Class 11th

6. Let us denote B1, B2 to be the boys and G1, G2 to be the girls in Room X and B3 to be the boy and G3, G4, G5 to the girls in Room Y. Hence the desired sample space is

S = { (X, B1), (X, B2), (X, G1), (X, G2), (Y, B3), (Y, G3), (Y, G4), (Y, G5)}

5. When a coin is tossed we have the outcomes of a head or a tail and when a dieice is rolled we have the outcome 1, 2, 3, 4, 5 or 6. So, the desired sample space is

S = {T, (H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6)}

4. The possible outcomes of throwing a coin and rolling a dice are Head and tail and 1, 2, 3, 4, 5 and 6 respective. So, the desired sample space is

S = { (H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}

3. On tossing a coin the possible outcomes are Head and Tail. The sample space of tossing a coin four times is

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

2. When a dice is thrown, we have 36 the possible outcome.

1. On tossing a coin the possible outcomes are that of a head or a tail. So, the angle space of tossing a coin three times is

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

76. 

If point P be the junction between the lines

2x – 3y + 4 = 0 ______ (1)

3x + 4y – 5 = 0 ______ (2)

Solving (1) and (2) using 3 * (1) – 2 * (2) we get,

6x – 9y + 12 – (6x + 8y – 10) = 0

–17y + 22 = 0

y = $\frac{22}{17}$

And 2x = 3y– 4

=> 2x = 3 * $\frac{22}{17}$ – 4

x = $\frac{33}{17}$ – 2 = $\frac{33-34}{17}$ = $-\frac{1}{17}$

Hence, the co-ordinate of the junction is P $\left(-\frac{1}{17},\frac{22}{17}\right)$

The eqⁿ of the path to be reach is

6x – 7y + 8 = 0 _____ (3)

Then, least distance will be perpendicular path.

So, slope of ⊥ path = $-1$
$\frac{\mathrm{}\mathrm{slope\; of \; line\; (3)}}{}$

$=\frac{-1}{(6/7)}=\frac{-7}{6}$

Hence eqⁿ of shortest/least distance path from P $\left(-\frac{1}{17},\frac{22}{17}\right)$is

$y-\frac{22}{17}=\frac{-7}{6}\left(x+\frac{1}{17}\right).$

$\Rightarrow 6y-\frac{132}{17}=-7x-\frac{7}{17}.$

$\Rightarrow 7x+6y-\frac{132}{17}+\frac{7}{17}=0$

$\Rightarrow 7x+6y-\frac{125}{17}=0.$

119x + 102y – 125 = 0

Kindly go through the solution

Let A have the co-ordinate (x, o)

By laws of reflection

∠PAB = ∠ QAB = θ

And ∠ CAQ + θ = 90°

As normal is ⊥ to surface (x-axis)

⇒ ∠CAQ = 90° - θ

and ∠CAP = 90° + θ

73. The given eqⁿ of limes are.

9x + 6y – 7 = 0 ______ (1)

3x + 2y + b = 0 ______ (2)

Let P (x₀, y₀) be a point equidistant from (1) and (2) so

9x₀ + 6y - 7 = ± 3 (3x₀ + 2y₀ + 6)

When, 9x₀ + 6y₀ – 7 = 3 (3x₀ + 2y₀ + 6)

⇒ 9x₀ + 6y₀ - 7 = 9x₀ + 6y₀ + 18

⇒ - 7 = 18 which in not true

So, 9x₀ + 6y₀ - 7 = -3 (3x₀ + 2y₀ + 6)

⇒ 9x₀ + 6y₀ -7 = -9x₀ -6y₀ -18

⇒ 18x₀ + 12y₀ + 11= 0.

Hence, the required eqn of line through (x₀, y₀) & equidistant  from parallel line 9x + 6y - 7 = 0

and 3x + 2y + 6 = 0 is 18x + 12y + 11 = 0.