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Ncert Solutions Maths class 12th

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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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Ncert Solutions Maths class 12th Probability

48. If A and B are two events such that A ⊂ B and P (B) ≠ 0, then which of the following is correct: $\begin{array}{l} \end{array}$

$\begin{array}{l} \end{array}$
$\begin{array}{l} A . P (A | B) = \frac{P (B)}{P (A)} \\ B . P (A | B) < P (A) \\ C . P (A | B) \leq P (A) \\ D . N o n e o f t h e s e \end{array}$

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alok kumar singh

Contributor-Level 10

48. If $A \subset B,$ then $A \cap B = A$

$P (A \cap B) = P (A)$ also $P (A) < P (B)$

Consider,

$P (A / B) = \frac{P (A \cap B)}{P (B)} = \frac{P (A)}{P (B)} \neq \frac{P (B)}{P (A)}$

$\Rightarrow P (A / B) = \frac{P (A \cap B)}{P (B)} = \frac{P (A)}{P (B)}$ $_____(1)$

We know, $P (B) \leq 1$ $\Rightarrow \frac{1}{P (B)} \geq 1$

From eq. $(1)$ , we have

$\frac{P (A)}{P (B)} \geq P (A)$

$\Rightarrow P (A / B) \geq P (A)$ $_____(2)$

$\Rightarrow P (A / B)$ is not less than $P (A)$

Hence, from eq. $(2)$ it can be concluded that the relation given in alternative $(C)$ is correct.

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Ncert Solutions Maths class 12th Probability

47. Probability that A speaks truth is 4/5 A coin is tossed. A report that a head appears. The probability that actually there was head is:

(A) 4/5

(B) 1/2

(C) 1/5

(D) 2/5

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alok kumar singh

Contributor-Level 10

47. Let, $E_{1}$ and $E_{2}$ be the event such that

$E_{1} :$ $A$ speak truth

$E_{2} :$ $A$ speak false

$X :$ that head appears

$P (E_{1}) = \frac{4}{5}$ and $P (E_{2}) = 1 - P (E_{1})$ = $= 1 - \frac{4}{5} = \frac{1}{5}$

If a coin tossed, then it may result either head $(H)$ or tail $(T)$ . The probability of getting a head is $\frac{1}{2}$ whether $A$ speak truth or not.

$P (X / E_{1}) = P (X / E_{2}) = \frac{1}{2}$

The probability that there are actually a head is given by $P (E_{1} / X)$

$P (E_{1} / X) = \frac{P (E_{1}) . P (X / E_{1})}{P (E_{1}) . P (X / E_{1}) + P (E_{2}) . P (X / E_{2})}$

$= \frac{\frac{4}{5} * \frac{1}{2}}{\frac{4}{5} * \frac{1}{2} + \frac{4}{5} * \frac{1}{2}}$

$= \frac{\frac{4}{5} * \frac{1}{2}}{\frac{1}{2} (\frac{4}{5} + \frac{1}{5})} = \frac{4}{5}$

Option $(A)$ is correct $\frac{4}{5}$

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New answer posted

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Ncert Solutions Maths class 12th Probability

46. A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.

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alok kumar singh

Contributor-Level 10

46. Let, $E_{1} :$ event of choosing a diamond card

$E_{2} :$ event of choosing a card which is not diamond

$A :$ denote the lost card, out of $52$ cards

We know,

$13$ cards are diamond

$39$ cards are not diamond

$P (E_{1}) = \frac{13}{52} = \frac{1}{4}$

$P (E_{2}) = \frac{39}{52} = \frac{3}{4}$

When one diamond card is lost, there are $12$ diamond cards out of $51$ cards, two cards can be drawn out of $12$ diamond cards in $12$ $C_{2}$
ways. Similarly, $2$ diamond cards can be drawn out of $51$ cards in $51$ $C_{2}$
ways.

