第12类RD Sharma解–第32章随机变量的均值和方差–练习32.2 |套装1

计算以下频率分布的中位数的平均偏差：

问题1(i)：找出以下每种概率分布的均值和标准差：

西：2 3 4

圆周率：0.3 0.5 0.3

解决方案：

Mean of any probability distribution is given by Mean = ∑xipi

Standard Deviation is given by SD = √Variance where variance is given by:

Variance = ∑ xi²pi – (∑xipi)²

∴ first we need to find the products i.e. pixi and pixi² and add them to get mean and apply the above formula to get the variance.

Following table gives the required products:

xi	pi	xipi	xi2pi
2	0.2	0.4	0.8
3	0.3	1.5	4.5
4	0.5	1.2	4.8

∴ mean = 0.4 + 1.5 + 1.2 = 3.1

And variance = 0.8 + 4.5 + 4.8 – (3.1)² = 0.49

∴ Standard deviation = √ 0.49 = 0.7

问题1(ii)：求出以下每种概率分布的均值和标准差：

xi：1 3 4 5

圆周率：0.4 0.1 0.2 0.3

解决方案：

Mean of any probability distribution is given by Mean = ∑xipi

Standard Deviation is given by SD = √ Variance where variance is given by:

Variance = ∑ xi²pi – (∑xipi)²

∴ first we need to find the products i.e. pixi and pixi² and add them to get mean and apply the above formula to get the variance.

Following table gives the required products:

xi	pi	xipi	xi2pi
1	0.4	0.4	0.4
3	0.1	0.3	0.9
4	0.2	0.8	3.2
5	0.3	1.5	7.5

∴ mean = 0.4 + 0.3+0.8+1.5= 3.0

And variance =0.4+0.9+3.2+7.5 – (3.0)² = 3

∴ Standard deviation = √ 3= 1.732

问题1(iii)：求出以下每种概率分布的均值和标准差：

xi：-5 -4 1 2

圆周率：1/4 1/8 1/2 1/8

解决方案：

Mean of any probability distribution is given by Mean = ∑xipi

Standard Deviation is given by SD = √ Variance where variance is given by:

Variance = ∑ xi²pi – (∑xipi)²

∴ first we need to find the products i.e. pixi and pixi² and add them to get mean and apply the above formula to get the variance.

Following table gives the required products:

xi	pi	xipi	xi2pi
-5	1/4	-1.25	6.25
-4	1/8	-0.5	2
1	1/2	0.5	0.5
2	1/8	0.25	0.5

∴ mean = -1.25-0.5+0.5+0.25 = -1

And variance = 6.25+2+0.5+0.5 – (-1)² = 8.25

∴ Standard deviation = √8.25= 2.9

问题1(iv)：找到以下每种概率分布的均值和标准差：

xi：-1 0 1 2 3

圆周率：0.3 0.1 0.1 0.3 0.2

解决方案：

Mean of any probability distribution is given by Mean = ∑xipi

Standard Deviation is given by SD = √ Variance where variance is given by:

Variance = ∑ xi²pi – (∑xipi)²

∴ first we need to find the products i.e. pixi and pixi² and add them to get mean and apply the above formula to get the variance.

Following table gives the required products:

xi	pi	xipi	xi2pi
-1	0.3	-0.3	0.3
0	0.1	0	0
1	0.1	0.1	0.1
2	0.3	0.6	1.2
3	0.2	0.6	1.8

∴ mean = -0.3 + 0 + 0.1 + 0.6 + 0.6 = 1.0

And variance =0.3 + 0 + 0.1 + 1.2 + 1.8 – (1)² = 2.4

∴ Standard deviation = √2.4 = 1.5

问题1(v)：找出以下每种概率分布的均值和标准差：

xi：1 2 3 4

圆周率：0.4 0.3 0.2 0.1

解决方案：

Mean of any probability distribution is given by Mean = ∑xipi

Standard Deviation is given by SD = √Variance where variance is given by:

Variance = ∑ xi²pi – (∑xipi)²

∴ first we need to find the products i.e. pixi and pixi² and add them to get mean and apply the above formula to get the variance.

