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New answer posted•a year 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?

7 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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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New answer posted•a year 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.

30 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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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New answer posted•a year 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

$i.$ $B a g I I$

$ii. B a g I I I$

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

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New answer posted•a year 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?

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 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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New question posted•a year ago

Probability Delhi Pharmaceutical Sciences and Research University

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

4 Views|Posted a year ago

Asked by Jyoti

New answer posted•a year 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

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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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New answer posted•a year 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)

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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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New answer posted•a year 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

10 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

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