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Class 12th Probability

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

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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New answer posted

a year ago

Class 12th Probability

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

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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New answer posted

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Class 12th Probability

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

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

Sol:

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

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New answer posted

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Class 12th Probability

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

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

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New answer posted

a year ago

Class 12th Probability

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

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

Sol:

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

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New answer posted

a year ago

Class 12th Probability

A die is tossed twice. A 'success' is getting an even number on a toss. Find the variance of the number of successes.

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alok kumar singh

Contributor-Level 10

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

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

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New answer posted

a year ago

Class 12th Probability

Two cards are drawn successively without replacement from a well-shuffled deck of cards. Find the mean and standard deviation of the random variable $X$ , where $X$ is the number of aces

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alok kumar singh

Contributor-Level 10

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

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

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New answer posted

a year ago

Class 12th Probability

A bag contains $(2 n + 1)$ coins. It is known that $n$ of these coins have a head on both sides, while the rest of the coins are fair. A coin is picked at random from the bag and tossed. If the probability that the toss results in a head is $\frac{31}{42}$ , determine the value of $n$ .

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t n c o i n s a r e t w o h e a d e d c o i n s a n d t h e r e m a i n i n g (n + 1) c o i n s a r e f a i r . \\ L e t E_{1} : t h e e v e n t t h a t u n f a i r c o i n i s s e l e c t e d . \\ E_{2} : t h e e v e n t t h a t f a i r c o i n i s s e l e c t e d . \\ E_{3} : t h e e v e n t t h a t t h e t o s s r e s u l t i s a h e a d . \\ ∴ P (E_{1}) = \frac{n}{2 n + 1} a n d P (E_{2}) = \frac{n + 1}{2 n + 1} \\ P (E / E_{1}) = 1 (s u r e e v e n t) a n d P (E / E_{2}) = \frac{1}{2} \\ ∴ P (E) = P (E_{1}) . P (E / E_{1}) + P (E_{2}) . P (E / E_{2}) \\ = \frac{n}{2 n + 1} . 1 + \frac{n + 1}{2 n + 1} . \frac{1}{2} = \frac{1}{2 n + 1} (n + \frac{n + 1}{2}) \\ = \frac{1}{2 n + 1} (\frac{2 n + n + 1}{2}) = \frac{3 n + 1}{2 (2 n + 1)} \\ B u t P (E) = \frac{31}{42} (g i v e n) \\ ∴ \frac{3 n + 1}{2 (2 n + 1)} = \frac{31}{42} \Rightarrow \frac{3 n + 1}{2 n + 1} = \frac{31}{21} \\ \Rightarrow 63 n + 21 = 62 n + 31 \\ \Rightarrow n = 10 \\ H e n c e, t h e r e q u i r e d v a l u e o f n i s 10 . \end{array}$