Differentiation (Part-38) | S N De

Random Variable and Its Distribution (Class 12): Short and Long Answer Questions

Complete solutions to 15 Short Answer questions (Q3–Q17, 4 marks) and 2 Long Answer questions (5 marks) from the chapter Random Variable and Its Distribution. They cover probability distributions of coin, dice, card, ball and watch problems, finding unknown constants, mean, variance, standard deviation and expectation, and a fair-game problem. Each solution has the distribution table, the working and a boxed final answer. Useful for CBSE Class 12 board exam, JEE and other competitive exam preparation.

Short Answer Q3–17Long Answer Q1–2

Part A: Short Answer Type Questions (Q3–Q17)

Q34 Marks

Examine which one of the following represents the probability distribution of a random variable \(X\).

(i) \(X\)	\(-2\)	\(-1\)	\(0\)	\(1\)	\(2\)
\(P(X)\)	0.14	0.24	0.32	0.28	0.12

(ii) \(X\)	\(1\)	\(3\)	\(5\)	\(7\)	\(9\)
\(f(X)\)	0.25	0.36	\(-0.26\)	0.38	0.27

(iii) \(X\)	\(-3\)	\(-2\)	\(-1\)	\(0\)	\(1\)
\(f(X)\)	0.12	0.16	0.24	0.32	0.16

Solution

A table is a probability distribution only if (a) every probability satisfies \(p_i\ge0\), and (b) \(\sum p_i=1\).

(i)\[0.14+0.24+0.32+0.28+0.12=1.10\ne1.\]

Condition (b) fails, so (i) is not a probability distribution.

(ii)

\(f(5)=-0.26<0\), so condition (a) fails. (The sum is 1, but a probability cannot be negative.) (ii) is not a probability distribution.

(iii)

All values are non-negative and

\[0.12+0.16+0.24+0.32+0.16=1.00.\]

Both conditions hold.

Answer: Only (iii) represents a probability distribution.

Q44 Marks

(i) A random experiment consists of three independent tosses of a fair coin. Write down the sample space. Let \(X\) be the number of heads obtained. Obtain the probability distribution of \(X\) and calculate its expectation and variance.
(ii) Four unbiased coins are tossed simultaneously. Let \(X\) be the random variable denoting the number of heads obtained. Obtain the probability distribution of \(X\). Find \(E(X)\) and \(\mathrm{Var}(X)\). [Symbols have their usual meanings.]

Solution

(i) Three tosses

Sample space: \(S=\{HHH,\ HHT,\ HTH,\ THH,\ HTT,\ THT,\ TTH,\ TTT\}\), 8 equally likely outcomes.

\(X=0\) for \(TTT\); \(X=1\) for \(HTT,THT,TTH\); \(X=2\) for \(HHT,HTH,THH\); \(X=3\) for \(HHH\).

\(X\)	\(0\)	\(1\)	\(2\)	\(3\)
\(P(X)\)	\(\frac18\)	\(\frac38\)	\(\frac38\)	\(\frac18\)

\[E(X)=0\cdot\tfrac18+1\cdot\tfrac38+2\cdot\tfrac38+3\cdot\tfrac18=\frac{0+3+6+3}{8}=\frac32.\]\[E(X^2)=0+1\cdot\tfrac38+4\cdot\tfrac38+9\cdot\tfrac18=\frac{3+12+9}{8}=3,\qquad \mathrm{Var}(X)=3-\left(\tfrac32\right)^2=\frac34.\]

Answer (i): \(E(X)=\dfrac32\), \(\mathrm{Var}(X)=\dfrac34\).

(ii) Four coins

There are \(2^4=16\) equally likely outcomes, and \(P(X=r)=\dfrac{\binom4r}{16}\).

\(X\)	\(0\)	\(1\)	\(2\)	\(3\)	\(4\)
\(P(X)\)	\(\frac1{16}\)	\(\frac4{16}\)	\(\frac6{16}\)	\(\frac4{16}\)	\(\frac1{16}\)

\[E(X)=\frac{0+4+12+12+4}{16}=2,\qquad E(X^2)=\frac{0+4+24+36+16}{16}=5,\qquad \mathrm{Var}(X)=5-2^2=1.\]

Answer (ii): \(E(X)=2\), \(\mathrm{Var}(X)=1\).

