Class 12th

Kindly go through the solution

(i) True, as vector quantity $\overrightarrow{a}$ and - $\overrightarrow{a}$ are parallel to same line.

(ii) False, as collinear vector are those vectors that are parallel to same line, but it is not necessary that they are equal also.

(iii) False, as two vectors having same magnitude may have different directions, so they are not collinear.

(iv) False, as two collinear vectors having same magnitude are not equal whey they are opposite in direction.

(a) Vector $\overrightarrow{a}$ and $\overrightarrow{d}$ are co initial same initial point.

(b) $\overrightarrow{b}$ and $\stackrel{?}{d}$ same magnitude & direction.

(c) $\overrightarrow{a}$ and $\overrightarrow{c}$ are collinear but not equal they are parallels their direction are not same.

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

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