Class 11th

In a large metropolitan area, the probabilities are 0.87, 0.36, 0.30 that a family (randomly chosen for a sample survey) owns a colour television set, a black and white television set, or both kinds of sets. What is the probability that a family owns either any one or both kinds of sets?

0 Follower 4 Views

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} L e t E_{1} b e t h e e v e n t t h a t a f a m i l y o w n s c o l o u r t e l e v i s i o n s e t . \\ a n d E_{2} b e t h e e v e n t t h a t a f a m i l y o w n s b l a c k a n d w h i t e t e l e v i s i o n s e t . \\ G i v e n t h a t P (E_{1}) = 0.87, P (E_{2}) = 0.36 \\ a n d P (E_{1} \cap E_{2}) = 0.30 \\ ∴ T h e p r o b a b i l i t y t h a t a f a m i l y o w n s e i t h e r c o l o u r o r b l a c k a n d w h i t e t e l e v i s i o n s e t . \\ ∴ P (E_{1} \cup E_{2}) = P (E_{1}) + P (E_{2}) - P (E_{1} \cap E_{2}) \\ = 0.87 + 0.36 - 0.30 = 0.93 \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y = 0.93 \end{array}$

0 0

New answer posted

a year ago

A die is loaded in such a way that each odd number is twice as likely to occur as each even number. Find $P (G)$ , where $G$ is the event that a number greater than 3 occurs on a single roll of the die.

0 Follower 6 Views

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t p r o b a b i l i t y o f e v e n n u m b e r s = \frac{1}{2} * p r o b a b i l i t y o f o d d n u m b e r s \\ \Rightarrow P (O d d) : P (E v e n) = 2 : 1 \\ ∴ P (o d d n u m b e r) = \frac{2}{2 + 1} = \frac{2}{3} \\ a n d P (e v e n n u m b e r) = \frac{1}{2 + 1} = \frac{1}{3} \\ A l s o, g i v e n t h a t, G t h e e v e n t t h a t a n u m b e r g r e a t e r t h a n 3 o c c u r s i n a single t h r o w o f d i e . \\ ∴ T h e p o s s i b l e o u t c o m e a r e 4, 5 a n d 6 o u t o f w h i c h t w o a r e e v e n a n d o n e i s o d d . \\ ∴ R e q u i r e d p r o b a b i l i t y = P (G) \\ = 2 * P (e v e n) * P (o d d) \\ = 2 * \frac{1}{3} * \frac{2}{3} = \frac{4}{9} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y = \frac{4}{9} . \end{array}$

0 0

New answer posted

a year ago

An experiment consists of rolling a die until a 2 appears.

(i) How many elements of the sample space correspond to the event that the 2 appears on the $k^{t h}$ roll of the die?

(ii) How many elements of the sample space correspond to the event that the 2 appears not later than the $k^{t h}$ roll of the die?

0 Follower 6 Views

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} N u m b e r o f S a m p l e s p a c e = 6 \\ (i) G i v e n t h a t 2 a p p e a r s o n t h e k t h r o l l o f t h e d i e . \\ S o f i r s t (k - 1) t h r o l l h a v e 5 o u t c o m e s e a c h a n d k t h r o l l r e s u l t s 2 i . e . o n l y 1 o u t c o m e . \\ ∴ N u m b e r o f e l e m e n t o f s a m p l e s p a c e c o r r e s p o n d t o t h e e v e n t t h a t 2 a p p e a r s o n t h e k t h \\ r o l l o f t h e d i e = 5^{k - 1} . \\ (i i) I n t h i s c a s e, 2 a p p e a r s n o t l a t e r t h a n k t h r o l l o f t h e d i e, t h e n i t i s p o s s i b l e t h a t 2 c o m e s \\ i n f i r s t t h r o w t h e n o u t c o m e s w i l l b e 5 a n d 2 o u t c o m e s i n second t h r o w i . e . 1 o u t c o m e . \\ ∴ P o s s i b l e o u t c o m e = 5 * 1 = 5 \\ S i m i l a r l y, i f 2 d o e s n o t a p p e a r i n second t h r o w a n d a p p e a r s i n t h i r d t h r o w \\ ∴ P o s s i b l e o u t c o m e = 5 * 5 * 1 \\ N o w w e h a v e t h e s e r i e s : \\ = 1 + 5 + 5 * 5 + 5 * 5 * 5 + \dots + 5^{k - 1} \\ = 1 + 5 + 5^{2} + 5^{3} + \dots + 5^{k - 1} \\ = \frac{1 . (r^{k} - 1)}{r - 1} = \frac{5^{k - 1}}{5 - 1} \\ = \frac{5^{k - 1}}{4} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y = \frac{5^{k - 1}}{4} . \end{array}$

