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Ncert Solutions Maths class 12th

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Ncert Solutions Maths class 12th Probability

64. In a meeting 70% of the members favour a certain proposal, 30% being opposed. A member is selected at random and we let X = 0 if the opposed and X = 1 if he is in favour. Find E (X) and Var (X).

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alok kumar singh

Contributor-Level 10

64. It is given that P (X = 0) = 30% = 30/100 = 0.3

P (X = 1) = 70% = 70/100 = 0.7

Therefore, the probability distribution is as follows.

X	0	1
P (X)	0.3	0.7

$\begin{array}{l} T h e n, E (X) = \sum_{}^{} X_{i} P (X_{i}) \\ = 0 * 0.3 + 1 * 0.7 \\ = 0.7 \end{array}$

$\begin{array}{l} E (X^{2}) = \sum_{}^{} {X^{2}}_{i} P (X_{i}) \\ = 0^{2} * 0.3 + {(1)}^{2} * 0.7 \\ = 0.7 \end{array}$

It is know that, $V a r (X) = E (X^{2}) - {[E (X)]}^{2}$

= 0.7 − (0.7)²

= 0.7 − 0.49

= 0.21

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Ncert Solutions Maths class 12th Probability

63. A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years. One student is selected in such a manner that each has the same chance of being chosen and the age X of the selected student is recorded. What is the probability distribution of the random variable X? Find mean, variance and standard deviation of X.

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alok kumar singh

Contributor-Level 10

63. There are 15 students in the class. Each student has the same chance to be chosen. Therefore, the probability of each student to be selected is 1/15.

The given information can be compiled in the frequency table as follows.

14	15	16	17	18	19	20	21
2	1	2	3	1	2	3	1

P(X = 14) =2/15, P(X = 15) =1/15, P(X = 16) =2/15, P(X = 16) =3/15,

P(X = 18) =115, P(X = 19) =2/15, P(X = 20) =3/15, P(X = 21) =1/15

Therefore, the probability distribution of random variable X is as follows.

X	14	15	16	17	18	19	20	21
f	2/15	1/15	2/15	3/15	1/15	2/15	3/15	1/15

Then, mean of X = E(X)

$\begin{array}{l} = \sum_{}^{} X_{i} P (X_{i}) \\ = 14 * \frac{2}{15} + 15 * \frac{1}{15} + 16 * \frac{2}{15} + 17 * \frac{3}{15} + 18 * \frac{1}{15} + 19 * \frac{2}{15} + 20 * \frac{3}{15} + 21 * \frac{1}{15} \\ = \frac{1}{15} (28 + 15 + 32 + 51 + 18 + 38 + 60 + 21) \\ = \frac{263}{15} \\ = 17.53 \end{array}$

$\begin{array}{l} E (X^{2}) = \sum_{}^{} {X^{2}}_{i} P (X_{i}) \\ = {(14)}^{2} . \frac{2}{15} + {(15)}^{2} . \frac{1}{15} + {(16)}^{2} . \frac{2}{15} + {(17)}^{2} . \frac{3}{15} + {(18)}^{2} . \frac{1}{15} + {(19)}^{2} . \frac{2}{15} + {(20)}^{2} . \frac{3}{15} + {(21)}^{2} . \frac{1}{15} \\ = \frac{1}{15} . (392 + 225 + 512 + 867 + 324 + 722 + 1200 + 441) \\ = \frac{4683}{15} \\ = 312.2 \end{array}$

$\begin{array}{l} ∴ V a r i a n c e (X) = E (X^{2}) - {[E (X)]}^{2} \\ = 312.2 - {(\frac{263}{15})}^{2} \\ = 312.2 - 307.4177 \\ = 4.7823 \\ \approx 4.78 \end{array}$

Standard derivation $\begin{array}{l} =\sqrtvariance (X) \\ =\sqrt4.78 \\ = 2.186 \approx 2.19 \end{array}$

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Ncert Solutions Maths class 12th Probability

62. Let X denotes the sum of the numbers obtained when two fair dice are rolled. Find the variance and standard deviation of X.

