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Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

Three events A, B, and C have probabilities $P (A) = \frac{2}{5}$ , $\frac{1}{3}$ and $\frac{1}{2}$ , respectively. Given that $P (A \cap C) = \frac{1}{5}$ and $P (B \cap C) = \frac{1}{4}$ , find the values of $P (C ? B)$ and $P (A^{'} \cap C^{'})$ .

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Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} W e h a v e P (A) = \frac{2}{5}, P (B) = \frac{1}{3} a n d P (C) = \frac{1}{2} \\ P (A \cap C) = \frac{1}{5} a n d P (B \cap C) = \frac{1}{4} \\ ∴ P (C / B) = \frac{P (B \cap C)}{P (B)} = \frac{1 / 4}{1 / 3} = \frac{3}{4} \\ P (A' \cap C') = 1 - P (A \cup C) = 1 - [P (A) + P (C) - P (A \cap C)] \\ = 1 - [\frac{2}{5} + \frac{1}{2} - \frac{1}{5}] = 1 - \frac{7}{10} = \frac{3}{10} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t i e s a r e \frac{3}{4} a n d \frac{3}{10} . \end{array}$

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Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

<p><strong>Three events A, B, and C have probabilities <math><mrow><mi>P</mi><mrow><mo>(</mo><mrow><mi>A</mi></mrow><mo>)</mo></mrow><mo>=</mo><mfrac><mrow><mn>2</mn></mrow><mrow><mn>5</mn></mrow></mfrac></mrow></math> , <math><mrow><mo> </mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>3</mn></mrow></mfrac></mrow></math> and <math><mrow><mfrac><mrow><mn>1</mn></mrow><mrow><mn>2</mn></mrow></mfrac></mrow></math> , respectively. Given that <math><mrow><mi>P</mi><mrow><mo>(</mo><mrow><mi>A</mi><mo>∩</mo><mi>C</mi></mrow><mo>)</mo></mrow><mo>=</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>5</mn></mrow></mfrac></mrow></math> and <math><mrow><mi>P</mi><mrow><mo>(</mo><mrow><mi>B</mi><mo>∩</mo><mi>C</mi></mrow><mo>)</mo></mrow><mo>=</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>4</mn></mrow></mfrac></mrow></math> , find the values of <math><mrow><mi>P</mi><mrow><mo>(</mo><mrow><mi>C</mi><mo>?</mo><mi>B</mi></mrow><mo>)</mo></mrow></mrow></math> and <math><mrow><mi>P</mi><mrow><mo>(</mo><mrow><msup><mi>A</mi><mo>′</mo></msup><mo>∩</mo><msup><mi>C</mi><mo>′</mo></msup></mrow><mo>)</mo></mrow></mrow></math> .</strong></p>

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Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

A and B are two events such that $P (A) = \frac{1}{2}$ , $P (B) = \frac{1}{3}$ , and $P (A \cap B) = \frac{1}{4}$ . Find:

$i. P(A ∣ B$

$ii. (B∣ A)$

$iii. (A' ∣ B)$

$iv. (A' ∣ B')$

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Asked by Alok Kumar Singh

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Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} W e h a v e P (A) = \frac{1}{2}, P (B) = \frac{1}{3} a n d P (A \cap B) = \frac{1}{4} \\ P (A') = 1 - \frac{1}{2} = \frac{1}{2}, P (B') = 1 - \frac{1}{3} = \frac{2}{3} \\ P (A' \cap B') = 1 - P (A \cup B) = 1 - [P (A) + P (B) - P (A \cap B)] \\ = 1 - [\frac{1}{2} + \frac{1}{3} - \frac{1}{4}] = 1 - [\frac{6 + 4 - 3}{12}] = 1 - \frac{7}{12} = \frac{5}{12} \\ (i) P (A / B) = \frac{P (A \cap B)}{P (B)} = \frac{1 / 4}{1 / 3} = \frac{3}{4} \\ (i i) P (B / A) = \frac{P (A \cap B)}{P (A)} = \frac{1 / 4}{1 / 2} = \frac{1}{2} \\ (i i i) P (A' / B) = \frac{P (A' \cap B)}{P (B)} = \frac{P (B) - P (A \cap B)}{P (B)} = 1 - \frac{P (A \cap B)}{P (B)} \\ = 1 - \frac{1 / 4}{1 / 3} = 1 - \frac{3}{4} = \frac{1}{4} \\ (i v) P (A' / B') = \frac{P (A' \cap B')}{P (B')} = \frac{5 / 12}{2 / 3} = \frac{5}{12} \times \frac{3}{2} = \frac{5}{8} \end{array}$

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Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

Explain why the experiment of tossing a coin three times is said to have binomial distribution.

