NCERT Solutions for Class 12 Maths Chapter 13 Probability (Ex 13.4) Exercise 13.4

NCERT Solutions for Class 12 Maths Chapter 13 Probability (Ex 13.4) Exercise 13.4

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Access NCERT Solutions for Class 12 Maths Chapter 13 - Probability

Exercise 13.4

1. State which of the following are not the probability distribution of a  random variable. Give reasons for your answer.

(i)

 X 0 1 2 P(X) 0.4 0.4 0.2

Ans: For any distribution to be probability distribution ∑ P (x ) =1 and   0 ≤ p(x)  ≤  1.

∴ 0.4 + 0.4 + 0.2 = 1

Therefore it is a probability distribution.

(ii)

 X 0 1 2 3 4 P(X) 0.1 0.5 0.2 −0.1 0.3

Ans: For any distribution to be probability distribution ∑ P (x ) =1 and 0 ≤ p(x)  ≤  1.

∴  −0.1does not lie in the range

Therefore it is not a probability distribution.

(iii)

 Y −1 0 1 P(Y) 0.6 0.1 0.2

Ans: For any distribution to be probability distribution  ∑ P (x ) =1 and 0 ≤ p(x)  ≤  1.

∴   0.6 +  0.1 +  0.2 +≠ 1

Therefore it is not a probability distribution.

(iv)

 Z 3 2 1 0 −1 P(Z) 0.3 0.2 0.4 0.1 0.05

Ans: For any distribution to be probability distribution   ∑ P (x ) =1 and 0 ≤ p(x)  ≤  1.

∴  0.3+ 0.2+ 0.1+ 0.05  ≠ 1

Therefore it is not a probability distribution.

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

Ans: let us have following notations

B: represents black ball

R: represents red ball

The sample space can be represented as

S = {(BB) ,( BR) , (RB) , (RR)}

X represent the number of black ball

Thus possible values for X is 0,1,2

Also    ∑ P (x ) =1 it can be clearly seen.

3. Let X represents the difference between the number of heads and the  number of tails obtained when a coin is tossed 6 times. What are possible  values of X ?

Ans: Given that a coin is tossed six times and X denotes the difference between  number of heads and that of tails

Mathematically we can write as shown

X (6H,0T) =  6

X (5H,1T) =  4

X (4H,2T) = 2

X (3H,3T) =  0

X (2H,4T) =  2

X (1H,5T) =  4

X (0H,6T) =   6

The possible values of X are 0,2,4,6

4. Find the probability distribution of

(i) number of heads in two tosses of a coin

Ans: the sample space is given by

S  = {(HH) , (HT) , (TH) , (TT) }

Let Y represents the number of heads then

Y (HH) =  2 , Y (HT) =  1 , Y (TH) =  1 , Y (TT) =  0

Also since the coins are unbiased therefore each of them has probability of $\frac{1}{4}$We have

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

P(Y = 1) = $\frac{1}{4}$+ $\frac{1}{4}$ = $\frac{1}{2}$

P(Y = 2) =  $\frac{1}{4}$

Therefore table made is as shown

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

(ii) Number of tails in the simultaneous tosses of three coins

Ans: The sample space is given by

S ={ (HHH) , (HTH) , (HTT) , (THH) , (THT) , (TTH) , (TTT)}

Let Y represents the number of tails then it can take values 0,1,2, or 3 Also since the coins are unbiased therefore each of them has probability of $\frac{1}{4}$ We have

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

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

P(Y = 2) =  $\frac{3}{8}$

P(Y = 3) =  $\frac{1}{8}$

Therefore table made is as shown

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

(iii) Number of heads in four tosses of a coin

Ans: the sample space is given by

S = { (HHHH) , (HHHT) , (HHTT) , (HTHT) , (HTHH) , (HTTH) , (HTTT) (THHH) , (THHT) , (THTH) , (THTT) , (TTHH) , (TTHT) , (TTTH) , (TTTT) }

