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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen

Let S = {1, 2, 3, 4, 5, 6} and E = {1, 3, 5}. Then E is ___.

5 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : S = {1, 2, 3, 4, 5, 6} \\ E = {1, 3, 5} \\ ∴ \bar{E} = S - E = {2, 4, 6} \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s {2, 4, 6} . \end{array}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen

The probability that a student will pass his examination is 0.73, the probability of the student getting a compartment is 0.13, and the probability that the student will either pass or get a compartment is 0.96.

3 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} L e t E b e t h e e v e n t t h a t t h e s t u d e n t w i l l p a s s \\ a n d F b e t h e e v e n t t h a t h e w i l l g e t c o m p a r t m e n t \\ ∴ P (E) = 0.73, P (F) = 0.13 a n d P (E \cup F) = 0.96 \\ ∴ P (E \cup F) = P (E) + P (F) - P (E \cap F) \\ = 0.73 + 0.13 - 0 [? P (E \cap F) = 0] \\ = 0.86 \\ B u t P (E \cup F) = 0.96 \\ H e n c e, t h e g i v e n s t a t e m e n t i s' F a l s e' . \end{array}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen

The probability that a person visiting a zoo will see the giraffe is 0.72, the probability that he will see the bears is 0.84, and the probability that he will see both is 0.52.

5 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : \\ P (t o s e e g i r a f f e e) = 0.72 \\ P (t o s e e b e a r) = 0.84 \\ P (t o s e e b o t h g i r a f f e e a n d b e a r) = 0.52 \\ ∴ P (t o s e e g i r a f f e e o r b e a r) = P (t o s e e g i r a f f e e) + P (t o s e e b e a r) - P (t o s e e b o t h) \\ = 0.72 + 0.84 - 0.52 \\ = 1.04 w h i c h i s n o t p o s s i b l e . \\ H e n c e, t h e g i v e n s t a t e m e n t i s' F a l s e' . \end{array}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen

If M and N are any two events, the probability that at least one of them occurs is:

(a) P(M) + P(N) – 2P(M ∩ N)

(b) P(M) + P(N) – P(M ∩ N)

(c) P(M) + P(N) + P(M ∩ N)

(d) P(M) + P(N) + 2P(M ∩ N)

3 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} I f M a n d N a r e a n y t w o e v e n t s, t h e n \\ P (M \cup N) = P (M) + P (N) - P (M \cap N) \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen

The probability that at least one of the events A and B occurs is 0.6. If A and B occur simultaneously with a probability of 0.2, then P(A) + P(B) is:

(a) 0.4

(b) 0.8

(c) 1.2

(d) 1.6

3 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : P (A \cup B) = 0.6 \\ P (A \cap B) = 0.2 \\ ∴ P (A \cup B) = P (A) + P (B) - P (A \cap B) \\ \Rightarrow 0.6 = P (A) + P (B) - 0.2 \\ \Rightarrow P (A) + P (B) = 0.6 + 0.2 = 0.8 \\ a n d 1 - P (\bar{A}) + 1 - P (\bar{B}) = 0.8 \\ \Rightarrow P (\bar{A}) + P (\bar{B}) = 2 - 0.8 = 1.2 \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen

If the probabilities for A to fail in an examination are 0.2 and for B are 0.3, then the probability that either A or B fails is:

(a) > 0.5

(b) 0.5

(c) ≤ 0.5

(d) 0

4 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : P (A f a i l s) = 0.2 \\ P (B f a i l s) = 0.3 \\ ∴ P (e i t h e r A o r B f a i l s) \leq P (A f a i l s) + P (B f a i l s) \\ \leq 0.2 + 0.3 \\ \leq 0.5 \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen

A single letter is selected at random from the word “PROBABILITY.” The probability that it is a vowel is:

(a) $\frac{1}{3}$ 

(b) $\frac{1}{4}$ 

(c) $\frac{2}{11}$ 

(d) None of these.

4 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T o t a l n u m b e r o f a l p h a b e t s i n p r o b a b i l i t y = 11 \\ N u m b e r o f v o w e l s = 4 \\ ∴ Required p r o b a b i l i t y = \frac{4}{11} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen

Six boys and six girls sit in a row at random. The probability that all the girls sit together is:

(a) $\frac{1}{432}$ 

(b) $\frac{12}{431}$ 

(c) $\frac{1}{132}$ 

(d) None of these.

11 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} I f a l l t h e g i r l s s i t t o g e t h e r, t h e n w e c o n s i d e r i t a s 1 g r o u p \\ ∴ T o t a l n u m b e r o f a r r a n g e m e n t o f 6 + 1 = 7 p e r s o n s i n a r o w = 7! \\ a n d t h e g i r l s a l s o t h e i r p l a c e s w i t h 6! w a y s . \\ ∴ Required p r o b a b i l i t y = \frac{6! 7!}{12!} = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 7!}{12 \times 11 \times 10 \times 9 \times 8 \times 7!} = \frac{1}{132} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

$\begin{array}{l} . \end{array}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

Let $A$ and $B$ be two events. If $P (A ? B) = P (A)$ , then $A$ is _____ of $B$ .

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Fill in the Blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} ? P (A / B) = \frac{P (A \cap B)}{P (B)} \\ \Rightarrow P (A) = \frac{P (A \cap B)}{P (B)} \\ \Rightarrow P (A \cap B) = P (A) . P (B) \\ S o, A i s i n d e p e n d e n t o f B . \end{array}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

Let $X$ be a random variable taking values $x_{1}, x_{2}, \dots, x_{n}$ with probabilities $p_{1}, p_{2}, \dots, p_{n}$ , respectively. Then $(X) =$ _____

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Fill in the Blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} V a r (X) = E (X^{2}) - {[E (X)]}^{2} \\ = {\sum_{}^{} X^{2} P (X) - [\sum_{}^{} X . P (X)]}^{2} = \sum_{}^{} p_{i} x_{i}^{2} - {(\sum_{}^{} p_{i} x_{i})}^{2} \\ H e n c e, V a r (X) i s e q u a l t o \sum_{}^{} p_{i} x_{i}^{2} - {(\sum_{}^{} p_{i} x_{i})}^{2} . \end{array}$

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