Ncert Solutions Maths class 12th

(i) Time period involves only magnitude. So, it is scalar quantity.

(ii) Distance involves only magnitude. So, it is scalar quantity.

(iii) Force involves both magnitude and direction. So, it is vector quantity.

(iv) Velocity involves both magnitude and direction. So, it is vector quantity.

(v) Work done involves only magnitude. So, it is scalar quantity.

(i) 10kg involves only magnitude. So, it is scalar quantity.

(ii) 2 meters north-west involves both magnitude and direction. So, it is vector quantity.

(iii) 40⁰  involves only magnitude. So, it is scalar quantity.

(iv) 40⁰  watts involves only magnitude. So, it is scalar quantity.

(v) 10^-19 coulomb involves only magnitude. So, it is scalar quantity.

(vi) 20m/s^-2 involves magnitude and direction. So, it is vector quantity.

94.

$\begin{array}{l}P\left(A|B\right)>P\left(A\right)\\ \Rightarrow \frac{P\left(A\cap B\right)}{P\left(B\right)}>P\left(A\right)\\ \Rightarrow P\left(A\cap B\right)>P\left(A\right).P\left(B\right)\\ \Rightarrow \frac{P\left(A\cap B\right)}{P\left(A\right)}>P\left(B\right)\\ \Rightarrow P\left(B|A\right)>P\left(B\right)\end{array}$

Therefore, option (C) is correct.

$40km,30^0$ east of north.

93. $P\left(A\right)\ne 0$ and $P\left(B|A\right)=1$

$\begin{array}{l}P\left(B\cap A\right)=\frac{P\left(B\cap A\right)}{P\left(A\right)}\\ 1=\frac{P\left(B\cap A\right)}{P\left(A\right)}\\ P\left(A\right)=P\left(B\cap A\right)\\ \Rightarrow A\subset B\end{array}$

Therefore, option (A) is correct.

92. Let E₁ = Ball transferred from Bag I to Bag II is red

E₂ = Ball transferred from Bag I to Bag Ii is black

A = Ball drawn from Bag II is red in colour

P (E₁) =3/7 and P (E₂) = 4/7

Let A be the event that the ball drawn is red.

When a red ball is transferred from bag I to II,

P (A | E₁) = 5/10 = 1/2

When a black ball is transferred from bag I to II,

P (A | E₂) = 4/10 = 2/5

$\begin{array}{l}\therefore P\left(E_2|A\right)=\frac{P\left(E_2\right)P\left(A|E_2\right)}{P\left(E_1\right)P\left(A|E_1\right)+P\left(E_2\right)P\left(A|E_2\right)}\\ =\frac{\frac{4}{7}*\frac{2}{5}}{\frac{3}{7}*\frac{1}{2}+\frac{4}{7}*\frac{2}{5}}\\ =\frac{16}{31}\end{array}$

91. Let the event in which A fails and B fails be denoted by E_A and E_B.

P (EA) = 0.2

P (E_A ∩ E_B) = 0.15

P (B fails alone) = P (E_B) − P (E_A ∩ E_B)

⇒ 0.15 = P (E_B) − 0.15

⇒ P (E_B) = 0.3

$P\left(E_A|E_B\right)=\frac{P\left(E_A\cap E_B\right)}{P\left(E_B\right)}=\frac{0.15}{0.3}=0.5$

(ii) P (A fails alone) = P (E_A) − P (E_A ∩ E_B)

= 0.2 − 0.15

= 0.05

90. There are four entries in a determinant of 2 x 2 order. Each entry may be filled up in two ways with 0 or 1.

 Number of determinants that can be formed = (2)⁴ = 16

The value of determinants is positive in the following cases:

$\left|\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right|\left|\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right|\left|\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right|$

Therefore, the probability that the determinant is positive = 3/16

88. Let R be the event of drawing the red marble.

Let EA, EB, and EC respectively denote the events of selecting the box A, B, and C.

Total number of marbles = 40

Number of red marbles = 15

P (R) = 15/40 = 3/8

Probability of drawing the red marble from box A is given by P (EA|R).

$?P\left(E_A|R\right)=\frac{P\left(E_A?R\right)}{P\left(R\right)}=\frac{\frac{1}{40}}{\frac{3}{8}}=\frac{1}{15}$

Probability that the red marble is from box B is P (EB|R).

$?P\left(E_B|R\right)=\frac{P\left(E_B?R\right)}{P\left(R\right)}=\frac{\frac{6}{40}}{\frac{3}{8}}=\frac{2}{5}$

Probability that the red marble is from box C is P (EC|R).

$?P\left(E_C|R\right)=\frac{P\left(E_C?R\right)}{P\left(R\right)}=\frac{\frac{8}{40}}{\frac{3}{8}}=\frac{8}{15}$