0 0

New answer posted

a year ago

Suppose an integer from 1 through 1000 is chosen at random. Find the probability that the integer is a multiple of 2 or a multiple of 9.

0 Follower 4 Views

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} W e h a v e m u l t i p l e s o f 2, f r o m 1 t o 1000 a r e 2, 4, 6, 8, \dots, 1000 \\ L e t n b e t h e n u m b e r o f t e r m s \\ H e r e a = 2, d = 2, a_{n} = 1000 \\ a_{n} = a + (n - 1) d \\ 1000 = 2 + (n - 1) 2 \\ \Rightarrow 1000 = 2 n \Rightarrow n = 500 \\ N o w m u l t i p l e s o f 9, f r o m 1 t o 1000 a r e 9, 18, 27, \dots, 999 \\ H e r e a = 9, d = 9, a_{m} = 999 [m b e t h e n u m b e r o f t e r m s] \\ a_{m} = a + (m - 1) d \\ \Rightarrow 999 = 9 + (m - 1) 9 \\ \Rightarrow 999 = 9 m \Rightarrow m = 111 \\ N o w m u l t i p l e s o f 2 a n d 9 a r e 18, 36, 54, \dots, 990 \\ H e r e a = 18, d = 18, a_{p} = 999 [p b e t h e n u m b e r o f t e r m s] \\ ∴ a_{p} = a + (p - 1) d \\ \Rightarrow 990 = 18 + (p - 1) 18 \\ \Rightarrow 990 = 18 p \Rightarrow p = 55 \\ ∴ N u m b e r o f m u l t i p l e s o f 2 o r 9 = 500 + 111 - 55 = 556 \\ ∴ R e q u i r e d p r o b a b i l i t y = \frac{P (E)}{n (E)} = \frac{556}{1000} = 0.556 \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y = 0.556 \end{array}$

0 0

New answer posted

a year ago

Six new employees, two of whom are married to each other, are to be assigned six desks that are lined up in a row. If the assignment of employees to desks is made randomly, what is the probability that the married couple will have nonadjacent desks?

[Hint: First find the probability that the couple has adjacent desks, and then subtract it from 1.]

0 Follower 12 Views

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar $\begin{array}{l} W o r d A L G O R I T H M h a s 9 l e t t e r s . \\ N u m b e r o f d e s k o c c u p i e d b y o n e c o u p l e = 1 \\ O n l y (4 + 1) = 5 p e r s o n s t o b e a s s i g n e d . \\ ∴ N u m b e r o f w a y s o f a s s i g n i n g t h e s e 5 p e r s o n s = 5! * 2! \\ T o t a l n u m b e r o f w a y s o f a s s i g n i n g t h e s e 6 p e r s o n s = 6! \\ ∴ Probability t h a t a c o u p l e h a s a d j a c e n t d e s k = \frac{5! * 2!}{6!} = \frac{1}{3} \\ S o, t h e p r o b a b i l i t y t h a t t h e m a r r i e d c o u p l e w i l l h a v e n o a d j a c e n t d e s k s = 1 - \frac{1}{3} = \frac{2}{3} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y = \frac{2}{3} . \end{array}$

0 0

New answer posted

a year ago

If the letters of the word ALGORITHM are arranged at random in a row, what is the probability that the letters GOR must remain together as a unit?

