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2 months ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

If a point A(x, y) lies in the region bounded by the y-axis, straight lines 2y + x = 6 and 5x – 6y = 30, then the probability that y < 1 is

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Payal Gupta

Contributor-Level 10

The required probability

$= \frac{A r e a o f R e g i o n P Q C A P}{A r e a o f R e g i o n A B C A}$

$= \frac{\frac{1}{2} * 8 * 6 - \frac{1}{2} * 2 * 4}{\frac{1}{2} * 8 * 6}$

$= \frac{5}{6}$

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2 months ago

Probability Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen

The sum of probabilities of two students getting distinction in their final examinations is 1.2.

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} Since, t h e t w o g i v e n e v e n t s a r e n o t r e l a t e d t o t h e s a m e S a m p l e s p a c e . \\ ∴ T h e s u m o f p r o b a b i l i t i e s o f t w o s t u d e n t s g e t t i n g d i s t i n c t i o n i n t h e i r f i n a l \\ m a y b e 1.2 \\ H e n c e, t h e g i v e n s t a t e m e n t i s' T r u e' \end{array}$

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New answer posted

2 months ago

Probability Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen

The probability of the occurrence of event A is 0.7 and that of the occurrence of event B is 0.3, but the probability of the occurrence of both is 0.4.

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} H e r e, P (A) = 0.7, P (B) = 0.3 \\ ∴ P (A \cap B) = P (A) * P (B) \\ = 0.7 * 0.3 = 0.21 \\ B u t t h e g i v e n p r o b a b i l i t y i s 0.4 \\ H e n c e, t h e g i v e n s t a t e m e n t i s' F a l s e' \end{array}$

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New answer posted

2 months ago

Probability Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen

The probability of the intersection of two events A and B is always less than or equal to the probability of A.

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} H e r e, P (A \cap B) \leq P (A) W h i c h i s a l w a y s t r u e . \\ H e n c e, t h e g i v e n s t a t e m e n t i s' T r u e' \end{array}$

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New answer posted

2 months ago

Probability Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen

If A and B are two candidates seeking admission in an engineering college, the probability that A is selected is 0.5, and the probability that both A and B are selected is at most 0.3. Is it possible that the probability of B getting selected is 0.7?

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} G i v e n t h a t P (A) = 0.5 a n d P (A \cap B) \leq 0.3 \\ N o w P (A) * P (B) \leq 0.3 \\ \Rightarrow 0.5 * P (B) \leq 0.3 \\ \Rightarrow P (B) \leq \frac{0.3}{0.5} \\ \Rightarrow P (B) \leq 0.6 \\ H e n c e, t h e g i v e n s t a t e m e n t i s' F a l s e' \end{array}$

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New answer posted

2 months ago

Probability Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen

The probabilities that a typist will make 0, 1, 2, 3, 4, or 5 or more mistakes in typing a report are, respectively, 0.12, 0.25, 0.36, 0.14, 0.08, 0.11.

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} S u m o f a l l p r o b a b i l i t i e s = 1 \\ ∴ P (0) + P (1) + P (2) + P (3) + P (4) + P (5) = 0.12 + 0.25 + 0.36 + 0.14 + 0.08 + 0.11 \\ = 1.06 > 1 \\ H e n c e, t h e g i v e n s t a t e m e n t i s' F a l s e' . \end{array}$

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New answer posted

2 months 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.

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Payal Gupta

Contributor-Level 10

This is a True or False 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

2 months 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.

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Payal Gupta

Contributor-Level 10

This is a True or False 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

2 months ago

Probability Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen

Match the following

(a) If E₁ and E₂ are the two mutually (i) E₁ ∩ E₂ = E₁ exclusive events

(b) If E₁ and E₂ are the mutually (ii) (E₁ – E₂) ∪ (E₁ ∩ E₂) = E₁

exclusive and exhaustive events

(c) If E₁ and E₂ have common

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} (a) I f E_{1} a n d E_{2} a r e m u t u a l l y e x c l u s i v e e v e n t s, t h e n E_{1} \cap E_{2} = ? . \\ (b) I f E_{1} a n d E_{2} a r e m u t u a l l y e x c l u s i v e a n d e x h a u s t i v e e v e n t s, t h e n \\ E_{1} \cap E_{2} = ? a n d E_{1} \cup E_{2} = S . \\ (c) I f E_{1} a n d E_{2} h a v e c o m m o n o u t c o m e s, t h e n (E_{1} - E_{2}) \cup (E_{1} \cap E_{2}) = E_{1} \\ (d) I f E_{1} a n d E_{2} a r e t w o e v e n t s s u c h t h a t E_{1} \subset E_{2} \Rightarrow E_{1} \cap E_{2} = E_{1} \\ H e n c e, (a) \leftrightarrow (i v), (b) \leftrightarrow (i i i), (c) \leftrightarrow (i i), (d) \leftrightarrow (i) \end{array}$

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New answer posted

2 months ago

Probability Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen

Match the proposed probability under Column C₁ with the appropriate written description under column C₂ :

C₁ C₂

Probability &n

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} (a) 0.95 = V e r y l i k e l y t o h a p p e n, s o i t i s c l o s e t o 1 . \\ (b) 0.02 = V e r y l i t t l e c h a n c e o f h a p p e n i n g a s t h e p r o b a b i l i t y i s v e r y l o w . \\ (c) - 0.3 = A n i n c o r r e c t a s s i g n m e n t b e c a u s e p r o b a b i l i t y i s n e v e r n e g a t i v e . \\ (d) 0.5 = A s m u c h c h a n c e o f h a p p e n i n g a s n o t b e c a u s e s u m o f c h a n c e s o f h a p p e n i n g \\ a n d n o t h a p p e n i n g i s o n e . \\ (e) 0 = N o c h a n c e o f h a p p e n i n g . \\ H e n c e, (a) \leftrightarrow (i v), (b) \leftrightarrow (v), (c) \leftrightarrow (i), (d) \leftrightarrow (i i i), (e) \leftrightarrow (i i) \end{array}$

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