The probability of getting two cards, when one diamond card is lost, is given by $P (A / E_{1}),$

$P (A / E_{1}) = \frac{{}_{12}{C_{2}}}{{}_{51}{C_{2}}} = \frac{12!}{2! * 10!} * \frac{2! * 49!}{51!}$

$= \frac{11 * 12}{50 * 51} = \frac{22}{425}$

By using Baye's theorem,

$P (E_{1} / A) =$ probability that the lost card is diamond, given that the card is l

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46. Let, $E_{1} :$ event of choosing a diamond card

$E_{2} :$ event of choosing a card which is not diamond

$A :$ denote the lost card, out of $52$ cards

We know,

$13$ cards are diamond

$39$ cards are not diamond

$P (E_{1}) = \frac{13}{52} = \frac{1}{4}$

$P (E_{2}) = \frac{39}{52} = \frac{3}{4}$

The probability of getting two cards, when one diamond card is lost, is given by $P (A / E_{1}),$

$P (A / E_{1}) = \frac{{}_{12}{C_{2}}}{{}_{51}{C_{2}}} = \frac{12!}{2! * 10!} * \frac{2! * 49!}{51!}$

$= \frac{11 * 12}{50 * 51} = \frac{22}{425}$

By using Baye's theorem,

$P (E_{1} / A) =$ probability that the lost card is diamond, given that the card is lost

$P (E_{1} / A) = \frac{P (E_{1}) . P (A / E_{1})}{P (E_{1}) . P (A / E_{1}) + P (E_{2}) . P (A / E_{2})}$

$= \frac{\frac{1}{4} * \frac{22}{425}}{\frac{1}{4} * \frac{22}{425} + \frac{3}{4} * \frac{26}{425}}$

$= \frac{\frac{1}{4} * \frac{22}{425}}{\frac{1}{425} (\frac{22}{4} + \frac{26 * 3}{4})}$

$= \frac{11}{2} * \frac{4}{100} = \frac{11}{50}$

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New answer posted

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Ncert Solutions Maths class 12th Probability

45. A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, whereas the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B on the job for 30% of the time and C on the job for 20% of the time. A defective item is produced, what is the probability that it was produced by A?

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alok kumar singh

Contributor-Level 10

45. Let, $E_{1} :$ event of time consume by $A$

$E_{2} :$ event of time consume by $B$

$E_{3} :$ event of time consume by $C$

$P (E_{1}) = 50 % = \frac{50}{100} = \frac{1}{2}$

$P (E_{2}) = 30 % = \frac{30}{100} = \frac{3}{10}$

$P (E_{3}) = 20 % = \frac{20}{100} = \frac{1}{5}$

Let, $A :$ event of producing the defective item. Therefore,

$P (A / E_{1}) = 1 % = \frac{1}{100}$

$P (A / E_{2}) = 5 % = \frac{5}{100}$

$P (A / E_{3}) = 7 % = \frac{7}{100}$

Therefore, by Baye's theorem,

$P (E_{1} / A) =$ probability that the defective item was produced by $A$ ,

$P (E_{1} / A) = \frac{P (E_{1}) . P (A / E_{1})}{P (E_{1}) . P (A / E_{1}) + P (E_{2}) . P (A / E_{2}) + P (E_{3}) . P (A / E_{3})}$

$= \frac{\frac{1}{2} * \frac{1}{100}}{\frac{1}{2} * \frac{1}{100} + \frac{3}{10} * \frac{5}{100} + \frac{1}{5} * \frac{7}{100}}$

$= \frac{\frac{1}{2} * \frac{1}{100}}{\frac{1}{100} (\frac{1}{2} + \frac{15}{10} + \frac{7}{5})} = \frac{\frac{1}{2} * \frac{1}{100}}{\frac{1}{100} (\frac{5 + 15 + 14}{10})} = \frac{\frac{1}{2}}{\frac{34}{10}}$

$= \frac{1}{2} * \frac{10}{34} = \frac{5}{34}$

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