0 Follower 4 Views

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} W o r d A L G O R I T H M h a s 9 l e t t e r s . \\ I f G O R r e m a i n t o g e t h e r, t h e n i t w i l l r e m a i n t o g e t h e r . \\ ∴ N u m b e r o f l e t t e r s = A L G O R I T H M = 6 + 1 = 7 \\ N u m b e r o f w o r d s = 7! \\ a n d t h e t o t a l n u m b e r o f w o r d s f r o m A L G O R I T H M = 9! \\ S o, t h e r e q u i r e d p r o b a b i l i t y = \frac{7!}{9!} = \frac{7!}{9.8.7!} = \frac{1}{72} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y = \frac{1}{72} . \end{array}$

0 0

New answer posted

a year ago

The equation of the line through the intersection of the lines $2 x – 3 y = 0$ and $4 x – 5 y = 2$ and

Column C1Column C2

through the point (2, 1) is (i) $2 x – y = 4$

perpendicular to the line $x + 2 y + 1 = 0$ is (ii) $x + y – 5 = 0$

parallel to the line $3 x – 4 y + 5 = 0$ is (iii) $x – y – 1 = 0$

equally inclined to the axes is (iv) $3 x – 4 y – 1 = 0$

0 Follower 4 Views

Contributor-Level 10

Sol:

$\begin{array}{l} (a) G i v e n ? ? e q u a t i o n s ? ? a r e ? ? 2 x ? 3 y = 0 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? \dots (i) \\ ? ? ? ? ? ? ? a n d ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 4 x ? 5 y = 2 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? \dots (i i) \\ ? ? ? ? ? ? E q u a t i o n s ? ? o f ? ? l i n e ? ? passing ? ? t h r o u g h ? ? e q n . (i) ? ? a n d ? ? (i i) ? ? w e ? ? g e t \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? (2 x ? 3 y) + k (4 x ? 5 y ? 2) = 0 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? \dots (i i i) \\ ? ? ? ? ? ? I f ? ? e q n . (i i i) ? ? p a s s e s ? ? t h r o u g h ? ? (2, 1), ? ? w e ? ? g e t \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? (2 * 2 ? 3 * 1) + k (4 * 2 ? 5 * 1 ? 2) = 0 \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? (4 ? 3) + k (8 ? 5 ? 2) = 0 \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 1 + k (8 ? 7) = 0 \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? k = ? 1 \\ ? ? ? ? ? S o, ? ? t h e ? ? r e q u i r e d ? ? e q u a t i o n ? ? i s \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? (2 x ? 3 y) ? 1 (4 x ? 5 y ? 2) = 0 \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 2 x ? 3 y ? 4 x + 5 y + 2 = 0 \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 2 x + 2 y + 2 = 0 \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? x ? y ? 1 = 0 \\ ? ? ? ? ? ? H e n c e, ? ? (a) ? (i i i) \\ (b) E q u a t i o n ? ? o f ? ? a n y ? ? l i n e ? ? passing ? ? t h r o u g h ? ? t h e ? ? point ? ? o f ? ? intersection ? ? o f ? ? t h e ? ? l i n e \\ ? ? ? ? ? ? ? 2 x ? 3 y = 0 ? ? a n d ? ? 4 x ? 5 y = 2 ? ? i s \\ ? ? ? ? ? ? ? ? ? ? (2 x ? 3 y) + k (4 x ? 5 y ? 2) = 0 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? \dots (i) \\ ? ? ? (2 + 4 k) x + (? 3 ? 5 k) y ? 2 k = 0 \\ ? ? ? ? ? ? ? ? S l o p e = \frac{? (2 + 4 k)}{? 3 ? 5 k} = \frac{2 + 4 k}{3 + 5 k} \\ ? ? ? ? ? ? ? ? S l o p e ? ? o f ? ? t h e ? ? g i v e n ? ? l i n e ? ? x + 2 y + 1 = 0 ? ? i s ? ? ? \frac{1}{2} . \\ ? ? ? ? ? ? ? ? I f ? ? t h e y ? ? a r e ? ? p e r p e n d i c u l a r ? ? t o ? ? e a c h ? ? o t h e r ? ? t h e n \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? \frac{1}{2} (\frac{2 + 4 k}{3 + 5 k}) = ? 1 \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? \frac{1 + 2 k}{3 + 5 k} = 1 \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 1 + 2 k = 3 + 5 k \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 3 k = ? 2 ? ? ? ? ? ? ? ? ? k = \frac{? 2}{3} \\ ? ? ? ? ? ? P u t t i n g ? ? t h e ? ? v a l u e ? ? o f ? ? k ? ? i n ? ? e q n . (i) ? ? w e ? ? g e t \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? (2 x ? 3 y) ? \frac{2}{3} (4 x ? 5 y ? 2) = 0 \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 6 x ? 9 y ? 8 x + 10 y + 4 = 0 \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 2 x + y + 4 = 0 \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 2 x ? y = 4 \\ ? ? ? ? ? ? H e n c e, ? ? (b) ? (i) \end{array}$

