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9 months ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

A letter is known to have come either from TATA NAGAR or from CALCUTTA. On the envelope, just two consecutive letters "TA" are visible. What is the probability that the letter came from TATA NAGAR?

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} L e t E_{1} : T h e e v e n t t h a t t h e l e t t e r c o m e s f r o m T A T A N A G A R \\ a n d E_{2} : T h e e v e n t t h a t t h e l e t t e r c o m e s f r o m C A L C U T T A \\ A l s o, E_{3} : T h e e v e n t t h a t o n t h e l e t t e r, t w o l e t t e r s T A a r e v i s i b l e . \\ ∴ P (E_{1}) = \frac{1}{2}, P (E_{2}) = \frac{1}{2} a n d P (\frac{E_{3}}{E_{1}}) = \frac{2}{8} a n d P (\frac{E_{3}}{E_{2}}) = \frac{1}{7} \\ [? F o r T A T A N A G A R, t h e t w o consecutive l e t t e r s v i s i b l e a r e T A, A T, T A, A N, N A, A G, G A, A R] \\ ∴ P (E_{3} / E_{1}) = \frac{2}{8} \\ a n d [F o r C A L C U T T A, t h e t w o consecutive l e t t e r s v i s i b l e a r e C A, A L, L C, C U, U T, T T a n d T A] \\ S o, P (E_{3} / E_{2}) = \frac{1}{7} \\ N o w using B a y e s' T h e r o r m, w e h a v e \\ ∴ P (E_{1} / E_{3}) = \frac{P (E_{1}) . P (E_{3} / E_{1})}{P (E_{1}) . P (E_{3} / E_{1}) + P (E_{2}) . P (E_{3} / E_{2})} \\ = \frac{\frac{1}{2} . \frac{2}{8}}{\frac{1}{2} . \frac{2}{8} + \frac{1}{2} . \frac{1}{7}} = \frac{\frac{1}{8}}{\frac{1}{8} + \frac{1}{14}} = \frac{\frac{1}{8}}{\frac{7 + 4}{56}} = \frac{7}{11} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y i s \frac{7}{11} . \end{array}$

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9 months ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

A shopkeeper sells three types of flower seeds: $A_{1}$ , $A_{2}$ , and $A_{3}$ . They are sold as a mixture where the proportions are 4:4:2, respectively. The germination rates of the three types of seeds are 45%, 60%, and 35%. Calculate the probability:

(i) Of a randomly chosen seed to germinate

(ii) That it will not germinate given that the seed is of type $A_{3}$

(iii) That it is of type $A_{2}$ given that a randomly chosen seed does not germinate.

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t A_{1} : A_{2} : A_{3} = 4 : 4 : 2 \\ ∴ P (A_{1}) = \frac{4}{10}, P (A_{2}) = \frac{4}{10} a n d P (A_{3}) = \frac{2}{10} w h e r e A_{1}, A_{2} a n d A_{3} a r e t h e t h r e e t y p e s o f s e e d s . \\ L e t E b e t h e e v e n t s t h a t a s e e d a n d \bar{E} b e t h e e v e n t s t h a t a s e e d d o e s n o t \\ ∴ P (\frac{E}{A_{1}}) = \frac{45}{100}, P (\frac{E}{A_{2}}) = \frac{60}{100} a n d P (\frac{E}{A_{3}}) = \frac{35}{100} \\ a n d P (\frac{\bar{E}}{A_{1}}) = \frac{55}{100}, P (\frac{\bar{E}}{A_{2}}) = \frac{40}{100} a n d P (\frac{\bar{E}}{A_{3}}) = \frac{65}{100} \\ (i) P (E) = P (A_{1}) . P (\frac{E}{A_{1}}) + P (A_{2}) . P (\frac{E}{A_{2}}) + P (A_{3}) . P (\frac{E}{A_{3}}) \\ = \frac{4}{10} . \frac{45}{100} + \frac{4}{10} . \frac{60}{100} + \frac{2}{10} . \frac{35}{100} \\ = \frac{180}{1000} + \frac{240}{1000} + \frac{70}{1000} = \frac{490}{1000} = 0.49 \\ (i i) P (\bar{E} / A_{3}) = 1 - P (E / A_{3}) = 1 - \frac{35}{100} = \frac{65}{100} = 0.65 \\ (i i i), B a y e s' T h e o r e m, w e g e t \\ P (A_{2} / \bar{E}) = \frac{P (A_{2}) . P (\bar{E} / A_{2})}{P (A_{1}) . P (\bar{E} / A_{1}) + P (A_{2}) . P (\bar{E} / A_{2}) + P (A_{3}) . P (\bar{E} / A_{3})} \\ = \frac{\frac{4}{10} . \frac{40}{100}}{\frac{4}{10} . \frac{55}{100} + \frac{4}{10} . \frac{40}{100} + \frac{2}{10} . \frac{65}{100}} = \frac{\frac{160}{1000}}{\frac{220}{1000} + \frac{160}{1000} + \frac{130}{1000}} \\ = \frac{160}{220 + 160 + 130} = \frac{160}{510} = \frac{16}{51} = 0.314 \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y i s \frac{16}{51} o r 0.314 \end{array}$

