of the constant's value.

Variance measures the expected squared deviation from the mean: $\text{Var}(X)=E[(X-E[X])^2]$ .

For a constant $c$ (viewed as a "random variable" that always takes the value $3$ ), $E[c]=c=3$ , and $(c-E[c])^2=0$ always.

So $\text{Var}(3)=E[0]=0$ .

✓Final answer
$0$
CBSE 2019Set ANNUAL1 markMCQ
$X$
(a) $6$
(b) $8$
(c) $2$
(d) $4$
›Reveal solutionSolution
Since $\text{Var}(X)=E(X^2)-[E(X)]^2$ , with $\text{Var}(X)=4$ and $E(X)=2$ , we get $E(X^2)=8$ .

The variance formula is $\text{Var}(X)=E(X^2)-[E(X)]^2$ .

Given $\text{Var}(X)=4$ and $E(X)=2$ .

Substitute: $4=E(X^2)-(2)^2=E(X^2)-4$ .

Solve for $E(X^2)$ : $E(X^2)=4+4=8$ .

✓Final answer
$E(X^2)=8$
CBSE 2016Set ANNUAL1 markMCQ
$E(X+C)=8$
(a) $-2$
(b) $4$
(c) $-4$
(d) $2$
›Reveal solutionSolution
Solving the two expectation equations gives $C=-2$ .

By linearity of expectation, $E(X+C)=E(X)+C=8$ and $E(X-C)=E(X)-C=12$ .

Subtracting the second equation from the first: $(E(X)+C)-(E(X)-C)=8-12\Rightarrow 2C=-4$ .

So $C=-2$ .

Check: adding the two equations gives $2E(X)=20\Rightarrow E(X)=10$ ; then $E(X)+C=10-2=8$ (matches) and $E(X)-C=10+2=12$ (matches).

Options (b) $4$ , (c) $-4$ , (d) $2$ fail this check.

✓Final answer
$C=-2$

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Q.Consider a game where the player tosses a six-sided fair die. If the face that comes up is 666, the player wins ₹36; otherwise he loses ₹k2k^2k2, where kkk is the face that comes up, k∈{1,2,3,4,5}k\in\{1,2,3,4,5\}k∈{1,2,3,4,5}. The expected amount to win at this game in ₹ is

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Q.Consider a game where the player tosses a six-sided fair die. If the face that comes up is $6$ , the player wins ₹36; otherwise he loses ₹ $k^2$ , where $k$ is the face that comes up, $k\in\{1,2,3,4,5\}$ . The expected amount to win at this game in ₹ is

Unlock everything free for 14 days