Let Y represents the number of heads then it can take values 0,1,2,3, or 4 Also since the coins are unbiased therefore each of them has probability of $\frac{1}{4}$ We have

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

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

P(Y = 2) =  $\frac{6}{16}$

P(Y = 3) =  $\frac{4}{16}$

P(Y = 4) =  $\frac{1}{16}$

Therefore table made is as shown

 Y 0 1 2 3 4 P(Y) $\frac{1}{16}$ $\frac{4}{16}$ $\frac{6}{16}$ $\frac{4}{16}$ $\frac{1}{16}$

5. Find the probability distribution of the number of success in two tosses of  die, where a success is defined as

(i) Number greater than 4

Ans: Let Y be the random variable which represents the number of successes i.e  it represents number greater than 4

We can easily observe that Y can have three values as shown

0 for number less than or equal to 4 in the both tosses

1 for greater than 4and less than or equal to 4in either of tosses

2 for greater than 4in the both tosses

Now P( Y) =  0 means probability of number less than 4

∴p(Y= 0) = $\frac{4}{6} \times \frac{4}{6} = \frac{4}{9}$

p(Y= 1) = $\frac{4}{6} \times \frac{2}{6} + \frac{2}{6} \times \frac{4}{6} = \frac{4}{9}$

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

(ii) Six appears on at least one die

Ans: Let Y be the random variable which represents the number of successes i.e  it represents six appearing atleast on one die

We can easily observe that Y can have two values as shown

0 for when six doesn’t appear in any of the dices

1 for when six appears in either of the dice

Now P( Y) =  0 means probability six doesn’t appear in any of the dices

∴p(Y= 0) = $\frac{5}{6} \times \frac{5}{6} = \frac{25}{36}$

p(Y= 1) = $\frac{1}{6} \times \frac{5}{6} + \frac{5}{6} \times \frac{1}{6} = \frac{5}{18}$

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

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

Ans: Let us denote D to be the random variable that denotes the number of  defective bulbs

Number of defective bulbs is 30 - 6 = 24

Let us have following notations

D = 0 means all four selected are non defective

D = 1 means three selected are non defective and one is defective

D = 2 means two selected are non defective and two is defective

D = 3 means one selected are non defective and three is defective

D=  4 means all four selected are defective

∴P(D=0)= 4C0 $(\frac{4}{5})^4 = \frac{256}{625}$

∴P(D=1)= 4C1 $(\frac{4}{5})^3 (\frac{1}{5}) = \frac{256}{625}$

∴P(D=2)= 4C2 $(\frac{4}{5})^2 (\frac{1}{5})^2 = \frac{96}{625}$

∴P(D=3)= 4C1 $(\frac{4}{5}) (\frac{1}{5})^3 = \frac{16}{625}$

∴P(D=4)= 4C4 $(\frac{1}{5})^4 = \frac{1}{625}$

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

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

Ans: Given that the ratio of occurring heads is 3times that of tails  i.e P (H) =  3P (T)

Also we know that

⇒ P(T) = $\frac{1}{4}$

⇒ P(H) = $\frac{3}{4}$

The sample space for toss of two coins is given by

S = {(HH) ,( HT ) , (TH ) , (TT) }

Let K be the random variable which represents number of tails  Therefore possible values of K are 0,1,2

∴p(K= 0) = $\frac{3}{4} \times \frac{3}{4} = \frac{9}{16}$

p(K= 1) = $\frac{3}{4} \times \frac{1}{4} + \frac{1}{4} \times \frac{3}{4} = \frac{6}{16}$