Following table gives the required products :

xi	pi	xipi	xi2pi
1	0.4	0.4	0.4
2	0.3	0.6	1.2
3	0.2	0.6	1.8
4	0.1	0.4	1.6

∴ mean = 0.4+0.6+0.6+0.4 = 2.0

And variance = 0.4 +1.2 + 1.8 + 1.6– (2)² = 1.0

∴ Standard deviation = √1 = 1

问题1(vi)：求出以下每种概率分布的均值和标准差：

xi：0 1 3 5

圆周率：0.2 0.5 0.2 0.1

解决方案：

Mean of any probability distribution is given by Mean = ∑xipi

Standard Deviation is given by SD = √ Variance where variance is given by:

Variance = ∑ xi²pi – (∑xipi)²

∴ first we need to find the products i.e. pixi and pixi² and add them to get mean and apply the above formula to get the variance.

Following table gives the required products:

xi	pi	xipi	xi2pi
0	0.2	0	0
1	0.5	0.5	0.5
3	0.2	0.6	1.8
5	0.1	0.5	2.5

∴ mean = 0+0.5+0.6+0.5 = 1.6

And variance = 0 +0.5 + 1.8 + 2.5– (1.6)² = 2.24

∴ Standard deviation = √2.24 = 1.497

问题1(vii)：找到以下每种概率分布的均值和标准差：

xi：-2 -1 0 1 2

圆周率：0.1 0.2 0.4 0.2 0.1

解决方案：

Mean of any probability distribution is given by Mean = ∑xipi

Standard Deviation is given by SD = √ Variance where variance is given by:

Variance = ∑ xi²pi – (∑xipi)²

∴ first we need to find the products i.e. pixi and pixi² and add them to get mean and apply the above formula to get the variance.

Following table gives the required products:

xi	pi	xipi	xi2pi
-2	0.1	-0.2	0.4
-1	0.2	-0.2	0.2
0	0.4	0	0
1	0.2	0.2	0.2
2	0.1	0.2	0.4

∴ mean = -0.2-0.2+0+0.2+0.2 = 0

And variance = 0 +0.4+0.2+0.2+0.4– (0)² = 1.2

∴ Standard deviation = √1.2 = 1.095

问题1(viii)：找到以下每种概率分布的均值和标准差：

xi：-3 -1 0 1 3

圆周率：0.05 0.45 0.20 0.25 0.05

解决方案：

Mean of any probability distribution is given by Mean = ∑xipi

Standard Deviation is given by SD = √ Variance where variance is given by:

Variance = ∑ xi²pi – (∑xipi)²

∴ first we need to find the products i.e. pixi and pixi² and add them to get mean and apply the above formula to get the variance.

Following table gives the required products:

xi	pi	xipi	xi2pi
-3	0.05	-0.15	0.45
-1	0.45	-0.45	0.45
0	0.20	0	0
1	0.25	0.25	0.25
3	0.05	0.15	0.45

∴ mean = -0.15-0.45+0+0.25+0.15 = -0.2

And variance = 0 +0.45+0.25+0.45+0.45– (-0.2)² = 1.56

∴ Standard deviation = √1.56 = 1.248

问题1(ix)：求出以下每种概率分布的均值和标准差：

xi：0 1 2 3 4 5

pi：1/6 5/18 2/9 1/6 1/9 1/18

解决方案：

Mean of any probability distribution is given by Mean = ∑xipi

Standard Deviation is given by SD = √ Variance where variance is given by:

Variance = ∑ xi²pi – (∑xipi)²

∴ first we need to find the products i.e. pixi and pixi² and add them to get mean and apply the above formula to get the variance.

Following table gives the required products :

xi	pi	xipi	xi2pi
0	1/6	0	0
1	5/18	5/18	5/18
2	2/9	4/9	8/9
3	1/6	1/2	3/2
4	1/9	4/9	16/9
5	1/18	5/18	25/18

∴ Mean =  0+5/18+4/9+1/2+4/9+5/18 = 35/18

Variance =  0+5/18+8/9+3/2+16/9+25/18 -{35/18)² = 665/324  

∴ standard deviation = √ (665/324) = √665/18

问题2：离散随机变量X具有以下给出的概率分布：

X：0.5 1 1.5 2

P(X)：kk ² 2k ² k

(i)找出k的值。 (ii)确定分布的平均值。

解决方案：

To find the value of k we will be using the very basic idea of probability.

Note: We know that the sum of the probabilities of all random variables taken from a given sample space is equal to 1.

