Class 12th

95.

$\begin{array}{l}P\left(A\right)+P\left(B\right)-P\left(AandB\right)=P\left(A\right)\\ \Rightarrow P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)=P\left(A\right)\\ \Rightarrow P\left(B\right)-P\left(A\cap B\right)=0\\ \Rightarrow P\left(A\cap B\right)=P\left(B\right)\\ \therefore P\left(A|B\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}=\frac{P\left(B\right)}{P\left(B\right)}=1\end{array}$

Therefore, option (B) is correct.

(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