Maths

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.

38. 

2. The given system of inequalities are

3x+2y≤ 12- (1)

x≥ 1- (2)

y≥ 2- (3)

We draws the graphs of the lines 3x+2y=12 using points and as 3 * 0 + 2 * 0 ≤ 12

The solution is plane which includes the origin (0, 0).

0 ≤ 12

$\begin{array}{l}x\\ y\end{array}|\begin{array}{l}0\\ 6\end{array}|\begin{array}{l}4\\ 0\end{array}|$

and x = 1 and y = 2.

The inequality (1), (2) and (3) represents the region between these three lines including the points on the respective lines. So, every point on the shaded region in first quadrant represents a solution of the given system of inequalities.

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

7. Here,

r= length of pendulum.

r= 75 cm.

(i) Arc of length, l = 10 cm

Ø= $\frac{l}{r}$ = $\frac{10cm}{75cm}=\frac{2}{15}$ radian.

(ii) Arc of length, l = 15 cm.

So, Ø= $\frac{l}{r}$ = $\frac{15cm}{75cm}=\frac{1}{5}$ radian.

(iii) Arc for length, l= 21 cm.

So, Ø= $\frac{l}{r}=\frac{21cm}{75cm}=\frac{7}{25}$ radian.

6. Let r₁ and r₂ be the radii of two circles.

Then using relation

So, radius, r = $\frac{40}{2}$ cm = 20 cm

Length of chord (AB) = 20cm

In OAB

OA = OB=AB=20 cm

Hence, AOAB is equilateral triangle and end of the angle is 60°

:. Ø =60° = 60 *
$\frac{\mathrm{}\mathrm{\pi }}{180}$ radian =$\frac{\mathrm{\pi }}{3}$ radian

Hence, length of minor are of the chord, l=rØ.

l = 20 * $\frac{\mathrm{\pi }}{3}$ cm

l = $\frac{2\mathrm{0\pi }}{3}$ cm.

4. Here l = 22cm.

r =100cm.

Ø =?

Hence by r = $\frac{1}{\mathrm{Ø}}$

= Ø = $\frac{l}{r}$ = $\frac{22}{100}$ radian

= $\frac{22}{100}$ * $\mathrm{180°}$/π

$\frac{\mathrm{}\mathrm{}}{}$$\frac{}{}$= $\frac{22}{100}$ * 180° * $\frac{7}{22}$

= ${\left(\frac{63}{5}\right)}^0$

=12 $\frac{3°}{5}$ = 12° $\frac{3*60\text{'}}{5}=12°36\text{'}$

3. Given that a wheel makes 360 revolutions in one minute

Then, number of revolutions in one second = $\frac{360}{60}$ =6.

In 1 complete revolution the wheel turns 360°= 2π radian.

So, In 6 revolution, the wheel will turns 6*2π radian = 12π radian.

Hence, in one second the wheel will turn an angle of 12π radian.