NCERT Solutions for Exercise 13.4 Class 12 Maths Chapter 13

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NCERT Solutions for Exercise 13.4 Class 12 Maths Chapter 13 - Probability

Updated on 04 Dec 2023, 11:27 AM IST

NCERT Solutions For Class 12 Maths Chapter 13 Exercise 13.4

NCERT Solutions for Exercise 13.4 Class 12 Maths Chapter 13 Probability are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. NCERT solutions for exercise 13.4 Class 12 Maths chapter 13 you will learn some important concepts like random variable, the probability distribution of random variable, calculating mean and variance of a probability distribution. . The concept of the random variable is a very important concept in statistics. You are advised to be thorough with Class 12 Maths chapter 13 exercise 13.4 as this is very frequently asked in the board exams. There are 17 questions in Class 12th Maths chapter 13 exercise 13.4 which you should try to solve on your own. You can take help from these exercise 13.4 Class 12 Maths solutions if you are not able to solve them on your own. These NCERT book solutions are prepared in a detailed manner which could be understood very easily.

12th class Maths exercise 13.4 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise enumerated in NCERT Book together using the link provided below.

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Probability Class 12 Chapter 13-Exercise: 13.4

Question:1(i) State which the following are not the probability distributions of a random variable. Give reasons for your answer.

Answer:

As we know the sum of probabilities of a probability distribution is 1.

Sum of probabilities $=0.4+0.4+0.2=1$

$\therefore$ The given table is the probability distributions of a random variable.

Question:1(ii) State which of the following are not the probability distributions of a random variable. Give reasons for your answer.

Answer:

As we know probabilities cannot be negative for a probability distribution .

$P(3) = -0.1$

$\therefore$ The given table is not a the probability distributions of a random variable.

Question:1(iii) State which of the following are not the probability distributions of a random variable. Give reasons for your answer.

Answer:

As we know sum of probabilities of a probability distribution is 1.

Sum of probablities $=0.6+0.1+0.2=0.9\neq 1$

$\therefore$ The given table is not a the probability distributions of a random variable because sum of probabilities is not 1.

Question:1(iv) State which of the following are not the probability distributions of a random variable. Give reasons for your answer.

Answer:

As we know sum of probabilities of a probability distribution is 1.

Sum of probablities $=0.3+0.2+0.4+0.1+0.05=1.05\neq 1$

$\therefore$ The given table is not a the probability distributions of a random variable because sum of probabilities is not 1.

Question:2 An urn contains $5$ red and $2$ black balls. Two balls are randomly drawn. Let $X$ represent the number of black balls. What are the possible values of $X?$ Is $X$ a random variable ?

Answer:

B = black balls

R = red balls

The two balls can be selected as BR,BB,RB,RR.

X = number of black balls.

$\therefore$ $X(BB)=2$

$X(RB)=1$

$X(BR)=1$

$X(RR)=0$

Hence, possible values of X can be 0, 1 and 2.

Yes, X is a random variable.

Question:3 Let $X$ represent the difference between the number of heads and the number of tails obtained when a coin is tossed $6$ times. What are possibl valuess of $X$?

Answer:

The difference between the number of heads and the number of tails obtained when a coin is tossed $6$ times are :

$\therefore$ $X(6H,0T)=\left | 6-0 \right |=6$

$\Rightarrow \, \, X(5H,1T)=\left | 5-1 \right |=4$

$\Rightarrow \, \, X(4H,2T)=\left | 4-2 \right |=2$

$\Rightarrow \, \, X(3H,3T)=\left | 3-3 \right |=0$

$\Rightarrow \, \, X(2H,4T)=\left | 2-4\right |=2$

$\Rightarrow \, \, X(1H,5T)=\left | 1-5\right |=4$

$\Rightarrow \, \, X(0H,6T)=\left |0-6\right |=6$

Thus, possible values of X are 0, 2, 4 and 6.

Question:4(i) Find the probability distribution of

number of heads in two tosses of a coin.

Answer:

When coin is tossed twice then sample space $=\left \{ HH,HT,TH,TT \right \}$

Let X be number of heads.

$\therefore$ $X(HH)=2$

$X(HT)=1$

$X(TH)=1$

$X(TT)=0$

X can take values of 0,1,2.