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alok kumar singh

Contributor-Level 10

62. When two fair dice are rolled, 6 * 6 = 36 observations are obtained.

P(X = 2) = P(1, 1) = 1/36

P(X = 3) = P (1, 2) + P(2, 1) = 2/36 = 1/18

P(X = 4) = P(1, 3) + P(2, 2) + P(3, 1) = 3/36 = 1/12

P(X = 5) = P(1, 4) + P(2, 3) + P(3, 2) + P(4, 1) = 4/36 = 1/9

P(X = 6) = P(1, 5) + P (2, 4) + P(3, 3) + P(4, 2) + P(5, 1) = 5/36

P(X = 7) = P(1, 6) + P(2, 5) + P(3, 4) + P(4, 3) + P(5, 2) + P(6, 1) = 6/36 = 1/6

P(X = 8) = P(2, 6) + P(3, 5) + P(4, 4) + P(5, 3) + P(6, 2) = 5/36

P(X = 9) = P(3, 6) + P(4, 5) + P(5, 4) + P(6, 3) = 4/36 = 1/9

P(X = 10) = P(4, 6) + P(5, 5) + P(6, 4) = 3/36 = 1/12

P(X = 11) = P(5, 6) + P(6, 5) = 2/36 = 1/18

P(X = 12) = P(6,

...more

62. When two fair dice are rolled, 6 * 6 = 36 observations are obtained.

P(X = 2) = P(1, 1) = 1/36

P(X = 3) = P (1, 2) + P(2, 1) = 2/36 = 1/18

P(X = 4) = P(1, 3) + P(2, 2) + P(3, 1) = 3/36 = 1/12

P(X = 5) = P(1, 4) + P(2, 3) + P(3, 2) + P(4, 1) = 4/36 = 1/9

P(X = 6) = P(1, 5) + P (2, 4) + P(3, 3) + P(4, 2) + P(5, 1) = 5/36

P(X = 7) = P(1, 6) + P(2, 5) + P(3, 4) + P(4, 3) + P(5, 2) + P(6, 1) = 6/36 = 1/6

P(X = 8) = P(2, 6) + P(3, 5) + P(4, 4) + P(5, 3) + P(6, 2) = 5/36

P(X = 9) = P(3, 6) + P(4, 5) + P(5, 4) + P(6, 3) = 4/36 = 1/9

P(X = 10) = P(4, 6) + P(5, 5) + P(6, 4) = 3/36 = 1/12

P(X = 11) = P(5, 6) + P(6, 5) = 2/36 = 1/18

P(X = 12) = P(6, 6) = 1 x 36

Therefore, the required probability distribution is as follows.

X	2	3	4	5	6	7	8	9	10	11	12
P(X)	1/36	1/18	1/12	1/9	5/36	1/6	5/36	1/9	1/12	1/18	1/36

$\begin{array}{l} T h e n, E (X) = \sum_{}^{} X_{i} . P (X_{i}) \\ = 2 * \frac{1}{36} + 3 * \frac{1}{18} + 4 * \frac{1}{12} + 5 * \frac{1}{9} + 6 * \frac{5}{36} + 7 * \frac{1}{6} + 8 * \frac{5}{36} + 9 * \frac{1}{9} + 10 * \frac{1}{12} + 11 * \frac{1}{18} + 12 * \frac{1}{36} \\ = \frac{1}{18} + \frac{1}{6} + \frac{1}{3} + \frac{5}{9} + \frac{5}{6} + \frac{7}{6} + \frac{10}{9} + 1 + \frac{5}{6} + \frac{11}{18} + \frac{1}{3} \\ = 7 \end{array}$

$\begin{array}{l} E (X^{2}) - \sum_{}^{} {X^{2}}_{i} . P (X_{i}) \\ = 4 * \frac{1}{36} + 9 * \frac{1}{18} + 16 * \frac{1}{12} + 25 * \frac{1}{9} + 36 * \frac{5}{36} + 49 * \frac{1}{6} + 64 * \frac{5}{36} + 81 * \frac{1}{9} + 100 * \frac{1}{12} + 121 * \frac{1}{18} + 144 * \frac{1}{36} \\ = \frac{1}{9} + \frac{1}{2} + \frac{4}{3} + \frac{25}{9} + 5 + \frac{49}{6} + \frac{80}{9} + 9 + \frac{25}{3} + \frac{121}{18} + 4 \\ = \frac{987}{18} = \frac{329}{6} = 54.833 \end{array}$

