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Ncert Solutions Maths class 12th

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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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New answer posted

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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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Ncert Solutions Maths class 12th Probability

54. From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.

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alok kumar singh

Contributor-Level 10

54. It is given that out of 30 bulbs, 6 are defective.

⇒ Number of non-defective bulbs = 30 − 6 = 24

4 bulbs are drawn from the lot with replacement.

Let X be the random variable that denotes the number of defective bulbs in the selected bulbs.

$∴ P (X = 0) = P$ (4 non-defective and 0 defective) $=^{4} C_{0} . \frac{4}{5} . \frac{4}{5} . \frac{4}{5} . \frac{4}{5} = \frac{256}{625}$

$P (X = 1) = P$ (3 non-defective and 1 defective) $=^{4} C_{1} . (\frac{1}{5}) . {(\frac{4}{5})}^{3} = \frac{256}{625}$

$P (X = 2) = P$ (2 non-defective and 2 defective) $=^{4} C_{2} . {(\frac{1}{5})}^{2} . {(\frac{4}{5})}^{2} = \frac{96}{625}$

$P (X = 3) = P$ (1 non-defective and 3 defective) $=^{4} C_{3} . {(\frac{1}{5})}^{3} . (\frac{4}{5}) = \frac{16}{625}$

$P (X = 4) = P$ (0 non-defective and 4 defective) $=^{4} C_{4} . {(\frac{1}{5})}^{4} . {(\frac{4}{5})}^{0} = \frac{1}{625}$

Therefore, the required probability distribution is as follows.

X	0	1	2	3	4
P (X)	256/625	256/625	96/625	16/625	1/625

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Ncert Solutions Maths class 12th Probability

53. Find the probability distribution of the number of success in two tosses of a die where a success is defined as:

(i) Number greater than 4.

(ii) Six appears on at least one die.

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alok kumar singh

Contributor-Level 10

53. When a die is tossed two times, we obtain (6 * 6) = 36 number of observations.

Let X be the random variable, which represents the number of successes.

(i) Here, success refers to the number greater than 4.

P (X = 0) = P (number less than or equal to 4 on both the tosses) = 4/6 X 4/6 = 4/9

P (X = 1) = P (number less than or equal to 4 on first toss and greater than 4 on second toss) + P (number greater than 4 on first toss and less than or equal to 4 on second toss)

= 4/6 X 2/6 + 4/6 X 2/6 = 4/9

P (X = 2) = P (number greater than 4 on both the tosses)

= 2/6 X 2/6 = 1/9

Thus, the probability distribution is as follows.

X	0	1	2
P (X)	4/9	4/9	1/9

(ii) Here,

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Ncert Solutions Maths class 12th Probability

52. Find the probability distribution of:

(i) Number of heads in two tosses of a coin.

(ii) Number of tails in the simultaneous tosses of three coins.

(iii) Number of heads in four tosses of a coin.

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alok kumar singh

Contributor-Level 10

(i) When one coin is tossed twice, the sample space is

{HH, HT, TH, TT}

Let X represent the number of heads.

∴ X (HH) = 2, X (HT) = 1, X (TH) = 1, X (TT) = 0

Therefore, X can take the value of 0, 1, or 2.

It is known that,

P (HH) = P (HT) = P (TH) = P (TT) = 1/4

P (X = 0) = P (TT) = 1/4

P (X = 1) = P (HT) + P (TH) = 14 + 1/4 = 1/2

P (X = 2) = P (HH) = 1/4

Thus, the required probability distribution is as follows.

X	0	1	2
P (X)	1/4	1/2	1/4

(ii) When three coins are tossed simultaneously, the sample space is {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

Let X represent the number of tails.

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

P (X = 0) = P (

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Ncert Solutions Maths class 12th Probability

51. Let X represents the difference between the number of heads and the number of tails obtained when a coin is tossed 6 times. What are possible values of X?

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alok kumar singh

Contributor-Level 10

51. A coin is tossed six times and X represents the difference between the number of heads and the number of tails.

∴ X (6 H, 0T) = |6 - 0| = 6

X (5 H, 1 T) = |5 - 1| = 4

X (4 H, 2 T) = |4 - 2| = 2

X (3 H, 3 T) = 3 - 3| = 0

X (2 H, 4 T) = |2 - 4| = 2

X (1 H, 5 T) = |1 - 5| = 4

X (0H, 6 T) = | 0 - 6| = 6

Thus, the possible values of X are 6, 4, 2, and 0.

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Ncert Solutions Maths class 12th Probability

50. An urn contains 5 red and 2 black balls. Two balls are randomly selected. Let X represents the number of black balls. What are the possible values of X? Is X a random variable?

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alok kumar singh

Contributor-Level 10

50. There two balls may be selected as BR, RB, BR, BB, where R represents red ball and B represents black ball.

Variable X has the value 0, 1, 2, i.e., there may be no black ball, may be one black ball or both the balls are black.

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Ncert Solutions Maths class 12th Probability

49. State which of the following are not the probability distributions of a random variable. Give reasons for your answer.

(i)

X	0	1	2
P (X)	0.4	0.4	0.2

(ii)

X	0	1	2	3	4
P (X)	0.1	0.5	0.2	– 0.1	0.3

(iii)

Y	– 1	0	1
P (Y)	0.6	0.1	0.2

(iv)

Y	– 1	0	1
P (Y)	0.6	0.1	0.2

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alok kumar singh

Contributor-Level 10

49. It is known that the sum of all the probabilities in a probability distribution is one.

(i) Sum of the probabilities = 0.4 + 0.4 + 0.2 = 1

Therefore, the given table is a probability distribution of random variables.

(ii) It can be seen that for X = 3, P (X) = −0.1

It is known that probability of any observation is not negative. Therefore, the given table is not a probability distribution of random variables.

(iii) Sum of the probabilities = 0.6 + 0.1 + 0.2 = 0.9 ≠ 1

Therefore, the given table is not a probability distribution of random variables.

(iv) Sum of the probabilities = 0.3 + 0.2 + 0.4 + 0.1 + 0.05 = 1

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