∴ P(X=0.5) + P(X=1) + P(X=1.5) + P(X=2) = 1

∴ k + k² + 2k² + k = 1

⇒ 3k² + 2k – 1 = 0

⇒ 3k² + 3k – k – 1 = 0

⇒ 3k(k+1) – (k+1) = 0

⇒ (3k-1)(k+1) = 0

∴ k = 1/3 or k = -1

∵ k represents probability of an event. Hence 0≤P(X)≤1

∴ k = 1/3

Mean of any probability distribution is given by- Mean = ∑xipi

Now we have,

X: 0.5 1 1.5 2

P(X): 1/3 1/9 2/9 1/3

∴ first we need to find the product i.e. pixi and add them to get mean.

∴ Mean = 0.5 x (1/3) + 1 x (1/9) + 1.5 x (2/9) +2 x (1/3) = 23/18.

问题3：找出以下概率分布的均值方差和标准差

习：ab

Pi：pq

其中p + q = 1。

解决方案：

Mean of any probability distribution is given by Mean = ∑xipi

Standard Deviation is given by SD = √ Variance where variance is given by:

Variance = ∑ xi²pi – (∑xipi)²

∴ first we need to find the products i.e. pixi and pixi² and add them to get mean and apply the above formula to get the variance.

∴ p₁x₁ = ap and p₂x₂ = bq Similarly p₁x₁² = a²p and p₂x₂²= b²q

∴ Mean = ap + bq

Variance = a²p + b²q – (ap + bq)²

=a²pq + b²pq + 2abpq [p + q=1]

=pq(a-b)²

∴ SD = √{pq(a-b)² } = |a-b|√pq

问题4：找出硬币三掷后的尾数均值和方差。

解决方案：

When we toss a coin three times we have the following possibilities:

{HHH,HHT,HTH,THH,HTT,THT,TTH,TTT}

Let X be a random variable representing number of tails in 3 tosses of a coin.

∵ probability of getting a head or probability of getting a tail are independent events and P(GETTING A HEAD) = P(GETTING A TAIL) = 1/2

∴ P(Head in first toss) and P(Head in second toss) and P(head in third toss) can be given by their individual products.

Note: P(AՈB) = P(A)P(B) where A and B are independent events.

Thus,

P(X=0) = P(HHH) = P(H)P(H)P(H) = 1/2 x 1/2 x 1/2 = 1/8

P(X=1) = P(HHT or HTH or THH) = P(HHT)+P(HTH)+P(THH)

= P(H)P(H)P(T)+ P(H)P(T)P(H)+ P(T)P(H)P(H)

= 1/2 x 1/2 x 1/2 + 1/2 x 1/2 x 1/2 + 1/2 x 1/2 x 1/2

= 3/8

P(X=2) = P(HTT or THT or TTH) = P(HTT)+P(THT)+P(TTH)

= P(H)P(T)P(T)+ P(T)P(H)P(T)+ P(T)P(T)P(H)

= 1/2 x 1/2 x 1/2 + 1/2 x 1/2 x 1/2 + 1/2 x 1/2 x 1/2

= 3/8

P(X=3) = P(TTT) = P(T)P(T)P(T) = 1/2 x 1/2 x 1/2 = 1/8

Now we have pi and xi.

Let’s proceed to find mean and variance.

Mean of any probability distribution is given by Mean = ∑xipi

Variance is given by:

Variance = ∑ xi²pi – (∑xipi)²

∴ first we need to find the products i.e. pixi and pixi² and add them to get mean and apply the above formula to get the variance.

Following table gives the required products: 

xi	pi	xipi	xi2pi
0	1/8	0	0
1	3/8	3/8	3/8
2	3/8	3/4	3/2
3	1/8	3/8	9/8

∴ Mean = 0 + 3/8 + 3/8 + 1/8 = 3/2

Variance = 0 + 3/8 + 3/4 + 3/8 – (3/2)² = 3/4

问题5：从52张纸牌中同时抽出两张纸牌。计算国王数的均值和标准差。

解决方案：

In a deck of 52 cards there are 4 kings each of one suit respectively.

Let X be the random variable denoting the number of kings for an event when two cards are drawn simultaneously.

∴ X can take values 0 , 1 or 2. 

P(X=0) = 48C₂/52C₂    = 48×47/52×51 = 188/221

[For selecting 0 kings, we removed all 4 kings from deck and selected out of 48]

P(X=1) = 4C₁ x 48C₁/52C₂ = 48 x 4 x 2/52 x 51 = 32/221

[For selecting 1 king, we need to select and 1 out of 4 and not any other]

P(X=2) = 4C₂/52C₂ = 4 x 3/52 x 51 = 1/221

[For selecting 2 king, we need to select and 2 out of 4]

Now we have pi and xi.

Let’s proceed to find mean and standard deviation.