$P(HH)=P(HT)=P(TH)=P(TT)=\frac{1}{4}$

$P(X=0)=P(TT)=\frac{1}{4}$

$P(X=2)=P(HH)=\frac{1}{4}$

$P(X=1)=P(HT)+P(TH)=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$

Table is as shown :

X	0	1	2
P(X)	$\frac{1}{4}$	$\frac{1}{2}$	$\frac{1}{4}$

Question:4(ii) Find the probability distribution of

number of tails in the simultaneous tosses of three coins.

Answer:

When 3 coins are simultaneous tossed then sample space $=\left \{ HHH,HHT,TTH,TTT,HTH,THT,THH,HTT \right \}$

Let X be number of tails.

$\therefore$ X can be 0,1,2,3

X can take values of 0,1,2.

$P(X=0)=P(HHH)=\frac{1}{8}$

$P(X=1)=P(HHT)+P(HTH)+P(THH)=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8}$

$P(X=2)=P(THT)+P(HTT)+P(TTH)=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8}$

$P(X=3)=P(TTT)=\frac{1}{8}$

Table is as shown :

X	0	1	2	3
P(X)	$\frac{1}{8}$	$\frac{3}{8}$	$\frac{3}{8}$	$\frac{1}{8}$

Question:4(iii) Find the probability distribution of

number of heads in four tosses of a coin.

Answer:

When coin is tossed 4 times then sample space $=\left \{ HHHH,HHHT,TTTH,HTHH,THHT,TTHH,HHTT ,TTTT,HTTH,THTH,HTHT,HTTT,THHH,THTT,HHTH,TTHT\right \}$

Let X be number of heads.

$\therefore$ X can be 0,1,2,3,4

$P(X=0)=P(TTTT)=\frac{1}{16}$

$P(X=1)=P(HTTT)+P(TTTH)+P(THTT)=(TTHT)=\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}=\frac{4}{16}=\frac{1}{4}$

$P(X=2)=P(THHT)+P(HHTT)+P(HTTH)+P(TTHH)+P(HTHT)+P(THTH)=\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}=\frac{6}{16}=\frac{3}{8}$

$P(X=3)=P(HHHT)+P(THHH)+P(HHTH)+P(HTHH)=\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}=\frac{4}{16}=\frac{1}{4}$

$P(X=4)=P(HHHH)=\frac{1}{16}$

Table is as shown :

X	0	1	2	3	4
P(X)	$\frac{1}{16}$	$\frac{1}{4}$	$\frac{3}{8}$	$\frac{1}{4}$	$\frac{1}{16}$

Question:5(i) Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as

number greater than 4

Answer:

When a die is tossed twice , total outcomes = 36

Number less than or equal to 4 in both toss : $P(X=0)=\frac{4}{6} \times \frac{4}{6}=\frac{4}{9}$

Number less than or equal to 4 in first toss and number more than or equal to 4 in second toss + Number less than or equal to 4 in second toss and number more than or equal to 4 in first toss: $P(X=1)=\frac{4}{6} \times \frac{2}{6}+ \frac{2}{6}\times \frac{4}{6}=\frac{4}{9}$

Number less than 4 in both tosses : $P(X=2)=\frac{2}{6} \times \frac{2}{6}= \frac{1}{9}$

Probability distribution is as :

X	0	1	2
P(X)	$\frac{4}{9}$	$\frac{4}{9}$	$\frac{1}{9}$

Question:5(ii) Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as

six appears on at least one die.

Answer:

When a die is tossed twice , total outcomes = 36

Six does not appear on any of the die : $P(X=0)=\frac{5}{6} \times \frac{5}{6}=\frac{25}{36}$

Six appear on atleast one die : $P(X=1)=\frac{1}{6} \times \frac{5}{6}+ \frac{1}{6}\times \frac{5}{6}=\frac{5}{18}$

Probability distribution is as :

X	0	1
P(X)	$\frac{25}{36}$	$\frac{5}{18}$

Question:6 From a lot of $30$ bulbs which include $6$ defectives, a sample of $4$ bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.

Answer:

Total bulbs = 30

defective bulbs = 6

Non defective bulbs $=30-6=24$

$P(defective \, \, \, \, \, bulbs)=\frac{6}{30}=\frac{1}{5}$

$P(Non \, \, \, \, defective \, \, \, \, \, bulbs)=\frac{24}{30}=\frac{4}{5}$

$4$ bulbs is drawn at random with replacement.