Q54 Marks

Let \(X\) denote the sum of digits in the random experiment of throwing two balanced dice. Find the probability distribution of the random variable \(X\).

Solution

Two dice give \(6\times6=36\) equally likely outcomes. The sum \(X\) takes the values \(2,3,\dots,12\).

Sum \(X\)	Favourable outcomes	Number
2	(1,1)	1
3	(1,2),(2,1)	2
4	(1,3),(2,2),(3,1)	3
5	(1,4),(2,3),(3,2),(4,1)	4
6	(1,5),(2,4),(3,3),(4,2),(5,1)	5
7	(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)	6
8	(2,6),(3,5),(4,4),(5,3),(6,2)	5
9	(3,6),(4,5),(5,4),(6,3)	4
10	(4,6),(5,5),(6,4)	3
11	(5,6),(6,5)	2
12	(6,6)	1

Hence the probability distribution is:

\(X\)	2	3	4	5	6	7	8	9	10	11	12
\(P(X)\)	\(\frac1{36}\)	\(\frac2{36}\)	\(\frac3{36}\)	\(\frac4{36}\)	\(\frac5{36}\)	\(\frac6{36}\)	\(\frac5{36}\)	\(\frac4{36}\)	\(\frac3{36}\)	\(\frac2{36}\)	\(\frac1{36}\)

Check: \(1+2+3+4+5+6+5+4+3+2+1=36\).

Answer: \(P(X=s)=\dfrac{6-|s-7|}{36}\) for \(s=2,3,\dots,12\), as tabulated above.

Q64 Marks

Let \(X\) denote the number of tails preceding the first head when an unbiased coin is tossed repeatedly. Find the probability distribution of \(X\).

Solution

\(X=k\) means that the first \(k\) tosses are all tails and the \((k+1)\)th toss is the first head. Since the tosses are independent and \(P(H)=P(T)=\frac12\),

\[P(X=k)=\left(\frac12\right)^k\cdot\frac12=\left(\frac12\right)^{k+1},\qquad k=0,1,2,3,\dots\]

\(X\)	\(0\)	\(1\)	\(2\)	\(3\)	\(\cdots\)	\(k\)	\(\cdots\)
\(P(X)\)	\(\frac12\)	\(\frac14\)	\(\frac18\)	\(\frac1{16}\)	\(\cdots\)	\(\frac1{2^{k+1}}\)	\(\cdots\)

Check: the probabilities form an infinite geometric series with first term \(\frac12\) and ratio \(\frac12\), whose sum is \(\dfrac{1/2}{1-1/2}=1\).

Answer: \(P(X=k)=\left(\dfrac12\right)^{k+1}\) for \(k=0,1,2,\dots\)

Q74 Marks

In a box there are 5 watches of which 2 are known to be defective. Two watches are taken out at random. Let \(X\) denote the number of defective watches selected. Obtain the probability distribution of \(X\). Also calculate the mean of \(X\).

Solution

There are 2 defective and 3 good watches. Two can be chosen in \(\binom52=10\) ways. \(X\) can be \(0,1,2\).

\[P(X=0)=\frac{\binom32}{10}=\frac3{10},\quad P(X=1)=\frac{\binom21\binom31}{10}=\frac6{10},\quad P(X=2)=\frac{\binom22}{10}=\frac1{10}.\]

\(X\)	\(0\)	\(1\)	\(2\)
\(P(X)\)	\(\frac3{10}\)	\(\frac6{10}\)	\(\frac1{10}\)

\[E(X)=0\cdot\frac3{10}+1\cdot\frac6{10}+2\cdot\frac1{10}=\frac8{10}=\frac45.\]

Answer: Mean \(=\dfrac45=0.8\).

Q84 MarksCBSE '07 (part i)

(i) Five defective bulbs are accidentally mixed with 20 good ones. It is not possible to just look at a bulb and tell whether or not it is defective. Find the probability distribution of the number of defective bulbs in a draw of 4 bulbs.
(ii) From a lot of 25 articles containing 5 defectives a sample of 4 articles was drawn at random (a) without replacement (b) with replacement. Find the probability distribution of \(X\), the number of defectives.

Solution

(i) 4 bulbs drawn from 25 (5 defective, 20 good)

The 4 bulbs are drawn without replacement, so \(P(X=k)=\dfrac{\binom5k\binom{20}{4-k}}{\binom{25}4}\), with \(\binom{25}4=12650\).

