Chapter 32 of RD Sharma's Class 12 Mathematics textbook focuses on the mean and variance of random variables, and fundamental concepts in probability and statistics. Exercise 32.2 deals specifically with calculating these measures for discrete probability distributions. This set of solutions demonstrates various techniques for computing means, variances, and standard deviations, providing students with practical applications of probability theory.
Key Formulas and Theorems
Mean (Expected Value) of a discrete random variable:
E(X) = Σ x * P(X = x)
Variance of a discrete random variable:
Var(X) = E(X²) - [E(X)]²
where E(X²) = Σ x² * P(X = x)
Standard Deviation: σ = √Var(X)
Properties of Expected Value: E(aX + b) = aE(X) + b
Properties of Variance: Var(aX + b) = a²Var(X)
Solved Examples
Calculate the mean deviation from the median of the following frequency distribution :
Question 1(i): Find the mean and standard deviation of each of the following probability distributions:
xi: 2 3 4
pi: 0.3 0.5 0.3
Solution:
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
Question 1(ii): Find the mean and standard deviation of each of the following probability distributions:
xi: 1 3 4 5
pi: 0.4 0.1 0.2 0.3
Solution:
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
Question 1(iii): Find the mean and standard deviation of each of the following probability distributions:
xi: -5 -4 1 2
pi: 1/4 1/8 1/2 1/8
Solution:
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
Question 1(iv): Find the mean and standard deviation of each of the following probability distributions:
xi: -1 0 1 2 3
pi: 0.3 0.1 0.1 0.3 0.2
Solution:
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
Question 1(v): Find the mean and standard deviation of each of the following probability distributions:
xi: 1 2 3 4
pi: 0.4 0.3 0.2 0.1
Solution:
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
Question 1(vi): Find the mean and standard deviation of each of the following probability distributions:
xi: 0 1 3 5
pi: 0.2 0.5 0.2 0.1
Solution:
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
Question 1(vii): Find the mean and standard deviation of each of the following probability distributions:
xi: -2 -1 0 1 2
pi: 0.1 0.2 0.4 0.2 0.1
Solution:
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
Question 1(viii): Find the mean and standard deviation of each of the following probability distributions:
xi: -3 -1 0 1 3
pi: 0.05 0.45 0.20 0.25 0.05
Solution:
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
Question 1(ix): Find the mean and standard deviation of each of the following probability distributions:
xi: 0 1 2 3 4 5
pi: 1/6 5/18 2/9 1/6 1/9 1/18
Solution:
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
Question 2: A discrete random variable X has the probability distribution given below:
X: 0.5 1 1.5 2
P(X): k k2 2k2 k
(i) Find the value of k. (ii) Determine the mean of the distribution.
Solution:
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.
Question 3: Find the mean-variance and standard deviation of the following probability distribution
Xi: a b
Pi: p q
Where p+q=1.
Solution:
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
Question 4: Find the mean and variance of the number of tails in three tosses of a coin.
Solution:
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(AÕˆB) = 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
Question 5: Two cards are drawn simultaneously from a pack of 52 cards. Compute the mean and standard deviation of the number of kings.
Solution:
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 = 48x47/52x51 = 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
Question 6: Find the mean, variance and standard deviation of the number of tails in three tosses of a coin.
Solution:
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(AÕˆB) = 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
Question 7: Two bad eggs are accidentally mixed up with ten good ones. Three eggs are drawn at random with replacement from this lot. Compute the mean for the number of bad eggs drawn.
Solution:
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
Question 8: A pair of fair dice is thrown. Let X be the random variable which denotes the minimum of the two numbers which appear. Find the probability distribution, mean and variance of X.
Solution:
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
Question 9: A fair coin is tossed four times. Let X denote the number of heads occurring. Find the probability distribution, mean and variance of X.
Solution:
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 occurring 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(AÕˆB) = 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
Question 10: A fair die is tossed. Let X denote twice the number appearing. Find probability distribution, mean and variance of X.
Solution:
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.
Question 11: A fair die is tossed. Let X denote 1 or 3 according as an odd or an even number appears. Find the probability distribution, mean and variance of X.
Solution:
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.
Question 12: A fair coin is tossed four times. Let X denote the longest string of heads occurring. Find the probability distribution, mean and variance of X.
Solution:
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(AÕˆB) = 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
Practice Questions
1. A random variable X has the following probability distribution: P(X = 0) = 0.2, P(X = 1) = 0.3, P(X = 2) = 0.4, and P(X = 3) = 0.1. Find the expected value and variance of X.
2. The weights of a certain product follow a normal distribution with a mean of 20 grams and a standard deviation of 4 grams. Find the probability that a randomly selected product weighs more than 24 grams.
3. The number of defective items in a batch of 50 items follows a Poisson distribution with a mean of 2. Find the probability that the number of defective items is less than 4.
4. The lifetime of a certain type of light bulb is a random variable with a mean of 1200 hours and a standard deviation of 60 hours. Find the probability that a randomly selected light bulb lasts more than 1300 hours.
5. The number of customers arriving at a bank follows a Poisson distribution with a mean of 5 customers per hour. Find the probability that the number of customers arriving in a 20-minute period is at most 2.
6. The length of time a student spends on homework follows an exponential distribution with a mean of 2 hours. Find the probability that a student spends less than 1 hour on homework.
7. The number of typos in a document follows a Poisson distribution with a mean of 4 typos per page. Find the probability that a randomly selected page has at least 6 typos.
8. The heights of adults in a certain population follow a normal distribution with a mean of 160 cm and a standard deviation of 12 cm. Find the probability that a randomly selected adult is between 150 cm and 170 cm tall.
9. The number of accidents at a certain intersection follows a Poisson distribution with a mean of 3 accidents per month. Find the probability that the number of accidents in a given month is exactly 2.
10. The lifetime of a certain type of battery follows an exponential distribution with a mean of 600 hours. Find the probability that a randomly selected battery lasts more than 800 hours.
Summary
Exercise 32.2 (Set 1) in Chapter 32 (Mean and Variance of a Random Variable) of RD Sharma's Class 12 mathematics textbook focuses on finding the mean and variance of random variables, including discrete and continuous probability distributions. The exercise covers topics such as calculating the expected value and standard deviation of random variables, applying the properties of mean and variance, and solving real-world problems involving random variables. This set of problems helps students develop a comprehensive understanding of the concepts of mean and variance and their applications in probability and statistics.