Let X : number of defective bulbs

4 Non defective bulbs and 0 defective bulbs : $P(X=0)={4}\textrm{C}_0\frac{4}{5}.\frac{4}{5}.\frac{4}{5}.\frac{4}{5}=\frac{256}{625}$

3 Non defective bulbs and 1 defective bulbs : $P(X=1)={4}\textrm{C}_1\frac{1}{5}.\frac{4}{5}.\frac{4}{5}.\frac{4}{5}=\frac{256}{625}$

2 Non defective bulbs and 2 defective bulbs : $P(X=2)={4}\textrm{C}_2\frac{1}{5}.\frac{1}{5}.\frac{4}{5}.\frac{4}{5}=\frac{96}{625}$

1 Non defective bulbs and 3 defective bulbs : $P(X=3)={4}\textrm{C}_3\frac{1}{5}.\frac{1}{5}.\frac{1}{5}.\frac{4}{5}=\frac{16}{625}$

0 Non defective bulbs and 4 defective bulbs : $P(X=4)={4}\textrm{C}_4\frac{1}{5}.\frac{1}{5}.\frac{1}{5}.\frac{1}{5}=\frac{1}{625}$

the probability distribution of the number of defective bulbs is as :

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

Question:7 A coin is biased so that the head is $3$ times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of number of tails.

Answer:

the coin is tossed twice, total outcomes =4 $=\left \{ HH,TT,HT,TH \right \}$

probability of getting a tail be x.

i.e. $P(T)=x$

Then $P(H)=3x$

$P(T)+P(H)=x+3x=1$

$4x=1$

$x=\frac{1}{4}$

$P(T)=\frac{1}{4}$ and $P(H)=\frac{3}{4}$

Let X : number of tails

No tail : $P(X=0)=P(H).P(H)=\frac{3}{4}\times \frac{3}{4}=\frac{9}{16}$

1 tail : $P(X=1)=P(HT)+P(TH)=\frac{3}{4}\times \frac{1}{4}+\frac{1}{4}\times \frac{3}{4}=\frac{3}{8}$

2 tail : $P(X=2)=P(TT)=\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}$

the probability distribution of number of tails are

X	0	1	2
P(X)	$\frac{9}{16}$	$\frac{3}{8}$	$\frac{1}{16}$

Question:8(i) A random variable X has the following probability distribution:

X	0	1	2	3	4	5	6	7
P(X)	0	k	2k	2k	3k	k²	2k²	7k²+k

$k$

Answer:

X	0	1	2	3	4	5	6	7
P(X)	0	k	2k	2k	3k	k²	2k²	7k²+k

Sum of probabilities of probability distribution of random variable is 1.

$\therefore \, \, \, \, 0+k+2k+2k+3k+k^{2}+2k^{2}+7k^{2}+k=1$

$10k^{2}+9k-1=0$

$(10k-1)(k+1)=0$

$k=\frac{1}{10}\, \, and\, \, k=-1$

Question:8(ii) A random variable $X$ has the following probability distribution:

X	0	1	2	3	4	5	6	7
P(X)	0	k	2k	2k	3k	k²	2k²	7k²+k

$P(X< 3)$

Answer:

$P(X< 3)=P(X=0)+P(X=1)+P(X=2)$

$=0+K+2K$

$=3K$

$=3\times \frac{1}{10}$

$= \frac{3}{10}$

Question:8(iii) A random variable $X$ has the following probability distribution:

1655186794057

$P(X> 6)$

Answer:

$P(X> 6)=P(X=7)$

$=7K^2+K$

$=7\times (\frac{1}{10})^2+\frac{1}{10}$

$=\frac{7}{100}+\frac{1}{10}$

$=\frac{17}{100}$

Question:8(iv) A random variable X has the following probability distribution:

1655186819360 $P(0< X< 3)$

Answer:

$P(0< X< 3)=P(X=1)+P(X=2)$

$=k+2k$

$=3k$

$=3\times \frac{1}{10}$

$= \frac{3}{10}$

Question:9(a) The random variable X has a probability distribution P(X) of the following form, where k is some number :

$P(X)=\left\{\begin{matrix} k, if&i x=0\\ 2k, if& x=1\\ 3k, if& x=2\\ 0,& otherwise \end{matrix}\right.$

Determine the value of $k.$

Answer:

Sum of probabilities of probability distribution of random variable is 1.