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Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} W e k n o w t h a t r a n d o m variable X t a k e s v a l u e s 0, 1, 2, 3, \dots, n i s s a i d t o b e b i n o m i a l d i s t r i b u t i o n \\ h a v i n g p a r a m e t e r s n a n d p, i f t h e p r o b a b i l i t y i s g i v e n b y \\ P (X = r) = {}_{n}{C_{r}} p^{r} q^{n - r}, w h e r e q = 1 - p a n d r = 0, 1, 2, 3, \dots \\ S i m i l a r l y, i n c a s e o f t o s s i n g a c o i n 3 t i m e s, \\ n = 3 a n d X h a s t h e v a l u e s 0, 1, 2, 3, \dots w i t h p = \frac{1}{2}, q = \frac{1}{2} . \\ H e n c e, i t i s s a i d t o h a v e a b i n o m i a l d i s t r i b u t i o n . \end{array}$

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Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

Two dice are thrown together and the total score is noted. The events E, F, and G are ‘a total of 4’, ‘a total of 9 or more’, and ‘a total divisible by 5’, respectively.

Calculate $P (E)$ , $P (F)$ , and $P (G)$ and decide which pairs of events, if any, are independent.

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Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} T w o d i c e a r e t h r o w n t o g e t h e r \\ ∴ n (S) = 36 \\ E = A t o t a l o f 4 = {(2, 2), (1, 3), (3, 1)} ∴ n (E) = 3 \\ F = A t o t a l o f 9 o r m o r e \\ = {(3, 6), (6, 3), (5, 4), (4, 5), (5, 5), (4, 6), (6, 4), (5, 6), (6, 5), (6, 6)} ∴ n (F) = 10 \\ G = A t o t a l d i v i s i b l e b y 5 \\ = {(1, 4), (4, 1), (2, 3), (3, 2), (4, 6), (6, 4), (5, 5)} ∴ n (G) = 7 \\ H e r e, w e s e e t h a t (E \cap F) = ? a n d (E \cap G) = ? \\ a n d (F \cap G) = {(4, 6), (6, 4), (5, 5)} \\ ∴ n (F \cap G) = 3 a n d (E \cap F \cap G) = ? \\ ∴ P (E) = \frac{n (E)}{n (S)} = \frac{3}{36} = \frac{1}{12} \\ P (F) = \frac{n (F)}{n (S)} = \frac{10}{36} = \frac{5}{18} a n d P (G) = \frac{n (G)}{n (S)} = \frac{7}{36} \\ P (F \cap G) = \frac{3}{36} = \frac{1}{12} a n d P (F) . P (G) = \frac{5}{18} . \frac{7}{36} = \frac{35}{648} \\ Since, P (F \cap G) \neq P (F) . P (G) \\ H e n c e, t h e r e i s n o p a i r o f i n d e p e n d e n t e v e n t s . \end{array}$

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Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

A bag contains 5 red marbles and 3 black marbles. Three marbles are drawn one by one without replacement. What is the probability that at least one of the three marbles drawn be black, if the first marble is red?