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9 months ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

Refer to Question 41 above. If a white ball is selected, what is the probability that it came from

$i.$ $B a g I I$

$ii. B a g I I I$

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} Referring t o E x e r c i s e Q . 41, w e w i l l u s e h e r e, B a y e s' T h e o r e m \\ (i) P (E_{2} / F) = \frac{P (E_{2}) . P (F / E_{2})}{P (E_{1}) . P (F / E_{1}) + P (E_{2}) . P (F / E_{2}) + P (E_{3}) . P (F / E_{3})} \\ = \frac{\frac{2}{6} . \frac{1}{3}}{\frac{1}{6} . 0 + \frac{2}{6} . \frac{1}{3} + \frac{3}{6} . 1} = \frac{\frac{2}{18}}{\frac{2}{18} + \frac{3}{6}} = \frac{2}{18} * \frac{18}{11} = \frac{2}{11} \\ (i i) P (E_{3} / F) = \frac{P (E_{3}) . P (F / E_{3})}{P (E_{1}) . P (F / E_{1}) + P (E_{2}) . P (F / E_{2}) + P (E_{3}) . P (F / E_{3})} \\ = \frac{\frac{3}{6} . 1}{\frac{1}{6} . 0 + \frac{2}{6} . \frac{1}{3} + \frac{3}{6} . 1} = \frac{\frac{3}{6}}{\frac{2}{18} + \frac{3}{6}} = \frac{3}{6} * \frac{18}{11} = \frac{9}{11} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t i e s a r e \frac{2}{11} a n d \frac{9}{11} \end{array}$

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9 months ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

Three bags contain a number of red and white balls as follows:

Bag 1: 3 red balls, Bag 2: 2 red balls and 1 white ball, Bag 3: 3 white balls. The probability that bag $i$ will be chosen and a ball is selected from it is $\frac{i}{6}$ , where $i = 1, 2, 3$ . What is the probability that:

i. A red ball will be selected?

ii. A white ball is selected?

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t : \\ ? ? ? B a g I = 3 r e d b a l l s a n d n o w h i t e b a l l \\ B a g I I = 2 r e d b a l l s a n d 1 w h i t e b a l l \\ B a g I I I = n o r e d b a l l a n d 3 w h i t e b a l l s \\ L e t E_{1}, E_{2} a n d E_{3} b e t h e e v e n t s o f B a g I I, a n d B a g I I I r e s p e c t i v e l y a n d a b a l l i s \\ d r a w n f r o m i t . \\ ∴ P (E_{1}) = \frac{1}{6}, P (E_{2}) = \frac{2}{6} a n d P (E_{3}) = \frac{3}{6} \\ (i) L e t E b e t h e e v e n t t h a t r e d b a l l i s s e l e c t e d \\ ∴ P (E) = P (E_{1}) . P (E / E_{1}) + P (E_{2}) . P (E / E_{2}) + P (E_{3}) . P (E / E_{3}) \\ = \frac{1}{6} . \frac{3}{3} + \frac{2}{6} . \frac{2}{3} + \frac{3}{6} . 0 = \frac{3}{18} + \frac{4}{18} = \frac{7}{18} \\ (i i) L e t F b e t h e e v e n t t h a t w h i t e b a l l i s s e l e c t e d \\ ∴ P (F) = 1 - P (E) [P (E) + P (F) = 1] \\ = 1 - \frac{7}{18} = \frac{11}{18} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t i e s a r e \frac{7}{18} a n d \frac{11}{18} . \end{array}$

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10 months ago

Probability Delhi Pharmaceutical Sciences and Research University

I have gotten 208 in gpay exam . Is any probability to taken admission I dipsar college

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New answer posted

10 months ago

Probability Class 12th

95. If A and B are any two events such that P (A) + P (B) – P (A and B) = P (A), then

(A) P (B|A) = 1

(B) P (A|B) = 1

(C) P (B|A) = 0

(D) P (A|B) = 0

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alok kumar singh

Contributor-Level 10

95.