$\therefore \, \, \, \, k+2k+3k+0=1$

$6k=1$

$k=\frac{1}{6}$

Question:9(b) The random variable $X$ has a probability distribution $P(X)$ of the following form, where k is some number :

$P(X)=\left\{\begin{matrix} k,& if & x=0\\ 2k,& if& x=1\\ 3k,& if& x=2\\ 0,& otherwise& \end{matrix}\right.$

Find $P\left ( X< 2 \right ),P\left ( X\leq 2 \right ),P\left ( X\geq 2 \right ).$

Answer:

$P(X< 2)=P(X=0)+P(X=1)$

$=k+2k$

$=3k$

$=3\times \frac{1}{6}$

$= \frac{1}{2}$

$P(X\leq 2)=P(X=0)+P(X=1)+p(X=2)$

$=k+2k+3k$

$=6k$

$=6\times \frac{1}{6}$

$=1$

$P(X\geq 2)=P(X=2)+P(X> 2)$

$=3k+0$

$=3\times \frac{1}{6}=\frac{1}{2}$

Question:10 Find the mean number of heads in three tosses of a fair coin.

Answer:

Let X be the success of getting head.

When 3 coins are tossed then sample space $=\left \{ HHH,HHT,TTH,TTT,HTH,THT,THH,HTT \right \}$

$\therefore$ X can be 0,1,2,3

$P(X=0)=P(TTT)=\frac{1}{8}$

$P(X=1)=P(HTT)+P(TTH)+P(HTH)=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8}$

$P(X=2)=P(HHT)+P(HTH)+P(THH)=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8}$

$P(X=3)=P(HHH)=\frac{1}{8}$

The probability distribution is as

X	0	1	2	3
P(X)	$\frac{1}{8}$	$\frac{3}{8}$	$\frac{3}{8}$	$\frac{1}{8}$

mean number of heads :

$=0\times \frac{1}{8}+1\times \frac{3}{8}+2\times \frac{3}{8}+3\times \frac{1}{8}$

$= \frac{3}{8}+ \frac{6}{8}+ \frac{3}{8}$

$=\frac{12}{8}$

$=\frac{3}{2}=1.5$

Question:11 Two dice are thrown simultaneously. If $X$ denotes the number of sixes, find the expectation of $X$.

Answer:

$X$ denotes the number of sixes, when two dice are thrown simultaneously.

X can be 0,1,2.

$\therefore$ Not getting six on dice $P(X)=\frac{25}{36}$

Getting six on one time when thrown twice : $P(X=1)=2\times \frac{5}{6}\times \frac{1}{6}=\frac{10}{36}$

Getting six on both dice : $P(X=2)= \frac{1}{36}=\frac{1}{36}$

X	0	1	2
P(X)	$\frac{25}{36}$	$\frac{10}{36}$	$\frac{1}{36}$

Expectation of X = E(X)

$E(X)=0\times \frac{25}{36}+1\times \frac{10}{36}+2\times \frac{1}{36}$

$E(X)= \frac{12}{36}$

$E(X)= \frac{1}{3}$

Question:12 Two numbers are selected at random (without replacement) from the first six positive integers. Let $X$ denote the larger of the two numbers obtained. Find $E(X).$

Answer:

Two numbers are selected at random (without replacement) from the first six positive integers in $6\times 5=30$ ways.

$X$ denote the larger of the two numbers obtained.

X can be 2,3,4,5,6.

X=2, obsevations : $(1,2),(2,1)$

$P(X=2)=\frac{2}{30}=\frac{1}{15}$

X=3, obsevations : $(1,3),(3,1),(2,3),(3,2)$

$P(X=3)=\frac{4}{30}=\frac{2}{15}$

X=4, obsevations : $(1,4),(4,1),(2,4),(4,2),(3,4),(4,3)$

$P(X=4)=\frac{6}{30}=\frac{3}{15}$

X=5, obsevations : $(1,5),(5,1),(2,5),(5,2),(3,5),(5,3),(4,5),(5,4)$

$P(X=5)=\frac{8}{30}=\frac{4}{15}$

X=6, obsevations : $(1,6),(6,1),(2,6),(6,2),(3,6),(6,3),(4,6),(6,4),(5,6),(6,5)$

$P(X=6)=\frac{10}{30}=\frac{1}{3}$

Probability distribution is as follows:

X	2	3	4	5	6
P(X)	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{3}{15}$	$\frac{4}{15}$	$\frac{1}{3}$

$E(X)=2\times \frac{1}{15}+3\times \frac{2}{15}+4\times \frac{3}{15}+5\times \frac{4}{15}+6\times \frac{1}{3}$

$E(X)= \frac{2}{15}+ \frac{2}{5}+ \frac{4}{5}+ \frac{4}{3}+ \frac{2}{1}$

$E(X)= \frac{70}{15}$

$E(X)= \frac{14}{3}$

Question:13 Let $X$ denote the sum of the numbers obtained when two fair dice are rolled. Find the variance and standard deviation of$X$.