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Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} L e t r e d m a r b l e s b e r e p r e s e n t e d w i t h R a n d b l a c k m a r b l e s b e r e p r e s e n t e d w i t h B . \\ T h e f o l l o w i n g t h r e e c o n d i t i o n s a r e p o s s i b l e, i f a t l e a s t o n e o f t h e t h r e e m a r b l e s \\ d r a w n b e b l a c k a n d f i r s t m a r b l e i s r e d . \\ (i) E_{1} = I I b a l l i s b l a c k a n d I I I i s r e d \\ (i i) E_{2} = I I b a l l i s b l a c k a n d I I I i s a l s o b l a c k \\ (i i i) E_{3} = I I b a l l i s r e d a n d I I I i s b l a c k \\ ∴ P (E_{1}) = P (R_{1}) . P (B_{1} / R_{1}) . P (R_{2} / R_{1} B_{1}) = \frac{5}{8} . \frac{3}{7} . \frac{4}{6} = \frac{60}{336} = \frac{5}{28} \\ P (E_{2}) = P (R_{1}) . P (B_{1} / R_{1}) . P (B_{2} / R_{1} B_{1}) = \frac{5}{8} . \frac{3}{7} . \frac{2}{6} = \frac{30}{336} = \frac{5}{56} \\ a n d P (E_{3}) = P (R_{1}) . P (R_{2} / R_{1}) . P (B_{1} / R_{1} R_{2}) = \frac{5}{8} . \frac{4}{7} . \frac{3}{6} = \frac{60}{336} = \frac{5}{28} \\ ∴ P (E) = P (E_{1}) + P (E_{2}) + P (E_{3}) = \frac{5}{28} + \frac{5}{56} + \frac{5}{28} = \frac{25}{56} \\ H e n c e, t h e r e q u i r e d p r o b a b i l i t y i s \frac{25}{56} . \end{array}$

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Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

The probability that at least one of the two events A and B occurs is 0.6. If A and B occur simultaneously with probability 0.3, evaluate $P (\bar{A}$ ) $+ P (\bar{B}))$ .

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Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} W e k n o w t h a t : \\ A \cup B d e n o t e s t h a t a t l e a s t o n e o f t h e e v e n t s o c c u r s a n d \\ A \cap B d e n o t e s t h a t t h e t w o e v e n t s o c c u r s simultaneously . \\ S o, P (A \cup B) = P (A) + P (B) - P (A \cap B) \\ \Rightarrow 0.6 = P (A) + P (B) - 0.3 \\ \Rightarrow 0.9 = P (A) + P (B) \\ \Rightarrow 0.9 = 1 - P (\bar{A}) + 1 - P (\bar{B}) \\ \Rightarrow P (\bar{A}) + P (\bar{B}) = 2 - 0.9 = 1.1 \\ H e n c e, t h e r e q u i r e d a n s w e r i s 1.1 \end{array}$

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Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

Refer to Exercise 1 above. If the die were fair, determine whether or not the events A and B are independent.

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Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} A c c o r d i n g t o t h e s o l u t i o n o f Q . 1, w e h a v e \\ A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)} \\ ∴ n (A) = 6 a n d n (S) = 6 \times 6 = 36 \\ S o, P (A) = \frac{n (A)}{n (S)} = \frac{6}{36} = \frac{1}{6} \\ a n d B = {(4, 6), (6, 4), (5, 5), (5, 6), (6, 5), (6, 6)} \\ n (B) = 6 a n d n (S) = 6 \times 6 = 36 \\ ∴ P (B) = \frac{n (B)}{n (S)} = \frac{6}{36} = \frac{1}{6} \\ A \cap B = {(5, 5), (6, 6)} \\ ∴ P (A \cap B) = \frac{2}{36} = \frac{1}{18} \\ T h e r e f o r e, i f A a n d B a r e i n d e p e n d e n t, t h e n \\ P (A \cap B) = P (A) . P (B) \\ \Rightarrow \frac{1}{18} \neq \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} \\ \Rightarrow \frac{1}{18} \neq \frac{1}{36} \\ H e n c e, A a n d B a r e n o t i n d e p e n d e n t e v e n t s . \end{array}$

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Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

For a loaded die, the probabilities of outcomes are given as under:

$P (1) = P (2) = 0.2, P (3) = P (5) = P (6) = 0.1$ and $P (4) = 0.3$ .

The die is thrown two times. Let A and B be the events, ‘same number each time’, and ‘a total score is 10 or more’, respectively. Determine whether or not A and B are independent.

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Alok Kumar Singh | Contributor-Level 10

a year ago

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

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Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

There are 5 cards numbered 1 to 5, one number on each card. Two cards are drawn at random without replacement. Let $X$ denote the sum of the numbers on the two cards drawn. Find the mean and variance of $X$ .