\(k\)	Favourable ways	\(P(X=k)\)
0	\(\binom{20}4=4845\)	\(\frac{969}{2530}\)
1	\(5\times\binom{20}3=5700\)	\(\frac{114}{253}\)
2	\(10\times\binom{20}2=1900\)	\(\frac{38}{253}\)
3	\(10\times20=200\)	\(\frac{4}{253}\)
4	\(\binom54=5\)	\(\frac{1}{2530}\)

Check: \(4845+5700+1900+200+5=12650\).

Answer (i): \(X\): 0, 1, 2, 3, 4 with \(P(X)\): \(\frac{969}{2530},\ \frac{114}{253},\ \frac{38}{253},\ \frac{4}{253},\ \frac{1}{2530}\).

(ii)(a) Without replacement

This is exactly the situation of part (i): 25 articles, 5 defective, 4 drawn without replacement.

\(X\)	\(0\)	\(1\)	\(2\)	\(3\)	\(4\)
\(P(X)\)	\(\frac{969}{2530}\)	\(\frac{114}{253}\)	\(\frac{38}{253}\)	\(\frac{4}{253}\)	\(\frac{1}{2530}\)

(ii)(b) With replacement

The draws are independent, with \(p=\frac5{25}=\frac15\) and \(q=\frac45\). So \(X\sim B\left(4,\frac15\right)\) and \(P(X=k)=\binom4k\left(\frac15\right)^k\left(\frac45\right)^{4-k}\).

\(X\)	\(0\)	\(1\)	\(2\)	\(3\)	\(4\)
\(P(X)\)	\(\frac{256}{625}\)	\(\frac{256}{625}\)	\(\frac{96}{625}\)	\(\frac{16}{625}\)	\(\frac{1}{625}\)

Check: \(256+256+96+16+1=625\).

Answer (ii): (a) the same distribution as in (i); (b) \(P(X=k)=\dbinom4k\left(\dfrac15\right)^k\left(\dfrac45\right)^{4-k}\) as tabulated.

Q94 MarksNCERT

Two cards are drawn simultaneously (or successively without replacement) from a well-shuffled pack of 52 cards. Find the mean, variance and standard deviation of the number of kings.

Solution

Let \(X\) be the number of kings. The pack has 4 kings and 48 other cards, and \(\binom{52}2=1326\).

\[P(X=0)=\frac{\binom{48}2}{1326}=\frac{1128}{1326}=\frac{188}{221},\quad P(X=1)=\frac{4\times48}{1326}=\frac{32}{221},\quad P(X=2)=\frac{\binom42}{1326}=\frac1{221}.\]

\(X\)	\(0\)	\(1\)	\(2\)
\(P(X)\)	\(\frac{188}{221}\)	\(\frac{32}{221}\)	\(\frac{1}{221}\)

Mean\[E(X)=0+1\cdot\frac{32}{221}+2\cdot\frac{1}{221}=\frac{34}{221}=\frac{2}{13}.\]Variance and standard deviation\[E(X^2)=0+1\cdot\frac{32}{221}+4\cdot\frac1{221}=\frac{36}{221}.\]\[\mathrm{Var}(X)=\frac{36}{221}-\left(\frac2{13}\right)^2=\frac{36}{221}-\frac4{169}=\frac{468-68}{2873}=\frac{400}{2873}.\]\[\text{S.D.}=\sqrt{\frac{400}{2873}}=\frac{20}{\sqrt{2873}}\approx0.373.\]

Answer: Mean \(=\dfrac2{13}\), Variance \(=\dfrac{400}{2873}\), S.D. \(=\dfrac{20}{\sqrt{2873}}\approx0.37\).

Q104 Marks

(i) A discrete random variable \(x\) has the following probability distribution:

\(x\)	0	1	2	3	4	5	6	7
\(p(x)\)	\(a\)	\(4a\)	\(3a\)	\(7a\)	\(8a\)	\(10a\)	\(6a\)	\(9a\)

(a) Determine the value of \(a\). (b) Find \(P(x<3)\), \(P(x\ge4)\), \(P(0
(ii) A random variable \(X\) has the following probability function:

\(x\)	\(-2\)	\(-1\)	0	1	2	3
\(p(x)\)	0.1	\(k\)	0.2	\(2k\)	0.3	\(3k\)

(a) Calculate \(k\). (b) Find \(P(x<2)\), \(P(x\ge2)\), \(P(-20.36\).