Answer:

$X$ denote the sum of the numbers obtained when two fair dice are rolled.

Total observations = 36

X can be 2,3,4,5,6,7,8,9,10,11,12

$P(X=2)=P(1,1)=\frac{1}{36}$

$P(X=3)=P(1,2)+P(2,1)=\frac{2}{36}=\frac{1}{18}$

$P(X=4)=P(1,3)+P(3,1)+P(2,2)=\frac{3}{36}=\frac{1}{12}$

$P(X=5)=P(1,4)+P(4,1)+P(2,3)+P(3,2)=\frac{4}{36}=\frac{1}{9}$

$P(X=6)=P(1,5)+P(5,1)+P(2,4)+P(4,2)+P(3,3)=\frac{5}{36}$

$P(X=7)=P(1,6)+P(6,1)+P(2,5)+P(5,2)+P(3,4)+P(4,3)=\frac{6}{36}=\frac{1}{6}$

$P(X=8)=P(2,6)+P(6,2)+P(3,5)+P(5,3)+P(4,4)=\frac{5}{36}$$P(X=9)=P(3,6)+P(6,3)+P(4,5)+P(5,4)=\frac{4}{36}=\frac{1}{9}$

$P(X=10)=P(4,6)+P(6,4)+P(5,5)=\frac{3}{36}=\frac{1}{12}$

$P(X=11)=P(5,6)+P(6,5)=\frac{2}{36}=\frac{1}{18}$

$P(X=12)=P(6,6)=\frac{1}{36}$

Probability distribution is as follows :

X	2	3	4	5	6	7	8	9	10	11	12
P(X)	$\frac{1}{36}$	$\frac{1}{18}$	$\frac{1}{12}$	$\frac{1}{9}$	$\frac{5}{36}$	$\frac{1}{6}$	$\frac{5}{36}$	$\frac{1}{9}$	$\frac{1}{12}$	$\frac{1}{18}$	$\frac{1}{36}$

$E(X)=2\times \frac{1}{36}+3\times \frac{1}{18}+4\times \frac{1}{12}+5\times \frac{1}{9}+6\times \frac{5}{36}+7\times \frac{1}{6}+8\times \frac{5}{36}+9\times \frac{1}{9}+10\times \frac{1}{12}+11\times \frac{1}{18}+12\times \frac{1}{36}$

$E(X)=\frac{1}{18}+\frac{1}{6}+\frac{1}{3}+\frac{5}{9}+\frac{5}{6}+\frac{7}{6}+\frac{10}{9}+1+\frac{5}{6}+\frac{11}{18}+\frac{1}{3}$

$E(X)=7$

$E(X^2)=4\times \frac{1}{36}+9\times \frac{1}{18}+16\times \frac{1}{12}+25\times \frac{1}{9}+36\times \frac{5}{36}+49\times \frac{1}{6}+64\times \frac{5}{36}+81\times \frac{1}{9}+100\times \frac{1}{12}+121\times \frac{1}{18}+144\times \frac{1}{36}$

$E(X^2)=\frac{987}{18}=\frac{329}{6}=54.833$

$Variance = E(X^2)-(E(X))^2$

$=54.833-7^2$

$=54.833-49$

$=5.833$

Standard deviation =$=\sqrt{5.833}=2.415$

Question:14 A class has $15$ students whose ages are $14,17,15,14,21,17,19,20,16,18,20,17,16,19$ and $20$ years. One student is selected in such a manner that each has the same chance of being chosen and the age X of the selected student is recorded. What is the probability distribution of the random variable $X?$ Find mean, variance and standard deviation of $X$.