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Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} H e r e, S a m p l e s p a c e S = {\begin{cases} (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2), (1, 4), (4, 1), (1, 5), (5, 1), (2, 4) \\ (4, 2), (2, 5), (5, 2), (3, 4), (4, 3), (3, 5), (5, 3), (5, 4), (4, 5) \end{cases}} \\ ∴ n (S) = 20 \\ L e t X b e t h e r a n d o m v a r i a b l e d e n o t i n g t h e s u m o f t h e n u m b e r s o n t w o c a r d s d r a w n . \\ ∴ X = 3, 4, 5, 6, 7, 8, 9 \\ S o, P (X = 3) = \frac{2}{20} P (X = 4) = \frac{2}{20} \\ P (X = 5) = \frac{4}{20} P (X = 6) = \frac{4}{20} \\ P (X = 7) = \frac{4}{20} P (X = 8) = \frac{2}{20} \\ P (X = 9) = \frac{2}{20} \\ ∴ t h e m e a n E (X) = \sum_{i = 1}^{n} X_{i} P_{i} \\ = 3 \times \frac{2}{20} + 4 \times \frac{2}{20} + 5 \times \frac{4}{20} + 6 \times \frac{4}{20} + 7 \times \frac{4}{20} + 8 \times \frac{2}{20} + 9 \times \frac{2}{20} \\ = \frac{6}{20} + \frac{8}{20} + \frac{20}{20} + \frac{24}{20} + \frac{28}{20} + \frac{16}{20} + \frac{18}{20} = \frac{120}{20} = 6 \\ ∴ E (X^{2}) = \sum_{i = 1}^{n} P_{i} X_{i}^{2} \\ = 9 \times \frac{2}{20} + 16 \times \frac{2}{20} + 25 \times \frac{4}{20} + 36 \times \frac{4}{20} + 49 \times \frac{4}{20} + 64 \times \frac{2}{20} + 81 \times \frac{2}{20} \\ = \frac{18}{20} + \frac{32}{20} + \frac{100}{20} + \frac{144}{20} + \frac{196}{20} + \frac{128}{20} + \frac{162}{20} = \frac{780}{20} = 39 \\ ∴ V a r i a n c e (X) = E (X^{2}) - {[E (X)]}^{2} \\ = 39 - {(6)}^{2} = 39 - 36 = 3 \end{array}$

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Three events A, B, and C have probabilities $P (A) = \frac{2}{5}$ , $\frac{1}{3}$ and $\frac{1}{2}$ , respectively. Given that $P (A \cap C) = \frac{1}{5}$ and $P (B \cap C) = \frac{1}{4}$ , find the values of $P (C ? B)$ and $P (A^{'} \cap C^{'})$ .

A and B are two events such that $P (A) = \frac{1}{2}$ , $P (B) = \frac{1}{3}$ , and $P (A \cap B) = \frac{1}{4}$ . Find:

$i. P(A ∣ B$

$ii. (B∣ A)$

$iii. (A' ∣ B)$

$iv. (A' ∣ B')$

Explain why the experiment of tossing a coin three times is said to have binomial distribution.

Two dice are thrown together and the total score is noted. The events E, F, and G are ‘a total of 4’, ‘a total of 9 or more’, and ‘a total divisible by 5’, respectively.

Calculate $P (E)$ , $P (F)$ , and $P (G)$ and decide which pairs of events, if any, are independent.

A bag contains 5 red marbles and 3 black marbles. Three marbles are drawn one by one without replacement. What is the probability that at least one of the three marbles drawn be black, if the first marble is red?

The probability that at least one of the two events A and B occurs is 0.6. If A and B occur simultaneously with probability 0.3, evaluate $P (\bar{A}$ ) $+ P (\bar{B}))$ .

Refer to Exercise 1 above. If the die were fair, determine whether or not the events A and B are independent.

There are 5 cards numbered 1 to 5, one number on each card. Two cards are drawn at random without replacement. Let $X$ denote the sum of the numbers on the two cards drawn. Find the mean and variance of $X$ .