∴p(K= 2) = $\frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$

Hence the distributions is

 K 0 1 2 P(K) $\frac{9}{16}$ $\frac{6}{16}$ $\frac{1}{16}$

8. A random variable X has the following probability distribution.

 Y 0 1 2 3 4 5 6 7 P(Y) 0 k 2k 2k 3k K2 2k2 7K2+K

Determine

(i) k

Ans: We know that the sum of probabilities is 1

⇒K + 2K + 2K + 3K + K2 + 2k2  + 7k2 +k = 1

⇒ 10 k2  + 9k -1 = 0

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

⇒k = -1, $\frac{1}{10}$

(ii)P (X) < 3

Ans: we know that

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

P (X< 3) =  0+ k + 2k=  3k

P (X< 3)  = $\frac{3}{10}$

(iii) P (X) > 6

Ans: Since there are 7possible values for random variable

∴P(X > 6) = P (X = 7)

⇒  P (X = 7)  = 7k2 + k

⇒  P (X = 7) = 7$(\frac{1}{10})^2 + \frac{1}{10}$

⇒  P (X > 6) = $\frac{17}{100}$

(iv)P( 0< X< 3)

Ans: we know that

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

⇒ P( 0< X< 3)  = K + 2K

⇒P ( 0< X< 3) = $\frac{3}{10}$

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

$\mathbf{\begin{Bmatrix}k, &if \;x=0 \\2k, &if \; x =1 \\3k, &if \; x =2 \\0, &otherwise \end{Bmatrix}}$

1. Determine the value of k.

Ans: We know that total sum of probabilities equal to

K + 2k +3= 1

⇒$\frac{1}{6}$

1. Find P( X < 2) ,P (X ≤ 2) , P (X≥ 2 ).

Ans: We know that

P (X <2) =  P( X  = 0) + P (X  = 1)

⇒ P (X <2) = $\frac{1}{6}$

Similarly,

P (X≤2) =  P( X  = 0) + P (X  = 1)  + P(X = 2)

⇒ P (X≤2) = $\frac{6}{6}$

Similarly,

P (X≥2) =  P(X = 2) + 0

⇒ P (X≥2) = $\frac{3}{6}$

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

Ans: The sample is given by

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

It is clearly visible that K can take values 0,1,2,3

P( K =  0 ) =  P (TTT )

P( K =  0) =  P( T) P( T )P( T )

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

Similarly,

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

P (K = 2) = $\frac{3}{8}$

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

The probability distribution is

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

Mean is given by

E(K ) = Σ ki p(ki )

⇒E(K) = 0 $\times \frac{1}{8} + 1 \times \frac{3}{8} + 2 \times \frac{3}{8} + 3 \times \frac{1}{8}$

Thus mean value is 1.5

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

Ans: Here, we take X to represent the number of sixes obtained when two dice  are thrown simultaneously.

Hence, it is easily visible that possible values for X 0,1,2

0represents number of zero sixes in any of the dices

1 represents number of sixes in any one of the dices

2 represents number of zero sixes in both of the dices

$\therefore P (X= 0) = (\frac{5}{6})^2 = \frac{25}{36}\\\\\therefore P(X=1) = 2 (\frac{5}{6}) (\frac{1}{6}) = \frac{10}{36} \\ \\ \therefore P(X=2) = (\frac{1}{6})^2 = \frac{1}{36}$

Thus, the required probability distribution is as follows.

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

Mean or Expectation is given by

E(K ) = Σ ki p(ki )

⇒E(K) = 0 $\times \frac{25}{36} + 1 \times \frac{10}{36} + 2 \times \frac{3}{8} + 3 \times \frac{1}{8}$

Thus mean value is$\frac{1}{3}$

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

Ans: The ways in which two positive integers can be selected from first six  without replacement is 6 5 30 × =ways

Let K denote the larger of two numbers obatained

Clearly it is visible that K can take values 2,3,4,5,6

For K =  2 possible observations are (1,2 )and (2,1 )

∴ P(K=2 ) = $\frac{2}{30}$

For K  = 3 possible observations are (1,3) ,( 3,1) , (3,2) ,( 2,3 )