Answer:

Total students = 15

probability of selecting a student :

$=\frac{1}{15}$

The information given can be represented as frequency table :

X	14	15	16	17	18	19	20	21
f	2	1	2	3	1	2	3	1

$P(X=14)=\frac{2}{15}$ $P(X=15)=\frac{1}{15}$ $P(X=16)=\frac{2}{15}$

$P(X=17)=\frac{3}{15}=\frac{1}{5}$ $P(X=18)=\frac{1}{15}$ $P(X=19)=\frac{2}{15}$

$P(X=20)=\frac{3}{15}=\frac{1}{5}$ $P(X=21)=\frac{1}{15}$

Probability distribution is as :

X	14	15	16	17	18	19	20	21
P(X)	$\frac{2}{15}$	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{1}{5}$	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{1}{5}$	$\frac{1}{15}$

$E(X)=14\times \frac{2}{15}+15\times \frac{1}{15}+16\times \frac{2}{15}+17\times \frac{1}{5}+18\times \frac{1}{15}+19\times \frac{2}{15}+20\times \frac{1}{5}+21\times \frac{1}{15}$

$E(X)=\frac{263}{15}=17.53$

$E(X^2)=14^2\times \frac{2}{15}+15^2\times \frac{1}{15}+16^2\times \frac{2}{15}+17^2\times \frac{1}{5}+18^2\times \frac{1}{15}+19^2\times \frac{2}{15}+20^2\times \frac{1}{5}+21^2\times \frac{1}{15}$

$E(X^2)=\frac{4683}{15}=312.2$

$Variance =E(X^2)-(E(X))^2$

$Variance =312.2-(17.53)^2$

$Variance =312.2-307.42$

$Variance =4.78$

$Standard\, \, deviation =\sqrt{4.78}=2.19$

Question:15 In a meeting, $70^{o}/_{o}$ of the members favour and $30^{o}/_{o}$ oppose a certain proposal. A member is selected at random and we take $X=0$. if he opposed, and $X=1$ if he is in favour. Find $E(X)$ and Var $(X)$.

Answer:

Given :

$P(X=0)=30\%=\frac{30}{100}=0.3$

$P(X=1)=70\%=\frac{70}{100}=0.7$

Probability distribution is as :

X	0	1
P(X)	0.3	0.7

$E(X)=0\times 0.3+1\times 0.7$

$E(X)= 0.7$

$E(X^2)=0^2\times 0.3+1^2\times 0.7$

$E(X^2)= 0.7$

$Variance = E(X^2)-(E(X))^2$

$Variance = 0.7-(0.7)^2$

$Variance = 0.7-0.49=0.21$

Question:16 The mean of the numbers obtained on throwing a die having written 1 on three faces, $2$ on two faces and $5$ on one face is, Choose the correct answer in the following:

(A) $1$

(B) $2$

(D) $\frac{8}{3}$

Answer:

X is number representing on die.

Total observations = 6

$P(X=1)=\frac{3}{6}=\frac{1}{2}$

$P(X=2)=\frac{2}{6}=\frac{1}{3}$

$P(X=5)=\frac{1}{6}$

X	1	2	5
P(X)	$\frac{1}{2}$	$\frac{1}{3}$	$\frac{1}{6}$

$E(X)=1\times \frac{1}{2}+2\times \frac{1}{3}+5\times \frac{1}{6}$

$E(X)= \frac{1}{2}+ \frac{2}{3}+ \frac{5}{6}$

$E(X)=\frac{12}{6}=2$

Option B is correct.

Question:17 Suppose that two cards are drawn at random from a deck of cards. Let$X$ be the number of aces obtained. Then the value of $E(X)$ is Choose the correct answer in the following:

(A) $\frac{37}{221}$

(B) $\frac{5}{13}$

(D) $\frac{2}{13}$

Answer:

X be number od aces obtained.

X can be 0,1,2

There 52 cards and 4 aces, 48 are non-ace cards.

$P(X=0)=P(0 \, ace\, and\, 2\, non\, ace\, cards)=\frac{^4C_2.^4^8C_2}{^5^2C_2}=\frac{1128}{1326}$

$P(X=1)=P(1 \, ace\, and\, 1\, non\, ace\, cards)=\frac{^4C_1.^4^8C_1}{^5^2C_2}=\frac{192}{1326}$

$P(X=2)=P(2 \, ace\, and\, 0\, non\, ace\, cards)=\frac{^4C_2.^4^8C_0}{^5^2C_2}=\frac{6}{1326}$

The probability distribution is as :

X	0	1	2
P(X)	$\frac{1128}{1326}$	$\frac{192}{1326}$	$\frac{6}{1326}$

$E(X)=0\times \frac{1128}{1326}+1\times \frac{192}{1326}+2\times \frac{6}{1326}$

$E(X)=\frac{204}{1326}$

$E(X)=\frac{2}{13}$

Option D is correct.

