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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

The probability of guessing correctly at least 8 out of 10 answers on a true-false type examination is

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 Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} H e r e, n = 10, p = \frac{1}{2}, q = 1 - \frac{1}{2} = \frac{1}{2} (f o r t r u e / f a l s e q u e s t i o n s) \\ a n d r \geq 8 i . e ., 8, 9, 10 \\ ∴ P (X \geq 8) = P (x = 8) + P (x = 9) + P (x = 10) \\ = {}_{10}{C_{8}} {(\frac{1}{2})}^{8} {(\frac{1}{2})}^{2} + {}_{10}{C_{9}} {(\frac{1}{2})}^{9} (\frac{1}{2}) + {}_{10}{C_{10}} {(\frac{1}{2})}^{10} {(\frac{1}{2})}^{0} \\ = 45 . {(\frac{1}{2})}^{10} + 10 . {(\frac{1}{2})}^{10} + {(\frac{1}{2})}^{10} = {(\frac{1}{2})}^{10} (45 + 10 + 1) \\ = 56 \times \frac{1}{1024} = \frac{7}{128} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

Two cards are drawn from a well-shuffled deck of 52 playing cards with replacement. The probability that both cards are queens is:

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 Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} P r o b a b i l i t y o f g e t t i n g Q u e e n = \frac{4}{52} \\ S o, t h e r e q u i r e d p r o b a b i l i t y = P (Q u e e n) . P (Q u e e n) \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{4}{52} . \frac{4}{52} = \frac{1}{13} . \frac{1}{13} (w i t h r e p l a c e m e n t) \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

Which one is not a requirement of a binomial distribution?

(a) There are 2 outcomes for each trial

(b) There is a fixed number of trials

(c) The outcomes must be dependent on each other

(d) The probability of success must be the same for all the trials

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 Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e k n o w t h a t f o r a B i n o m i a l d i s t r i b u t i o n, t h e o u t c o m e s m u s t n o t b e d e p e n d e n t o n e a c h o t h e r . \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

Two dice are thrown. If it is known that the sum of the numbers on the dice was less than 6, the probability of getting a sum of 3, is

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 Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t E_{1} b e t h e e v e n t s h o w i n g t h e s u m o f t h e n u m b e r s o n t h e t w o d i c e w a s l e s s t h a n 6 \\ ∴ E_{1} = {(1, 1), (1, 2), (2, 1), (1, 3), (3, 1), (1, 4), (4, 1), (2, 2), (2, 3), (3, 2)} \\ n (E_{1}) = 10 \\ . E_{2} b e t h e e v e n t t h a t t h e s u m o f t h e n u m b e r s i s 3 . \\ ∴ E_{2} = {(1, 2), (2, 1)} \\ n (E_{2}) = 2 a n d n (E_{1} \cap E_{2}) = 2 \\ ∴ P (E_{2} / E_{1}) = \frac{n (E_{1} \cap E_{2})}{n (E_{1})} = \frac{2}{10} = \frac{1}{5} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

Eight coins are tossed together. The probability of getting exactly 3 heads is

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol: $\begin{array}{l} \end{array}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

A flashlight has 8 batteries out of which 3 are dead. If two batteries are selected without replacement and tested, the probability that both are dead is

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 Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} p r o b a b i l i t y = P (d e a d) . P (d e a d) \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{3}{8} . \frac{2}{7} = \frac{3}{28} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

A box contains 3 orange balls, 3 green balls, and 2 blue balls. Three balls are drawn at random from the box without replacement. The probability of drawing 2 green balls and one blue ball is

17 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} P robability o f d r a w i n g 2 g r e e n a n d 1 b l u e b a l l s \\ = P (G) . P (G) . P (B) + P (G) . P (B) . P (G) + P (B) . P (G) . P (G) \\ = \frac{3}{8} . \frac{2}{7} . \frac{2}{6} + \frac{3}{8} . \frac{2}{7} . \frac{2}{6} + \frac{2}{8} . \frac{3}{7} . \frac{2}{6} = \frac{12}{336} + \frac{12}{336} + \frac{12}{336} = \frac{36}{336} = \frac{3}{28} \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

A die is thrown and a card is selected at random from a deck of 52 playing cards. The probability of getting an even number on the die and a spade card is

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 Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t E_{1} b e t h e e v e n t o f g e t t i n g e v e n n u m b e r o n t h e d i e . E_{2} b e t h e e v e n t o f s e l e c t i n g a s p a d e c a r d . \\ ∴ P (E_{1}) = \frac{3}{6} = \frac{1}{2} a n d P (E_{2}) = \frac{13}{52} = \frac{1}{4} \\ S o, P (E_{1} \cap E_{2}) = P (E_{1}) . P (E_{2}) = \frac{1}{2} . \frac{1}{4} = \frac{1}{8} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

Assume that in a family, each child is equally likely to be a boy or a girl. A family with three children is chosen at random. The probability that the eldest child is a girl, given that the family has at least one girl is

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 Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t G d e n o t e s t h e g i r l a n d B d e n o t e s t h e b o y o f t h e g i v e n f a m i l y . \\ S o, n (S) = {(B G G), (G B G), (G G B), (G B B), (B G B), (B B G), (B B B), (G G G)} \\ L e t E_{1} b e t h e e v e n t t h a t t h e f a m i l y h a s a t l e a s t o n e g i r l \\ ∴ E_{1} = {(B G G), (G B G), (G G B), (G B B), (B G B), (B B G), (G G G)} \\ . E_{2} b e t h e e v e n t t h a t t h e e l d e s t c h i l d i s a g i r l . \\ ∴ E_{2} = {(G B G), (G G B), (G B B), (G G G)} \\ (E_{1} \cap E_{2}) = {(G B B), (G G B), (G B G), (G G G)} \\ ∴ P (E_{2} / E_{1}) = \frac{P (E_{1} \cap E_{2})}{P (E_{1})} = \frac{4 / 8}{7 / 8} = \frac{4}{7} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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New answer posted•a year ago

Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

Three persons, A, B, and C, fire at a target in turn, starting with A. Their probability of hitting the target are 0.4, 0.3, and 0.2 respectively. The probability of exactly two hits is:

(a) 0.024

(b) 0.188

(c) 0.336

(d) 0.452

8 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : P (A) = 0.4, P (B) = 0.3, P (C) = 0.2 \\ A l s o, P (\bar{A}) = 1 - 0.4 = 0.6, P (\bar{B}) = 1 - 0.3 = 0.7, P (\bar{C}) = 1 - 0.2 = 0.8 \\ ∴ Probabilities o f t w o h i t s \\ = P (A) . P (B) . P (\bar{C}) + P (A) . P (\bar{B}) . P (C) + P (\bar{A}) . P (B) . P (C) \\ = 0.4 \times 0.3 \times 0.8 + 0.4 \times 0.7 \times 0.2 + 0.6 \times 0.3 \times 0.2 \\ = 0.096 + 0.056 + 0.036 = 0.188 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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