Mean of any probability distribution is given by Mean = ∑xipi

Standard Deviation is given by SD = √ Variance where variance is given by:

Variance = ∑ xi²pi – (∑xipi)²

∴ first we need to find the products i.e. pixi and pixi² and add them to get mean and apply the above formula to get the variance.

Following table gives the required products:

xi	pi	xipi	xi2pi
0	188/221	0	0
1	32/221	32/221	32/221
2	1/221	2/221	4/221

∴ mean = 0 + 32/221 + 2/221 = 34/221

Variance = 0 + 32/221 + 4/221 – (34/221) = 400/2873

∴ Standard deviation = √variance = √(400/2873) = 20/√2873

问题6：求出三枚硬币的尾数的均值，方差和标准差。

解决方案：

When we toss a coin three times we have the following possibilities:

{HHH,HHT,HTH,THH,HTT,THT,TTH,TTT}

Let X be a random variable representing number of tails in 3 tosses of a coin.

∵ probability of getting a head or probability of getting a tail are independent events and P(GETTING A HEAD) = P(GETTING A TAIL) = 1/2

∴ P(Head in first toss) and P(Head in second toss) and P(head in third toss) can be given by their individual products.

Note: P(AՈB) = P(A)P(B) where A and B are independent events.

Thus,

P(X=0) = P(HHH) = P(H)P(H)P(H) = 1/2 x 1/2 x 1/2 = 1/8

P(X=1) = P(HHT or HTH or THH) = P(HHT)+P(HTH)+P(THH)

= P(H)P(H)P(T)+ P(H)P(T)P(H)+ P(T)P(H)P(H)

= 1/2 x 1/2 x 1/2 + 1/2 x 1/2 x 1/2 + 1/2 x 1/2 x 1/2

= 3/8

P(X=2) = P(HTT or THT or TTH) = P(HTT)+P(THT)+P(TTH)

= P(H)P(T)P(T)+ P(T)P(H)P(T)+ P(T)P(T)P(H)

= 1/2 x 1/2 x 1/2 + 1/2 x 1/2 x 1/2 + 1/2 x 1/2 x 1/2

= 3/8

P(X=3) = P(TTT) = P(T)P(T)P(T) = 1/2 x 1/2 x 1/2 = 1/8

Now we have pi and xi.

Let’s proceed to find mean and variance.

Mean of any probability distribution is given by Mean = ∑xipi

Variance is given by:

Variance = ∑ xi²pi – (∑xipi)²

Standard Deviation is given by SD = √Variance

∴ first we need to find the products i.e. pixi and pixi² and add them to get mean and apply the above formula to get the variance.

Following table gives the required products :

xi	pi	xipi	xi2pi
0	1/8	0	0
1	3/8	3/8	3/8
2	3/8	3/4	3/2
3	1/8	3/8	9/8

∴ Mean = 0 + 3/8 + 3/4 + 3/8 = 3/2

Variance = 0 + 3/8 + 3/2 + 9/8  – (3/2)²  = 3/4

Standard Deviation = √(3/4) = 0.87

问题7：两个坏蛋不小心混入了十个好蛋。从该批次中随机抽取三个鸡蛋。计算抽取的坏蛋数的平均值。

解决方案：

As there are total of two bad eggs. Therefore while drawing 3 eggs we can draw 1 bad egg or 2 or 0 bad eggs.

Let X be the random variable denoting number of bad eggs that can be drawn in each draw.

Clearly X can take values 0,1 or 2

P(X=0) = P(all 3 are good eggs) = 2C₀ x 10C₃ /12C₃ = 120/220 = 6/11

[Since there are 10 good eggs so for selecting all good we took all three from 10 and 0 eggs from 2 bad ones. Total sample points are no of ways of selecting 3 eggs from total of 12 eggs]

Similarly,

P(X=1) = P(1 bad and 2 good eggs) =  2C₁ x 10C₂/12C₃  = 9/22 

P(X=2) = P(2 Bad eggs and 1 good egg) =2C₂ x 10C₁ /12C₃ = 1/22

Now we have pi and xi.

Let’s proceed to find mean

Mean of any probability distribution is given by Mean = ∑xipi

∴ first we need to find the products i.e. pixi and add them to get mean.

Following table gives the required products :

xi	pi	xipi
0	6/11	0
1	9/22	9/22
2	1/22	1/11

∴ mean = 0 + 9/22 + 1/11 = 1/2

问题8：掷出一对骰子。令X为随机变量，它表示出现的两个数字中的最小值。查找X的概率分布，均值和方差。

解决方案：

When a pair of fair dice is thrown there are total 36 possible outcomes.