∴  P(K = 3 ) = $\frac{4}{30}$

For  (K =  4) possible observations are (1,4) ,( 4,1) , (4,2) ,( 2,4 ) , (3,4 ) ,( 4,3 )

∴  P(K = 4 ) = $\frac{6}{30}$

For K = 5 possible observations are

(1,5) , (5,1) , (5,2) , (2,5) ,( 3,5 ) , (5,3) ,( 5,4) , (4,5 )

∴  P(K = 5 ) = $\frac{8}{30}$

For K 6 =possible observations are

(1,6) , (6,1) ,( 6,2 ) , (2,6 ), (3,6) , (6,3 ) ,( 6,4 ), (4,6) , (5,6) , (6,5 )

∴  P(K = 6 ) = $\frac{10}{30}$

Thus the required probability distribution is

 K 2 3 4 5 6 P(K) $\frac{2}{10}$ $\frac{4}{30}$ $\frac{6}{30}$ $\frac{8}{30}$ $\frac{10}{30}$

Hence, E(K ) = Σ ki p(ki )

⇒E(K) = 2 $\times \frac{2}{30} + 3 \times \frac{4}{30} + 4 \times \frac{6}{30} + 5 \times \frac{80}{30} + 6 \times \frac{10}{30}$

Thus expected value is $\frac{14}{3}$

13. Let X denotes the sum of the number obtained when two fair dice are  rolled. Find the variance and standard deviation of X.

Ans: Let K be the random variable that denote the sum of numbers obtained when  two dice rolled simultaneously

Clearly K can take values 2,3,4,5,6,7,8,9,10,11,12

We have

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

P(X = 3) = $\frac{2}{36}$

P(X = 4) = $\frac{3}{36}$

P(X = 5) = $\frac{4}{36}$

P(X = 6) = $\frac{5}{36}$

P(X = 7) = $\frac{6}{36}$

P(X = 8) = $\frac{5}{36}$

P(X = 9) = $\frac{4}{36}$

P(X = 10) = $\frac{3}{36}$

P(X = 11) = $\frac{2}{36}$

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

And hence the probability distribution is

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

Then , E(K ) = Σ ki p(ki )

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

⇒ E(K) = 7

Also,

⇒E(K2) = 2 $\times( \frac({1}{36}^2 + 3 \times (\frac{2}{36})^2 + 4 \times (\frac{3}{36})^2 + 5 \times (\frac{4}{36})^2 + 6 \times (\frac{5}{36})^2 + 7 \times (\frac{6}{36})^2 + 8\times (\frac{5}{36})^2 + 9 \times (\frac{4}{36})^2 + 10 \times (\frac{3}{36})^2 + 11 \times (\frac{2}{36})^2 + 12 \times (\frac{1}{36})^2$

Hence E(K2) = $\frac{329}{6}$

Now we know that variance is given by,

E(K2) - (E(K))2

= 54.833 -49

5.833

And standard deviation is given by,

$\sqrt{var(K)} = \sqrt{5.833}$

Hence we found variance to be 5.833and standard deviation to be 2.415

14. A class has 15 students whose ages are  14,17,15,14,21,17,19,16,18,2017,16,19 and 20years. One students 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.

Ans: It is given that probability of choosing each is same i.e $\frac{1}{15}$From the above data we can draw a frequency table as shown

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

We have following probabilities as shown

Hence probability distributions is as shown

 X 14 15 16 17 18 19 20 21 P(X) $\frac{2}{15}$ $\frac{1}{15}$ $\frac{2}{15}$ $\frac{3}{15}$ $\frac{1}{15}$ $\frac{2}{15}$ $\frac{3}{15}$ $\frac{1}{15}$

The mean is given by

E(K ) = Σ ki p(ki )

⇒E(K) = 14 $\times \frac{2}{15} + 15 \times \frac{1}{15} + 16 \times \frac{2}{15} + 17 \times \frac{3}{15} + 18 \times \frac{1}{15}+ 19 \times \frac{2}{15} + 20\times \frac{3}{15} + 21 \times \frac{1}{15}$

