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

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

Edited By Ramraj Saini | Updated on Dec 04, 2023 11:27 AM IST | #CBSE Class 12th

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.

b1

Answer:

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

b3

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.

b4

Answer:

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

b4

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.

b5

Answer:

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

b5

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.

b6

Answer:

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

b6

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:

X01234567
P(X)0k2k2k3kk22k27k2+k

k

Answer:

X01234567
P(X)0k2k2k3kk22k27k2+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:

X01234567
P(X)0k2k2k3kk22k27k2+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:

1655186819360P(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 ofX.

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

(C) 5

(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. LetX 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}

(C) \frac{1}{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.

More About NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13.4:-

A new concept called Random Variables and its Probability Distributions is introduced in NCERT solutions for Class 12 Maths chapter 13 exercise 13.4. There are 8 solved examples given before this exercise. You must solve these examples before moving to the NCERT exercise questions. It will help you understand the concept easily. There are 17 questions including 2 multiple-choice types questions given in NCERT syllabus exercise 13.4 Class 12 Maths.

Also Read| Probability Class 12th Notes

Benefits of NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13.4:-

  • There are 8 examples given before exercise 13.4 which you should solve first then try to solve exercise questions.
  • You are advised to solve more problems which will help you to get conceptual clarity.
  • Class 12 Maths chapter 13 exercise 13.4 solutions are helpful in the revision of the important concepts
  • NCERT solutions for Class 12 Maths chapter 13 exercise 13.4 can be used for reference.
  • Note- Don't buy any reference book as most of the questions in the board exams are asked directly from the NCERT textbook. Be thorough with the NCERT textbook.
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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.

Also see-

NCERT Solutions of Class 12 Subject Wise

Subject Wise NCERT Exampler Solutions

Happy learning!!!

Frequently Asked Questions (FAQs)

1. What is weightage of the probability in the CBSE Class 12?

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

2. What is the total marks for CBSE Class 12 Maths ?

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

3. What is the compound event ?

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

4. what is the probability ?

The mathematical measurement of uncertainty is called Probability.

5. Does CBSE provides NCERT exemplar solutions for Class 12 Mathematics ?

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

6. 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.

7. Where can I get the NCERT solutions ?

Check here for NCERT Solutions.

8. 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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quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

Read Complete Answer
Jefferson 25 September,2024
i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

  1. Re-evaluate Your Study Strategies:

    • Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
    • Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
    • Practice Regularly: Consistent practice is key to mastering chemistry.

  2. Consider Professional Help:

    • Tutoring: A tutor can provide personalized guidance and support.
    • Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.

  3. Explore Alternative Options:

    • Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
    • Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.

  4. Focus on NEET 2025 Preparation:

    • Stay Dedicated: Continue your NEET preparation with renewed determination.
    • Utilize Resources: Make use of study materials, online courses, and mock tests.

  5. Seek Support:

    • Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
    • Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.







Read Complete Answer
Kanishka kaushikii 17 September,2024
i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

Read Complete Answer
Yuvan 30 August,2024
i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Read Complete Answer
Ashvadha 12 July,2024
Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Read Complete Answer
kumardurgesh1802 10 July,2024
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