X denotes the minimum of two numbers which appear

∴ X can take values 1,2,3,4,5 and 6

P(X=1) = 11/36

[Possible Pairs: (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(3,1),(4,1),(5,1),(6,1)]

P(X=2) = 9/36

[Possible Pairs: (2,2),(3,2),(4,2),(5,2),(6,2),(2,6),(2,5),(2,4),(2,3)]

P(X=3) = 7/36

[Possible Pairs: (3,3),(3,4),(4,3),(5,3),(3,5),(3,6),(6,3)]

P(X=4) = 5/36

[Possible Pairs: (4,4),(5,4),(4,5),(4,6),(6,4)]

P(X=5) = 3/36

[Possible Pairs (5,5),(5,6),(6,5)]

P(X=6) = 1/36

[Possible Pairs: (6,6)]

Now we have pi and xi.

Let’s proceed to find mean and variance.

Mean of any probability distribution is given by Mean = ∑xipi

Variance is given by:

Variance = ∑ xi²pi – (∑xipi)²

Standard Deviation is given by SD = √Variance

∴ first we need to find the products i.e. pixi and pixi² and add them to get mean and apply the above formula to get the variance.

Following table gives the required products:

Required Probability distribution table:

xi	pi	xipi	xi2pi
1	11/36	11/36	11/36
2	9/36	18/36	1
3	7/36	21/36	63/36
4	5/36	20/36	80/36
5	3/36	15/36	75/36
6	1/36	6/36	1

∴ Mean = 11/36 + 18/36 + 21/ 36 + 20/36 + 15/36 + 6/36 = 91/36

Variance = 11/36 + 1 + 63/36 + 80/36 + 75/36 + 1 – (91/36)² =2555/1296

Standard deviation = √variance = 1.403

问题9：一个公平的硬币被扔了四次。令X表示出现的磁头数量。查找X的概率分布，均值和方差。

解决方案：

Say, H represents event of getting a head and T represents getting a tail.

When we toss a coin 4 times we have the following possibilities:

{HHHH,HHHT,HHTH,THHH,HTHH,THHT,TTHH,HHTT,THTH…………,TTTT}

A total of 24 = 16 possibilities.

Let X be a random variable representing number of heads occuring in 4 tosses of a coin.

∵ probability of getting a head or probability of getting a tail are independent events and P(GETTING A HEAD) = P(GETTING A TAIL) = 1/2

∴ P(Head in first toss) and P(Head in second toss) and P(head in third toss) and P(tail in 4th toss) can be given by their individual products.

Note: P(AՈB) = P(A)P(B) where A and B are independent events.

Thus,

P(X=0) = P(TTTT) = P(T)P(T)P(T)P(T) = 1/2 x 1/2 x 1/2 x 1/2 = 1/16

Selecting a coin out of 4 which will show head rest all showing tail

= 4C₁ x P(HHHT) = 4C₁ x(1/2) x (1/2) x (1/2) x (1/2) = 1/4

similarly ,

P(X=2) = 4C₂ x(1/2)⁴ = 3/8

P(X=3) = 4C₃ x (1/2)⁴ = 1/4

P(X=4) = P(HHHH) = 1/16

Now we have pi and xi.

Let’s proceed to find mean and variance.

Mean of any probability distribution is given by Mean = ∑xipi

Variance is given by:

Variance = ∑ xi²pi – (∑xipi)²

∴ first we need to find the products i.e. pixi and pixi² and add them to get mean and apply the above formula to get the variance.

Following table representing probability distribution gives the required products :

xi	pi	xipi	xi2pi
0	1/16	0	0
1	1/4	1/4	1/4
2	3/8	3/4	3/2
3	1/4	3/4	9/4
4	1/16	1/4	1

∴ Mean = 0+ 1/4 + 3/4 + 3/4 + 1/4 = 2

Variance = 0 + 1/4 + 3/2 + 9/4 + 1 – (2)²  = 1

问题10：一个公平的死者被扔了。令X代表出现的数字的两倍。查找X的概率分布，均值和方差。

解决方案：

When a fair dice is thrown there are total 6 possible outcomes.

∵ X denote twice the number appearing on die

∴ X can take values 2,4,6,8,10 and 12

As appearance of a number on a fair die is equally likely

i.e. P(appearing of 1) = P(appearing of 2) = P(appearing of 3) = P(appearing of 4) = P(appearing of 5) = P(appearing of 6) = 1/6

∴ appearance of twice of the number is also equally likely with a probability of 1/6.