(iii) For what value of \(k\) will the function \(f(x)=kx\), \(x=1,2,3,\dots,n\) be a probability distribution of a random variable?

Solution

(i)(a) Value of a

The probabilities add up to 1:

\[a+4a+3a+7a+8a+10a+6a+9a=48a=1\ \Rightarrow\ a=\frac1{48}.\](i)(b) Probabilities\[P(x<3)=P(0)+P(1)+P(2)=\frac{1+4+3}{48}=\frac8{48}=\frac16.\]\[P(x\ge4)=\frac{8+10+6+9}{48}=\frac{33}{48}=\frac{11}{16}.\]\[P(0(i)(c) Minimum m with P(x ≤ m) ≥ 0.6

The cumulative probabilities (in 48ths) are:

\(m\)	0	1	2	3	4	5
\(P(x\le m)\)	\(\frac1{48}\)	\(\frac5{48}\)	\(\frac8{48}\)	\(\frac{15}{48}\)	\(\frac{23}{48}\approx0.479\)	\(\frac{33}{48}\approx0.688\)

\(P(x\le4)=0.479<0.6\) but \(P(x\le5)=0.688\ge0.6\).

Answer (i): (a) \(a=\frac1{48}\); (b) \(\frac16,\ \frac{11}{16},\ \frac{11}{24}\); (c) \(m=5\).

(ii)(a) Value of k

\(0.1+k+0.2+2k+0.3+3k=1\Rightarrow 0.6+6k=1\Rightarrow k=\dfrac{0.4}{6}=\dfrac1{15}\).

So \(p(-1)=\frac1{15},\ p(1)=\frac2{15},\ p(3)=\frac3{15}=\frac15\).

(ii)(b) Probabilities\[P(x<2)=0.1+\tfrac1{15}+0.2+\tfrac2{15}=0.3+\tfrac3{15}=0.5.\]\[P(x\ge2)=0.3+\tfrac3{15}=0.5.\]\[P(-2(ii)(c) Condition on k

\(P(x\le1)=0.1+k+0.2+2k=0.3+3k\). The requirement \(P(x\le1)>0.36\) gives

\[0.3+3k>0.36\ \Rightarrow\ 3k>0.06\ \Rightarrow\ k>0.02.\]

So \(k\) must exceed \(0.02\). The value \(k=\frac1{15}\approx0.067\) from part (a) satisfies this, and then \(P(x\le1)=0.5>0.36\).

Answer (ii): (a) \(k=\frac1{15}\); (b) \(0.5,\ 0.5,\ 0.7\); (c) the condition is \(k>0.02\). The inequality is strict, so \(0.02\) is the lower limit for \(k\) and not itself allowed.

(iii) f(x) = kx for x = 1, 2, …, n

For a probability distribution, \(f(x)\ge0\) and \(\sum f(x)=1\):

\[k(1+2+3+\cdots+n)=k\cdot\frac{n(n+1)}{2}=1\ \Rightarrow\ k=\frac{2}{n(n+1)}.\]

This \(k\) is positive, so \(f(x)\ge0\) as well.

Answer (iii): \(k=\dfrac{2}{n(n+1)}\).

Q114 Marks

The probability distribution of a random variable \(X\) is as follows:

\(x\)	0	1	2	3	4
\(p(x)\)	0.20	0.25	0.35	0.14	0.06

Find the probability distribution of the variable \(Y\) where \(y=x^2+5\).

Solution

Check: \(0.20+0.25+0.35+0.14+0.06=1\). For each value of \(x\), compute \(y=x^2+5\):

\(x\)	0	1	2	3	4
\(y=x^2+5\)	5	6	9	14	21

The values of \(y\) are all different, so each keeps the probability of the corresponding \(x\).

\(Y\)	5	6	9	14	21
\(P(Y)\)	0.20	0.25	0.35	0.14	0.06

Answer: \(Y\) takes the values \(5,6,9,14,21\) with probabilities \(0.20,\ 0.25,\ 0.35,\ 0.14,\ 0.06\).