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Key Features Of NCERT Solutions for Exercise 13.4 Class 12 Maths Chapter 13

Comprehensive Coverage: The solutions encompass all the topics covered in ex 13.4 class 12, ensuring a thorough understanding of the concepts.
Step-by-Step Solutions: In this class 12 maths ex 13.4, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
Accuracy and Clarity: Solutions for class 12 ex 13.4 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
Conceptual Clarity: In this 12th class maths exercise 13.4 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
Inclusive Approach: Solutions for ex 13.4 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
Relevance to Curriculum: The solutions for class 12 maths ex 13.4 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

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Frequently Asked Questions (FAQs)

Q: What is weightage of the probability in the CBSE Class 12?

The weightage of probability is 8 marks in the final board exams.

Q: What is the total marks for CBSE Class 12 Maths ?

The total marks for CBSE Class 12 Maths are 100 marks.

Q: What is the compound event ?

A compound event is an aggregate of some elementary events and it is decomposable into simple events.

Q: what is the probability ?

The mathematical measurement of uncertainty is called Probability.

Q: Does CBSE provides NCERT exemplar solutions for Class 12 Mathematics ?

No,CBSE doesn’t provide NCERT exemplar solutions for any class.

Q: What is the probability of getting a tail when you toss a coin once ?

The probability of getting a tail when you toss a coin is 0.5.

Q: Where can I get the NCERT solutions ?

Check here for NCERT Solutions.

Q: What is uses of probability in daily life?

Probability is useful in Meteorologists, weather prediction, sports, gambling, etc.

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Questions related to CBSE Class 12th

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how may marks should i get in my 12th board exams to get fee concesion in sastra university

Hello Aspirant,

SASTRA University commonly provides concessions and scholarships based on merit in class 12 board exams and JEE Main purposes with regard to board merit you need above 95% in PCM (or on aggregate) to get bigger concessions, usually if you scored 90% and above you may get partial concessions. I suppose the exact cut offs may change yearly on application rates too.

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gopi

4 Sep'25

Which course is best for computer science students after 12th standard

Hello,

After 12th, if you are interested in computer science, the best courses are:

B.Tech in Computer Science Engineering (CSE) – most popular choice.
BCA (Bachelor of Computer Applications) – good for software and IT jobs.
B.Sc. Computer Science / IT – good for higher studies and research.
B.Tech in Information Technology (IT) – focuses on IT and networking.

All these courses have good career scope. Choose based on your interest in coding, software, hardware, or IT field.

Hope it helps !

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Prachi Kumari

4 Sep'25

I am a CBSE class 12 private candidate will my maksheet of CBSE supplement exam will be delivered on main exam form address for supplementary exam form address ,my both addresses are different please tell

Hello Vanshika,

CBSE generally forwards the marksheet for the supplementary exam to the correspondence address as identified in the supplementary exam application form. It is not sent to the address indicated in the main exam form. Addresses that differ will use the supplementary exam address.

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gopi

3 Sep'25

I gave my 12th exam this year in which I got less marks. Then I gave improvement exam in which also I could not score full marks. So can I become a private candidate and appear for the exam again next year or not?

Hello

Yes, if you’re not satisfied with your marks even after the improvement exam, many education boards allow you to reappear as a private candidate next year to improve your scores. This means you can register independently, study at your own pace, and take the exams without attending regular classes. It’s a good option to improve your results and open up more opportunities for higher studies or careers. Just make sure to check the specific rules and deadlines of your education board so you don’t miss the registration window. Keep your focus, and you will do better next time.

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SEJAL JAIN

20 Aug'25

i gave improvement 2025 as the roll no is different from my previous year marksheet so will i get new migration certificate

Hello Aspirant,

Yes, in the case that you appeared for the 2025 improvement exam and your roll number is different from what was on the previous year’s marksheet, the board will usually release a new migration certificate. This is because the migration certificate will reflect the most recent exam details, roll number and passing year. You can apply to get it from your board using the process prescribed by them either online or through your school/college.

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