$\begin{array}{l} T h e n, V a r (X) = E (X^{2}) - [E {(X)}^{2}] \\ = 54.833 - {(7)}^{2} \\ = 54.833 - 49 \\ = 5.833 \end{array}$

$∴$ Standard derivation $\begin{array}{l} =\sqrtvar(X) \\ =\sqrt5.833 \\ = 2.415 \end{array}$

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Ncert Solutions Maths class 12th Probability

61. Two numbers are selected at random (without replacement), from the first six positive integers. Let X denotes the larger of two numbers obtained. Find E (X).

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alok kumar singh

Contributor-Level 10

61. The two positive integers can be selected from the first six positive integers without replacement in 6 * 5 = 30 ways

X represents the larger of the two numbers obtained. Therefore, X can take the value of 2, 3, 4, 5, or 6.

For X = 2, the possible observations are (1, 2) and (2, 1).

P(X = 2) $= \frac{2}{30} = \frac{1}{15}$

For X = 3, the possible observations are (1, 3), (2, 3), (3, 1), and (3, 2).

P(X = 3) $= \frac{4}{30} = \frac{2}{15}$

For X = 4, the possible observations are (1, 4), (2, 4), (3, 4), (4, 3), (4, 2), and (4, 1).

P(X = 4) $= \frac{6}{30} = \frac{1}{5}$

For X = 5, the possible observations are (1, 5), (2, 5), (3, 5), (4, 5), (5, 4), (5, 3), (5, 2), and (5, 1).

P(X = 5) $= \frac{8}{30} = \frac{4}{15}$

For X = 6, the possible ob

...more

61. The two positive integers can be selected from the first six positive integers without replacement in 6 * 5 = 30 ways

X represents the larger of the two numbers obtained. Therefore, X can take the value of 2, 3, 4, 5, or 6.

For X = 2, the possible observations are (1, 2) and (2, 1).

P(X = 2) $= \frac{2}{30} = \frac{1}{15}$

For X = 3, the possible observations are (1, 3), (2, 3), (3, 1), and (3, 2).

P(X = 3) $= \frac{4}{30} = \frac{2}{15}$

For X = 4, the possible observations are (1, 4), (2, 4), (3, 4), (4, 3), (4, 2), and (4, 1).

P(X = 4) $= \frac{6}{30} = \frac{1}{5}$

For X = 5, the possible observations are (1, 5), (2, 5), (3, 5), (4, 5), (5, 4), (5, 3), (5, 2), and (5, 1).

P(X = 5) $= \frac{8}{30} = \frac{4}{15}$

For X = 6, the possible observations are (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6,5) , (6, 4), (6, 3), (6, 2), and (6, 1).

P(X = 6) $= \frac{10}{30} = \frac{1}{3}$

Therefore, the required probability distribution is as follows.

X	2	3	4	5	6
P(X)	1/15	2/15	1/5	4/15	1/3

$\begin{array}{l} T h e n, E (X) = \sum_{}^{} X_{i} P (X_{i}) \\ = 2 . \frac{1}{5} + 3 . \frac{2}{15} + 4 . \frac{1}{5} + 5 . \frac{4}{15} + 6 . \frac{1}{3} \\ = \frac{2}{15} + \frac{2}{5} + \frac{4}{5} + \frac{4}{3} + 3 \\ = \frac{70}{15} \\ = 14 \end{array}$

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Ncert Solutions Maths class 12th Probability

60. Two dice are thrown simultaneously. If X denotes the number of sixes, find expectation of X.

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alok kumar singh

Contributor-Level 10

60. Two dice thrown simultaneously is the same the die thrown 2 times.

Here, X represents the number of sixes obtained when two dice are thrown simultaneously. Therefore, X can take the value of 0, 1, or 2.