$\begin{array}{l} (c) G i v e n ? ? e q u a t i o n s ? ? a r e ? ? 2 x ? 3 y = 0 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? \dots (i) \\ ? ? ? ? ? ? ? a n d ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 4 x ? 5 y = 2 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? \dots (i i) \\ ? ? ? ? ? ? E q u a t i o n s ? ? o f ? ? l i n e ? ? passing ? ? t h r o u g h ? ? e q n . (i) ? ? a n d ? ? (i i) ? ? w e ? ? g e t \\ ? ? ? ? ? ? ? ? ? ? ? ? ? (2 x ? 3 y) + k (4 x ? 5 y ? 2) = 0 \\ ? ? ? ? ? (2 + 4 k) x + (? 3 ? 5 k) y ? 2 k = 0 \\ ? ? ? ? ? ? ? S l o p e = \frac{? (2 + 4 k)}{? 3 ? 5 k} = \frac{2 + 4 k}{3 + 5 k} \\ ? ? ? ? ? ? ? ? S l o p e ? ? o f ? ? t h e ? ? g i v e n ? ? l i n e ? ? 3 x ? 4 y + 5 = 0 ? ? i s ? ? \frac{3}{4} . \\ ? ? ? ? ? ? ? ? I f ? ? t h e ? ? t w o ? ? e q u a t i o n s ? ? a r e ? ? p a r a l l e l, ? ? t h e n \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? \frac{2 + 4 k}{3 + 5 k} = \frac{3}{4} \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 8 + 16 k = 9 + 15 k \\ ? ? ? ? ? ? ? ? ? ? ? ? ? 16 k ? 15 k = 9 ? 8 \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? k = 1 \\ ? ? ? ? ? S o, ? ? t h e ? ? r e q u i r e d ? ? e q u a t i o n ? ? o f ? ? l i n e ? ? i s \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? (2 x ? 3 y \end{array}$

0 0

New answer posted a year ago

The equation of the line through the intersection of the lines $2 x – 3 y = 0$ and $4 x – 5 y = 2$ and

Column C1Column C2

through the point (2, 1) is (i) $2 x – y = 4$

perpendicular to the line $x + 2 y + 1 = 0$ is (ii) $x + y – 5 = 0$

parallel to the line $3 x – 4 y + 5 = 0$ is (iii) $x – y – 1 = 0$

equally inclined to the axes is (iv) $3 x – 4 y – 1 = 0$

0 Follower 4 Views

Contributor-Level 10

Sol:

0 0

New answer posted a year ago

The value of $λ$ , if the lines $(2 x + 3 y + 4) + λ (6 x – y + 12) = 0$ are

Column C1 Column C2

i. Parallel to y-axis is &nb

...more

The value of $λ$ , if the lines $(2 x + 3 y + 4) + λ (6 x – y + 12) = 0$ are

Column C1 Column C2

i. Parallel to y-axis is (i) $λ = - \frac{3}{4}$

ii. Perpendicular to $7 x + y – 4 = 0$ is (ii) $λ = - \frac{1}{3}$

iii. passes through (1, 2) is (iii) $λ = - \frac{17}{41}$

iv. parallel to x-axis is (iv) $λ = 3$

less

0 Follower 3 Views

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} (a) G i v e n ? ? e q u a t i o n i s \\ (2 x + 3 y + 4) + λ (6 x - y + 12) = 0 \\ \Rightarrow (2 + 6 λ) x + (3 - λ) y + 4 + 12 λ = 0 \dots (i) \\ I f e q n . (i) i s p a r a l l e l t o y a x i s, t h e n \\ 3 - λ = 0 \Rightarrow λ = 3 \\ H e n c e, (a) \leftrightarrow (i v) \\ (b) G i v e n ? ? l i n e s a r e \\ (2 x + 3 y + 4) + λ (6 x - y + 12) = 0 \dots (i) \\ \Rightarrow (2 + 6 λ) x + (3 - λ) y + 4 + 12 λ = 0 \\ S l o p e = - (\frac{2 + 6 λ}{3 - λ}) \\ e q u a t i o n i s 7 x + y - 4 = 0 \dots (i i) \\ S l o p e = - 7 \\ I f e q n . (i) a n d e q n . (i i) a r e p e r p e n d i c u l a r t o e a c h o t h e r \\ ∴ (- 7) [- (\frac{2 + 6 λ}{3 - λ})] = - 1 \\ \Rightarrow \frac{14 + 42 λ}{3 - λ} = - 1 \\ \Rightarrow 14 + 42 λ = - 3 + λ \\ \Rightarrow 42 λ - λ = - 3 - 14 \\ \Rightarrow 41 λ = - 17 \\ \Rightarrow λ = - \frac{17}{41} \\ H e n c e, (b) \leftrightarrow (i i i) \\ (c) G i v e n ? ? e q u a t i o n i s (2 x + 3 y + 4) + λ (6 x - y + 12) = 0 \dots (i) \\ I f e q n . (i) t h r o u g h t h e g i v e n (1, 2) t h e n \\ (2 * 1 + 3 * 2 + 4) + λ (6 * 1 - 2 + 12) = 0 \\ \Rightarrow (2 + 6 + 4) + λ (6 - 2 + 12) = 0 \\ \Rightarrow 12 + 16 λ = 0 \\ \Rightarrow λ = \frac{- 12}{16} = \frac{- 3}{4} \\ H e n c e, (c) \leftrightarrow (i) \\ (d) T h e g i v e n ? ? e q u a t i o n i s (2 x + 3 y + 4) + λ (6 x - y + 12) = 0 \\ \Rightarrow (2 + 6 λ) x + (3 - λ) y + 4 + 12 λ = 0 \dots (i) \\ I f e q n . (i) i s p a r a l l e l t o x a x i s, t h e n \\ 2 + 6 λ = 0 \Rightarrow λ = \frac{- 1}{3} \\ H e n c e, (d) \leftrightarrow (i i) \end{array}$

0 0

New answer posted

a year ago

Match the following:

Column C1 Column C2

(a) The coordinates of the points P and Q on the (i) $(3, 1), (– 7, 11)$

line $x + 5 y = 13$ , which are at a distance of 2 units

from the line $12 x – 5 y + 26 = 0$ are

(b) The coordinates of the point on the (ii) $(3, 1), (– 7, 11)$

line $x + y = 4$ , which is at a unit distance

from the line $4 x + 3 y – 10 = 0$ .

joining A(–2, 5) and B(3, 1) such that

$A P = P Q = Q B$ are

less

0 Follower 4 Views