Q124 Marks

Let \(X\) denote the number of "heads" when 10 unbiased coins are tossed. Find the probability distribution of \(X\).

Solution

The 10 coins are independent and \(P(H)=\frac12\), so \(X\sim B\left(10,\frac12\right)\) takes the values \(0,1,\dots,10\):

\[P(X=r)=\binom{10}r\left(\frac12\right)^{10}=\frac{\binom{10}r}{1024}.\]

\(X\)	0	1	2	3	4	5	6	7	8	9	10
\(P(X)\)	\(\frac1{1024}\)	\(\frac{10}{1024}\)	\(\frac{45}{1024}\)	\(\frac{120}{1024}\)	\(\frac{210}{1024}\)	\(\frac{252}{1024}\)	\(\frac{210}{1024}\)	\(\frac{120}{1024}\)	\(\frac{45}{1024}\)	\(\frac{10}{1024}\)	\(\frac1{1024}\)

The numerators add up to \(2^{10}=1024\).

Answer: \(P(X=r)=\dfrac{\binom{10}r}{1024}\), \(r=0,1,2,\dots,10\).

Q134 Marks

In a given business venture a man can make a profit of ₹300 with probability 0.6 or incur a loss of ₹100 with probability 0.4. Calculate his expectation.

Solution

Let \(X\) be the gain in rupees. Then \(X=300\) with probability 0.6 and \(X=-100\) (a loss) with probability 0.4.

\(X\)	\(300\)	\(-100\)
\(P(X)\)	0.6	0.4

\[E(X)=300\times0.6+(-100)\times0.4=180-40=140.\]

Answer: His expectation is ₹140.

Q144 Marks

Find the mean of the sum of points on 2 unbiased dice.

Solution

Let \(X_1\) and \(X_2\) be the points on the two dice. For one die,

\[E(X_i)=\frac{1+2+3+4+5+6}{6}=\frac72.\]

The sum is \(X=X_1+X_2\), and the expectation of a sum is the sum of the expectations:

\[E(X)=E(X_1)+E(X_2)=\frac72+\frac72=7.\]

Check using the distribution of the sum (see Q5):

\[E(X)=\frac{2\cdot1+3\cdot2+4\cdot3+5\cdot4+6\cdot5+7\cdot6+8\cdot5+9\cdot4+10\cdot3+11\cdot2+12\cdot1}{36}=\frac{252}{36}=7.\]

Answer: The mean of the sum is 7.

Q154 Marks

(i) A box contains 4 white and 6 red balls. If 2 balls are drawn at random, find the expectation of the number of red balls.
(ii) A bag contains 5 white and 7 black balls. Find the expectation of a man who is allowed to draw two balls from the bag and who is to receive one rupee for each black ball and two rupees for each white ball drawn.

Solution

(i) Expected number of red balls

Let \(X\) be the number of red balls in 2 draws from 4 white and 6 red balls (\(\binom{10}2=45\) ways).

\[P(X=0)=\frac{\binom42}{45}=\frac6{45},\quad P(X=1)=\frac{4\times6}{45}=\frac{24}{45},\quad P(X=2)=\frac{\binom62}{45}=\frac{15}{45}.\]\[E(X)=0+1\cdot\frac{24}{45}+2\cdot\frac{15}{45}=\frac{54}{45}=\frac65.\]

Answer (i): \(\dfrac65=1.2\).

(ii) Expected winnings

There are 12 balls, so \(\binom{12}2=66\) ways to draw two. Let \(X\) be the amount received (in rupees):

Balls drawn	Ways	\(P\)	Amount \(X\)
2 white	\(\binom52=10\)	\(\frac{10}{66}\)	\(2+2=4\)
1 white, 1 black	\(5\times7=35\)	\(\frac{35}{66}\)	\(2+1=3\)
2 black	\(\binom72=21\)	\(\frac{21}{66}\)	\(1+1=2\)

\[E(X)=\frac{10\times4+35\times3+21\times2}{66}=\frac{40+105+42}{66}=\frac{187}{66}=\frac{17}{6}.\]

Answer (ii): \(₹\dfrac{17}{6}\approx ₹2.83\).

Q164 Marks

An urn contains 7 white and 3 red balls. Two balls are drawn together, at random, from this urn. Compute the probability that neither of them is white. Find also the probability of getting one white and one red ball. Hence compute the expected number of white balls.