∴ P (X = 0) = P (not getting six on any of the dice) = 25/36

P (X = 1) = P (six on first die and no six on second die) + P (no six on first die and six on second die)

=2 (1/6 x 5/6) = 10/36

P (X = 2) = P (six on both the dice) =1/36

Therefore, the required probability distribution is as follows.

X	0	1	2
P (X)	25/36	10/36	1/36

Then, expectation of X = E (X) =∑ X iP (Xi)

= 0 x 25/36 + 1 x 10/36 + 2 x 1/36

= 1/3

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Ncert Solutions Maths class 12th Probability

59. Find the mean number of heads in three tosses of fair coin.

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alok kumar singh

Contributor-Level 10

59. Let X denote the success of getting heads.

Therefore, the sample space is

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

It can be seen that X can take the value of 0, 1, 2, or 3

$\begin{array}{l} ∴ P (X = 0) = P (T T T) \\ = P (T) . P (T) . P (T) \\ = \frac{1}{2} * \frac{1}{2} * \frac{1}{2} \\ = \frac{1}{8} \end{array}$

$\begin{array}{l} ∴ P (X = 1) = P (H H T) + P (H T H) + P (T H H) \\ = \frac{1}{2} * \frac{1}{2} * \frac{1}{2} * \frac{1}{2} * \frac{1}{2} * \frac{1}{2} * \frac{1}{2} * \frac{1}{2} * \frac{1}{2} \\ = \frac{3}{8} \end{array}$

$\begin{array}{l} ∴ P (X = 2) = P (H H T) + P (H T H) + P (T H H) \\ = \frac{1}{2} * \frac{1}{2} * \frac{1}{2} * \frac{1}{2} * \frac{1}{2} * \frac{1}{2} * \frac{1}{2} * \frac{1}{2} * \frac{1}{2} \\ = \frac{3}{8} \end{array}$

$\begin{array}{l} ∴ P (X = 3) = P (H H H) \\ = \frac{1}{2} * \frac{1}{2} * \frac{1}{2} \\ = \frac{1}{8} \end{array}$

Therefore, the required probability distribution is as follows.

X	0	1	2	3
P(X)	1/8	3/8	3/8	1/8

Mean of $X E (X), μ = \sum_{}^{} X_{i} P (X_{i})$

$\begin{array}{l} = 0 * \frac{1}{8} + 1 * \frac{3}{8} + 2 * \frac{3}{8} + 3 * \frac{1}{8} \\ = \frac{3}{8} + \frac{3}{4} + \frac{3}{8} \\ = \frac{3}{2} \\ = 1.5 \end{array}$

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Ncert Solutions Maths class 12th Probability

58. The random variable X has a probability distribution P(X) of the following form, where k is some number:

$P (X) = {\begin{cases} k, i f, x = 0 \\ 2 k, i f, x = 1 \\ 3 k, i f, x = 2 \\ 0, o t h e r w w i s e \end{cases}$

(a) Determine the value of k.

(b) Find P(X < 2), P(X 2), P(X 2).

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alok kumar singh

Contributor-Level 10

58. (a) It is known that the sum of probabilities of a probability distribution of random variables is one.

∴ k + 2k + 3k + 0 = 1

⇒ 6k = 1

⇒ k =1/6

(b) P (X < 2) = P (X = 0) + P (X = 1)

= k + 2k

= 3k

=3/6 = 1/2

$\begin{array}{l} P (X \leq 2) = P (X = 0) + P (X = 1) + P (X = 2) \\ = k + 2 k + 3 k \\ = 6 k \\ = \frac{6}{6} \\ = 1 \end{array}$

$\begin{array}{l} P (X \geq 2) = P (X = 2) + P (X > 2) \\ = 3 k + 0 \\ = 3 k \\ = \frac{3}{6} \\ = \frac{1}{2} \end{array}$

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Ncert Solutions Maths class 12th Probability

57. A random variable X has the following probability distribution:

X	0	1	2	3	4	5	6	7
P (X)	0	k	2k	2k	3k	k²	2k²	7k²+ k

Determine:

(i) k

(ii) P (X < 3)

(iii) P (X > 6)

(iv) P (0 < X < 3)

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alok kumar singh

Contributor-Level 10

57. (i) Since, the sum of all the probabilities of a distribution is 1.