$\begin{array}{l} P (A) + P (B) - P (A a n d B) = P (A) \\ \Rightarrow P (A) + P (B) - P (A \cap B) = P (A) \\ \Rightarrow P (B) - P (A \cap B) = 0 \\ \Rightarrow P (A \cap B) = P (B) \\ ∴ P (A | B) = \frac{P (A \cap B)}{P (B)} = \frac{P (B)}{P (B)} = 1 \end{array}$

Therefore, option (B) is correct.

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10 months ago

Probability Class 12th

94. If P (A|B) > P (A), then which of the following is correct:

(A) P (B|A) < P (B)

(B) P (A ∩ B) < P (A).P (B)

(C) P (B|A) > P (B)

(D) P (B|A) = P (B)

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alok kumar singh

Contributor-Level 10

94.

$\begin{array}{l} P (A | B) > P (A) \\ \Rightarrow \frac{P (A \cap B)}{P (B)} > P (A) \\ \Rightarrow P (A \cap B) > P (A) . P (B) \\ \Rightarrow \frac{P (A \cap B)}{P (A)} > P (B) \\ \Rightarrow P (B | A) > P (B) \end{array}$

Therefore, option (C) is correct.

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10 months ago

Probability Class 12th

93. If A and B are two events such that P (A) ≠ 0 and P(B|A) = 1, then:

(A) A ⊂ B

(B) B ⊂ A

(C) B = Φ

(D) A = Φ

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alok kumar singh

Contributor-Level 10

93. $P (A) \neq 0$ and $P (B | A) = 1$

$\begin{array}{l} P (B \cap A) = \frac{P (B \cap A)}{P (A)} \\ 1 = \frac{P (B \cap A)}{P (A)} \\ P (A) = P (B \cap A) \\ \Rightarrow A \subset B \end{array}$

Therefore, option (A) is correct.

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10 months ago

Probability Class 12th

92. Bag I contains 3 Red and 4 Black balls and B II contains 4 Red and % Black balls. One ball is transferred from Bag I to bag II and then a ball is drawn from bag II. The ball so drawn is found to be Red in colour. Find the probability that the transferred ball is Black.

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alok kumar singh

Contributor-Level 10

92. Let E₁ = Ball transferred from Bag I to Bag II is red

E₂ = Ball transferred from Bag I to Bag Ii is black

A = Ball drawn from Bag II is red in colour

P (E₁) =3/7 and P (E₂) = 4/7

Let A be the event that the ball drawn is red.

When a red ball is transferred from bag I to II,

P (A | E₁) = 5/10 = 1/2

When a black ball is transferred from bag I to II,

P (A | E₂) = 4/10 = 2/5

$\begin{array}{l} ∴ P (E_{2} | A) = \frac{P (E_{2}) P (A | E_{2})}{P (E_{1}) P (A | E_{1}) + P (E_{2}) P (A | E_{2})} \\ = \frac{\frac{4}{7} * \frac{2}{5}}{\frac{3}{7} * \frac{1}{2} + \frac{4}{7} * \frac{2}{5}} \\ = \frac{16}{31} \end{array}$

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10 months ago

Probability Class 12th

91. An electronic assembly consists of two sub-systems say A and B. From previous testing procedures, the following probabilities are assumed to be known:

P (A fails) = 0.2

P (B fails alone) = 0.15

P (A and B fail) = 0.15

Evaluate the following probabilities.

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alok kumar singh

Contributor-Level 10

91. Let the event in which A fails and B fails be denoted by E_A and E_B.

P (EA) = 0.2

P (E_A∩ E_B) = 0.15

P (B fails alone) = P (E_B) − P (E_A ∩ E_B)

⇒ 0.15 = P (E_B) − 0.15

⇒ P (E_B) = 0.3

$P (E_{A} | E_{B}) = \frac{P (E_{A} \cap E_{B})}{P (E_{B})} = \frac{0.15}{0.3} = 0.5$

(ii) P (A fails alone) = P (E_A) − P (E_A∩ E_B)

= 0.2 − 0.15

= 0.05

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