Solution

Total balls \(=10\), and two can be drawn in \(\binom{10}2=45\) ways. Let \(X\) be the number of white balls.

Neither is white

Both are red:

\[P(X=0)=\frac{\binom32}{45}=\frac3{45}=\frac1{15}.\]One white and one red\[P(X=1)=\frac{7\times3}{45}=\frac{21}{45}=\frac7{15}.\]Expected number of white balls\[P(X=2)=\frac{\binom72}{45}=\frac{21}{45}=\frac7{15}.\]

\(X\)	\(0\)	\(1\)	\(2\)
\(P(X)\)	\(\frac1{15}\)	\(\frac7{15}\)	\(\frac7{15}\)

\[E(X)=0+1\cdot\frac7{15}+2\cdot\frac7{15}=\frac{21}{15}=\frac75.\]

Answer: \(P(\text{neither white})=\dfrac1{15}\), \(P(\text{one white, one red})=\dfrac7{15}\), expected number of white balls \(=\dfrac75=1.4\).

Q174 Marks

What is the expectation from a throw with a single die when a score of \(n\) is rewarded with a prize of \(n^2\) rupees?

Solution

Let \(X\) be the prize in rupees. When the score is \(n\) (for \(n=1,2,\dots,6\), each with probability \(\frac16\)), \(X=n^2\).

\(X\)	\(1\)	\(4\)	\(9\)	\(16\)	\(25\)	\(36\)
\(P(X)\)	\(\frac16\)	\(\frac16\)	\(\frac16\)	\(\frac16\)	\(\frac16\)	\(\frac16\)

\[E(X)=\frac{1^2+2^2+3^2+4^2+5^2+6^2}{6}=\frac{1+4+9+16+25+36}{6}=\frac{91}{6}.\]

Answer: The expectation is ₹\(\dfrac{91}{6}\) \(\approx\) ₹15.17.

Part B: Long Answer Type Questions (Q1–Q2)

LA Q15 Marks

If a person gains or loses an amount equal to the number appearing when a balanced die is rolled once, according to whether the number is even or odd, how much money can he expect per game in the long run?

Solution

Let \(X\) be the person's gain in rupees. If the number is even (2, 4, 6) he gains that amount, and if it is odd (1, 3, 5) he loses that amount. Each number has probability \(\frac16\).

Number shown	1	2	3	4	5	6
\(X\) (gain in ₹)	\(-1\)	\(2\)	\(-3\)	\(4\)	\(-5\)	\(6\)
\(P(X)\)	\(\frac16\)	\(\frac16\)	\(\frac16\)	\(\frac16\)	\(\frac16\)	\(\frac16\)

\[E(X)=\frac{(-1)+2+(-3)+4+(-5)+6}{6}=\frac{3}{6}=\frac12.\]

The expectation is positive, so in the long run the person gains.

Answer: He can expect to gain ₹\(\dfrac12\) (50 paise) per game in the long run.

LA Q25 Marks

A gambler rolls two unbiased dice and stands to lose ₹2 if he fails to throw a six, to win ₹4 if he throws one six and to win ₹10 if he throws two sixes. Is the game fair?

Solution

Two dice give 36 equally likely outcomes. Let \(X\) be the gambler's gain in rupees.

• No six: each die shows a non-six, so \(5\times5=25\) outcomes, \(P=\frac{25}{36}\), \(X=-2\).
• Exactly one six: \(2\times5=10\) outcomes, \(P=\frac{10}{36}\), \(X=4\).
• Two sixes: 1 outcome, \(P=\frac1{36}\), \(X=10\).

\(X\)	\(-2\)	\(4\)	\(10\)
\(P(X)\)	\(\frac{25}{36}\)	\(\frac{10}{36}\)	\(\frac1{36}\)

\[E(X)=(-2)\cdot\frac{25}{36}+4\cdot\frac{10}{36}+10\cdot\frac1{36}=\frac{-50+40+10}{36}=0.\]

Answer: Since \(E(X)=0\), the game is fair.

Formulas at a glance: \(E(X)=\sum x_ip_i\); \(\mathrm{Var}(X)=E(X^2)-\{E(X)\}^2\); a probability distribution needs \(p_i\ge0\) and \(\sum p_i=1\). A game is fair when the expected gain is 0.