⇒ E(K) = 17.53

Now

⇒E(K2) =$(14)^2 \times \frac{2}{15} + (15 )^2 \times \frac{1}{15} + (16)^2 \times \frac{2}{15} + (17)^2 \times \frac{3}{15} + (18)^2 \times \frac{1}{15}+ (19)^2 \times \frac{2}{15} + (20)^2 \times \frac{3}{15} + (21)^2 \times \frac{1}{15}$

Hence E(K2) = 312.2

Hence,  variance is given by,

E(K2) - (E(K))2

312.2 - 307. 4177

= 4.78

Hence standard deviation is $\sqrt{4.78}$ = 2.19

15. In a meeting, 70 percent of the members favour and 30 percent 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).

Ans: Given in the question that X 0 =means opposes the proposal and X = 1 means that he is in favour.

∴ P(X=0) = 0.3 and

P(X = 1) = 0.7

Thus the probability distribution is as shown

 X 0 1 P(X) 0.3 0.7

We know that The mean is given by

E(K ) = Σ ki p(ki )

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

⇒E(X) = 0.7

Also, E(X2) = 0.7

Now variance is given by

E(X2)  - E(X)

= 0.7 -0.49

=0.21

16. The mean of the numbers obtained on throwing a die having written 1 on three faces, 2on two faces and 5on one face is

Ans: Let us denote K to be the random variable representing number of faces in  which certain fixed number is to occur

According to the question

∴P(K =1) = $\frac{3}{6}$

P(K =2) = $\frac{2}{6}$

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

Thus the probability distribution is given by

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

Thus mean is given by

E(K ) = Σ ki p(ki )

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

⇒E(X) = 2

Hence we found that mean is 2

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

Ans: Let us denote K to be the random variable which measures the number of  aces occurred.

The possible values for K are 0,1,2

K = 0 means no ace is present

K = 1 means one ace is present

K = 2 means two aces are present

∴P(K = 0)= $\frac{ ^4C_0 \times ^48C_2}{ ^52C_2 } = \frac {1128}{1326}$   (since there are 4 aces  and 48 non aces cards )

∴P(K = 1)= $\frac{ ^4C_1 \times ^48C_1}{ ^52C_2} = \frac{192}{1326}$

∴P(K = 2)= $\frac{ ^4C_2\times ^48C_0}{ ^52C_2 } = \frac{6}{1326}$

Hence the probability distribution is given by

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

Thus mean is given by

E(K ) = Σ ki p(ki )

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

⇒E(X) = $\frac{2}{13}$

Hence we found that mean is $\frac{2}{13}$

NCERT Solution Class 12 Maths of Chapter 13 All Exercises

 Chapter 13 - Probability Exercises in PDF Format Exercise 13.1 17 Questions & Solutions (4 Short Answers, 13 Long Answers) 18 Questions & Solutions Exercise 13.3 14 Questions & Solutions 17 Questions & Solutions Exercise 13.5 15 Questions & Solutions

NCERT Solutions for Class 12 Maths Chapter 13 Probability Exercise 13.4

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Important Topics Covered in Exercise 13.4 of NCERT Solutions for Class 12 Maths Chapter 13 Probability

Exercise 13.4 of NCERT Solutions for Class 12 Maths Chapter 13 Probability mainly focuses on the below topics.

• The probability distribution of a random variable

• Mean of a random variable

• The variance of a random variable

Below are Some Key Points we Will Learn from this Chapter.

• In probability, a random variable can be defined as a real-value function whose domain is the sample space of a random experiment.

• The probability distribution of a random variable explains the probability of its value that is unknown.

• The mean or the expected value of a random variable (X) is the sum of the products of all the possible values of the variable (X) by their respective probabilities. It is represented by µ or E (X) = x1 p1+ x2 p2 + ... + xn pn.