P(X=2)=P(X=4)=P(X=6)=P(X=8)=P(X=10)=P(X=12)=1/6

Now we have pi and xi.

Let’s proceed to find mean and variance.

Mean of any probability distribution is given by Mean = ∑xipi

Variance is given by:

Variance = ∑xi²pi – (∑xipi)²

∴ first we need to find the products i.e. pixi and pixi² and add them to get mean and apply the above formula to get the variance.

Following table gives the required products:

Required Probability distribution table:

xi	pi	xipi	xi2pi
2	1/6	2/6	4/6
4	1/6	4/6	16/6
6	1/6	1	36/6
8	1/6	4/3	64/6
10	1/6	5/3	100/6
12	1/6	2	144/6

∴ Mean = 2/6 + 4/6 + 1 + 4/3 + 5/3 + 2 = 7

Variance = 4/6 + 16/6 + 36/6 + 64/6 + 100/6 + 144/6 – 7² = 70/6.

问题11：一个公平的死者被扔了。当出现奇数或偶数时，让X表示1或3。查找X的概率分布，均值和方差。

解决方案：

When a fair dice is thrown there are total 6 possible outcomes.

∵ X denote 1 or 3 according as an odd or an even number appears.

P(appearing of even number on a die) = 3/6 [favourable outcomes {2,4,6}]

P(appearing of an odd number on a die) = 3/6 [favourable outcomes {1,4,3}]

P(X=1) = 3/6 = 1/2

P(X=3) = 3/6 = 1/2

Now we have pi and xi.

Let’s proceed to find mean and variance.

Mean of any probability distribution is given by Mean = ∑xipi

Variance is given by:

Variance = ∑ xi²pi – (∑xipi)²

∴ first we need to find the products i.e. pixi and pixi² and add them to get mean and apply the above formula to get the variance.

Following table gives the required products :

Required Probability distribution table:-

xi	pi	xipi	xi2pi
1	1/2	1/2	1/2
3	1/2	3/2	9/2

∴ Mean = 1/2 + 3/2 = 2

Variance = 1/2 + 9/2 – (2)² = 1.

问题12：一个公平的硬币被扔了四次。令X表示出现的最长的字符串。查找X的概率分布，均值和方差。

解决方案：

Say, H represents event of getting a head and T represents getting a tail.

When we toss a coin 4 times we have the following possibilities:

{HHHH,HHHT,HHTH,THHH,HTHH,THHT,TTHH,HHTT,THTH…………,TTTT}

A total of 24 = 16 possibilities.

∵ probability of getting a head or probability of getting a tail are independent events and P(GETTING A HEAD) = P(GETTING A TAIL) = 1/2

∴ P(Head in first toss) and P(Head in second toss) and P(head in third toss) and P(tail in 4th toss) can be given by their individual products.

Note: P(AՈB) = P(A)P(B) where A and B are independent events.

As X is a random variable representing longest string of head occurring in 4 tosses.

∴ X can take following values:

X = 0 [ all tails (TTTT) ]

X = 1 [Longest string contains only 1 head e.g. (HTTT),(TTTH),(HTHT)..]

X = 2 [ Longest string contain only 2 head e.g. (HHTT),(HHTH),(THHT)…]

X = 3 [Longest string contain only 3 head e.g. ( HHHT) And (THHH)]

X = 4 [ Longest string contain 4 heads i.e. (HHHH) ]

Thus,

P(X=0) = 1/16

P(X=1) = 7/16 [by counting number of favourable outcomes as explained]

P(X=2) = 5/16

P(X=3) = 2/16

P(X=4) = 1/16

Now we have pi and xi.

Let’s proceed to find mean and variance.

Mean of any probability distribution is given by Mean = ∑xipi

Variance is given by:

Variance = ∑xi²pi – (∑xipi)²

∴ first we need to find the products i.e. pixi and pixi² and add them to get mean and apply the above formula to get the variance.

Following table representing probability distribution gives the required products :

xi	pi	xipi	xi2pi
0	1/16	0	0
1	7/16	7/16	7/16
2	5/16	10/16	20/16
3	2/16	6/16	18/16
4	1/16	1/4	1

∴ Mean = 0 + 7/16 + 10/16 + 6/16 + 1/4 = 1.7

Variance = 0 + 7/16 + 20/16 + 18/16 + 1 – (1.7)² = 0.935