Maths Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen Class 12th Probability Ncert Solutions Maths class 12th

A bag contains 5 red marbles and 3 black marbles. Three marbles are drawn one by one without replacement. What is the probability that at least one of the three marbles drawn be black, if the first marble is red?

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Alok Kumar Singh | Contributor-Level 10

7 months ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t r e d m a r b l e s b e r e p r e s e n t e d w i t h R a n d b l a c k m a r b l e s b e r e p r e s e n t e d w i t h B . \\ T h e f o l l o w i n g t h r e e c o n d i t i o n s a r e p o s s i b l e, i f a t l e a s t o n e o f t h e t h r e e m a r b l e s \\ d r a w n b e b l a c k a n d f i r s t m a r b l e i s r e d . \\ (i) E_{1} = I I b a l l i s b l a c k a n d I I I i s r e d \\ (i i) E_{2} = I I b a l l i s b l a c k a n d I I I i s a l s o b l a c k \\ (i i i) E_{3} = I I b a l l i s r e d a n d I I I i s b l a c k \\ ∴ P (E_{1}) = P (R_{1}) . P (B_{1} / R_{1}) . P (R_{2} / R_{1} B_{1}) = \frac{5}{8} . \frac{3}{7} . \frac{4}{6} = \frac{60}{336} = \frac{5}{28} \\ P (E_{2}) = P (R_{1}) . P (B_{1} / R_{1}) . P (B_{2} / R_{1} B_{1}) = \frac{5}{8} . \frac{3}{7} . \frac{2}{6} = \frac{30}{336} = \frac{5}{56} \\ a n d P (E_{3}) = P (R_{1}) . P (R_{2} / R_{1}) . P (B_{1} / R_{1} R_{2}) = \frac{5}{8} . \frac{4}{7} . \frac{3}{6} = \frac{60}{336} = \frac{5}{28} \\ ∴ P (E) = P (E_{1}) + P (E_{2}) + P (E_{3}) = \frac{5}{28} + \frac{5}{56} + \frac{5}{28} = \frac{25}{56} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y i s \frac{25}{56} . \end{array}$

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If $\vec{a}, \vec{b}, \vec{c}$  are non-zero and $\vec{b}$  and $\vec{c}$  are non-collinear. $\vec{a} + 5 \vec{b}$  is collinear with $\vec{c}$  and $\vec{b} + 6 \vec{c}$  is collinear with $\vec{a}$ . If $\vec{a} + α \vec{b} + β \vec{b} = 0$ , then find α + β.

Alok Kumar Singh

$\vec{a} + 5 \vec{b}$ is collinear with $\vec{c}$

⇒ $\vec{a} + 5 \vec{b}$ = $\vec{c}$ …(1)

$\vec{b} + 6 \vec{c}$ is collinear with $\vec{a}$

⇒ $\vec{b} + 6 \vec{c} = μ \vec{a}$ …(2)

From (1) and (2)

$\vec{b} + 6 \vec{c} = μ (λ \vec{c} - 5 \vec{b})$

-> $(1 + 5 μ) \vec{b} + (6 - λ μ) \vec{c} = 0$

$? \vec{b}$ and $\vec{c}$

$\int \frac{(s i n x - c o s x) s i n^{2} x}{t a n x (s i n^{3} x + c o s^{3} x)} d x$

$\int \frac{(s i n x - c o s x) s i n x c o s x}{s i n^{3} x + c o s^{3} x} d x$ , put sin³x + cos³x = t(3 sin²x×cosx – 3cos²xsinx) dx = dt

-> $\frac{1}{3} \int \frac{d t}{t}$

$= \frac{l n t}{3} + c$

$= \frac{l n | s i n^{3} x + c o s^{3} x |}{3} + c$

If $f (x) = \frac{(2^{x} + 2^{- x}) (t a n x) \sqrt[]{t a n^{- 1} (2 x^{2} - 3 x + 1)}}{{(7 x^{2} - 3 x + 1)}^{3}}$ , then f′(0) is equal to

Alok Kumar Singh

$f (x) = \frac{(2^{x} + 2^{- x}) t a n x \sqrt[]{t a n^{- 1} (2 x^{2} - 3 x + 1)}}{{(7 x^{2} - 3 x + 1)}^{3}}$

$f (x) = (2^{x} + 2^{- x}) . t a n x . \sqrt[]{t a n^{- 1} (2 x^{2} - 3 x + 1)} . {(7 x^{2} - 3 x + 1)}^{- 3}$

$f' (x) = (2^{x} + 2^{- x}) . s e c^{2} x . \sqrt[]{t a n^{- 1} (2 x^{2} - 3 x + 1)} . {(7 x^{2} - 3 x + 1)}^{- 3} + t a n x . (Q (x))$

$f' (0) = 2.1 \sqrt[]{\frac{π}{4}} . 1$

$= \sqrt[]{π}$

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