• Variance is used to calculate the variability in the values of a random variable. The variance of the random variable X is denoted by Var (X) and it is calculated using the formula Var (X) = E(X – µ)2 (where µ denotes the mean of X).

The questions provided in this exercise are mainly focused on finding the mean and variance of a random variable.

1. How many Questions are there in the Class 12 Maths Chapter 13 Exercise 13.4 of NCERT Textbook?

A total of 10 Questions/ Problems are there in the Class 12 Maths Chapter 13 Exercise 13.4 of NCERT textbook. However, this chapter contains four exercises in total.

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These NCERT Solutions for Maths Chapter 13 Exercise 13.4 play a crucial role in every Class 12 student’s life as these are curated by Vedantu’s subject matter experts who hold years of experience in the relevant industry. These solutions PDF are of excellent quality, hence any student can rely on these without any hesitation. The best part is that these Maths solutions are available in free PDF format so that students do not have to spend an extra penny in order to avail those from Vedantu’s website.

NCERT Solutions holds a very special place in every student’s life as the exercises are given at the end of every chapter helps the students to be thorough and well versed with the basic concepts. So that students can score an excellent grade in the exam.

Yes, these NCERT Solutions for Class 10 Maths by Vedantu are easily available online for absolutely free of cost and in PDF format. So that anyone can download these solutions as per their convenience and refer to these both online and offline. All they have to do is register on Vedantu site along with their email ID and contact. That’s all!

5. Is Class 12 Maths Chapter 13 Exercise 13.4 Easy?

Class 12 Maths Chapter 13 Exercise 13.4 may not be extremely simple but it's very important for the exams, both boards, and competitive exams. Therefore, it is crucial to know this chapter and clear out all queries. This implies that a guide proves to be helpful for understanding. Here is where NCERT solutions by Vedantu come into play. With its expert-made answers and accurate solutions to problems, it makes it easy to understand the concepts and also to get good marks in the exam.

6. Can you please brief the Class 12 Maths Exercise 13.4?

Exercise 13.4 of Class 12 Maths Chapter 13, Probability and its problems are based on the concept of independent events. Problems related to this topic have been carefully selected and solved in the NCERT solutions by Vedantu that are free to download. With the help of NCERT Solutions, you will be able to understand the concepts related to Exercise 13.4 of Class 12 Maths clearly so that it becomes easier for you to solve the same problems in the final board exams.

7. How can I understand Exercise 13.4 of Class 12 Maths probability?

The following concepts are important to understand the Exercise 13.4 of Class 12 Maths Chapter 13 Probability:

• Two events (denoted by E and F) are considered independent when

• P(E∩F) = P(E).P(F)

• Two given independent events having the non-parental probability of an event might not be mutually exclusive. In contrast, mutually exclusive events which will probably take place are not independent.

8. How can I study Class 12 Maths Chapter 13 Probability?

To study Class 12 Maths Chapter 13 Probability, it is important to understand these fundamental concepts:

• Probability And Statistics

• Multiplication theorem on Probability

• Bayes Theorem

• Bernoulli Trials

• Binomial Distribution

These concepts have been discussed in detail in the NCERT solutions by Vedantu to ensure that you get a clear picture of the chapter. These solutions are available on the Vedantu app and the website.

9. What are the most important formulas that I need to remember in Exercise 13.4?

When E and F are events associated with the sample space S, and the probability of either one of them happening does not get affected by the happening of another, then these events E and Event F are independent. This implies E and F are independent when,

• P(F | E) = P(F), where P (E) ≠ 0

• P(E | F) = P(E), where P (F) ≠ 0

In case of three events A, B, C, they are mutually independent when,

• P(A ∩ B) = P(A) P(B)

• P(A ∩ C) = P(A) P(C)

• P(B ∩ C) = P(B) P(C)

• P(A ∩ B ∩ C) = P(A) P(B) P(C)