$\begin{array}{l} \Rightarrow 0 + k + 2 k + 2 k + 3 k + k^{2} + 2 k^{2} + (7 k^{2} + k) = 1 \\ \Rightarrow 10 k^{2} + 9 k = 0 \\ \Rightarrow (10 k - 1) (k + 1) = 0 \\ \Rightarrow k = - 1, \frac{1}{4} \end{array}$

K=-1 is not possible as the probability of an event is never negative.

$∴ k = \frac{1}{10}$

(ii) $P (X < 3) = P (X = 0) + P (X = 1) + P (X = 2)$

$\begin{array}{l} = 0 + k + 2 k \\ = 3 k \\ = 3 * \frac{1}{10} \\ = \frac{3}{10} \end{array}$

(iii) $P (X > 6) = P (X = 7)$

$\begin{array}{l} = 7 k^{2} + k \\ = 7 * {(\frac{1}{10})}^{2} + \frac{1}{10} \\ = \frac{7}{100} + \frac{1}{10} \\ = \frac{17}{100} \end{array}$

(iv) $P (0 < X < 3) = P (X = 1) + P (X = 2)$

$\begin{array}{l} = k + 2 k \\ = 3 k \\ = 3 * \frac{1}{10} \\ = \frac{3}{10} \end{array}$

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Ncert Solutions Maths class 12th Probability

56. A coin is biased so that the head is 3 times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of number of tails.

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alok kumar singh

Contributor-Level 10

56. Let the probability of getting a tail in the biased coin be x.

∴ P (T) = x

⇒ P (H) = 3x

For a biased coin, P (T) + P (H) = 1

$\begin{array}{l} \Rightarrow x + 3 x = 1 \\ \Rightarrow 4 x = 1 \\ \Rightarrow x = \frac{1}{4} \\ ∴ P (T) = \frac{1}{4}, a n d, P (H) = \frac{3}{4} \end{array}$

When the coin is tossed twice, the sample space is {HH, TT, HT, TH}.

Let X be the random variable representing the number of tails.

∴ P (X = 0) = P (no tail) = P (H) * P (H) = 3/4 x 3/4 = 9/16

P (X = 1) = P (one tail) = P (HT) + P (TH)

= 3/4 . 1/4 + 1/4 . 3/4

= 3/16 + 3/16= 3/8

P (X = 2) = P (two tails) = P (TT) = 1/4 x 1/4 = 1/16

Therefore, the required probability distribution is as follows.

X	0	1	2
P (X)	9/16	3/8	1/16

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Ncert Solutions Maths class 12th Probability

55. A coin is biased so that the head is 3 times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of number of tails.

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alok kumar singh

Contributor-Level 10

55. Let the probability of getting a tail in the biased coin be x.

∴ P (T) = x

⇒ P (H) = 3x

For a biased coin, P (T) + P (H) = 1

$\begin{array}{l} \Rightarrow x + 3 x = 1 \\ \Rightarrow 4 x = 1 \\ \Rightarrow x = \frac{1}{4} \\ ∴ P (T) = \frac{1}{4}, a n d, P (H) = \frac{3}{4} \end{array}$

When the coin is tossed twice, the sample space is {HH, TT, HT, TH}.

Let X be the random variable representing the number of tails.

∴ P (X = 0) = P (no tail) = P (H) * P (H) = 3/4 x 3/4 = 9/16

P (X = 1) = P (one tail) = P (HT) + P (TH)

= 3/4 . 1/4 + 1/4 . 3/4

= 3/16 + 3/16= 3/8

P (X = 2) = P (two tails) = P (TT) = 1/4 x 1/4 = 1/16

Therefore, the required probability distribution is as follows.

X	0	1	2
P (X)	9/16	3/8	1/16

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