{"id":4541,"date":"2016-05-16T11:49:00","date_gmt":"2016-05-16T06:19:00","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/ncert-solutions-class-12-maths-exercise-13-4\/"},"modified":"2018-05-08T11:29:27","modified_gmt":"2018-05-08T05:59:27","slug":"ncert-solutions-class-12-maths-exercise-13-4","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-12-maths-exercise-13-4\/","title":{"rendered":"NCERT Solutions class 12 Maths Exercise 13.4"},"content":{"rendered":"

NCERT Solutions class 12 Maths Exercise 13.4 Class 12 Maths book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 12 Maths chapter wise NCERT solution for Maths part 1 and Maths part 2 for all the chapters can be downloaded from our website and myCBSEguide mobile app for free.<\/p>\n

Download\u00a0NCERT solutions for\u00a0Probability\u00a0<\/a>as PDF.<\/strong><\/p>\n

\"NCERT<\/p>\n

NCERT Solutions class 12 Maths Probability<\/h2>\n

1. State which of the following are not the probability distributions of a random variable. Give reasons for your answer.<\/strong><\/p>\n

(i) <\/strong><\/p>\n\n\n\n\n
\n

X<\/strong><\/p>\n<\/td>\n

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0<\/strong><\/p>\n<\/td>\n

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1<\/strong><\/p>\n<\/td>\n

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2<\/strong><\/p>\n<\/td>\n<\/tr>\n

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P (X)<\/strong><\/p>\n<\/td>\n

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0.4<\/strong><\/p>\n<\/td>\n

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0.4<\/strong><\/p>\n<\/td>\n

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0.2<\/strong><\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

(ii) <\/strong><\/p>\n\n\n\n\n
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X<\/strong><\/p>\n<\/td>\n

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0<\/strong><\/p>\n<\/td>\n

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1<\/strong><\/p>\n<\/td>\n

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2<\/strong><\/p>\n<\/td>\n

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3<\/strong><\/p>\n<\/td>\n

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4<\/strong><\/p>\n<\/td>\n<\/tr>\n

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P (X)<\/strong><\/p>\n<\/td>\n

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0.1<\/strong><\/p>\n<\/td>\n

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0.5<\/strong><\/p>\n<\/td>\n

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0.2<\/strong><\/p>\n<\/td>\n

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\u2013 0.1<\/strong><\/p>\n<\/td>\n

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0.3<\/strong><\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

(iii) <\/strong><\/p>\n\n\n\n\n
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Y<\/strong><\/p>\n<\/td>\n

\n

\u2013 1<\/strong><\/p>\n<\/td>\n

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0<\/strong><\/p>\n<\/td>\n

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1<\/strong><\/p>\n<\/td>\n<\/tr>\n

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P (Y)<\/strong><\/p>\n<\/td>\n

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0.6<\/strong><\/p>\n<\/td>\n

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0.1<\/strong><\/p>\n<\/td>\n

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0.2<\/strong><\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

(iv) <\/strong><\/p>\n\n\n\n\n
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Z<\/strong><\/p>\n<\/td>\n

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3<\/strong><\/p>\n<\/td>\n

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2<\/strong><\/p>\n<\/td>\n

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1<\/strong><\/p>\n<\/td>\n

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0<\/strong><\/p>\n<\/td>\n

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\u2013 1<\/strong><\/p>\n<\/td>\n<\/tr>\n

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P (Z)<\/strong><\/p>\n<\/td>\n

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0.3<\/strong><\/p>\n<\/td>\n

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0.2<\/strong><\/p>\n<\/td>\n

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0.4<\/strong><\/p>\n<\/td>\n

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0.1<\/strong><\/p>\n<\/td>\n

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0.05<\/strong><\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Ans. (i) <\/strong>P (0) + P (1) + P (2) = 0.4 + 0.4 + 0.2 = 1<\/p>\n

Therefore, it is a probability distribution.<\/p>\n

(ii)<\/strong> P (3) = \u20130.1 which is not possible.<\/p>\n

Therefore, it is not a probability distribution.<\/p>\n

(iii)<\/strong> P (\u20131) + P (0) + P (1) = 0.6 + 0.1 + 0.2 = 0.9 \u00a01<\/p>\n

Therefore, it is not a probability distribution.<\/p>\n

(iv)<\/strong> P (3) + P (2) + P (1) + P (0) + P (\u20131) = 0.3 + 0.2 + 0.4 + 0.1 + 0.05 = 1.05 \u00a01<\/p>\n

Therefore, it is not a probability distribution.<\/p>\n

\n

2. 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?<\/strong><\/p>\n

Ans.<\/strong> There two balls may be selected as BR, RB, BR, BB, where R represents red ball and B represents black ball.<\/p>\n

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.<\/p>\n

\n

3. 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?<\/strong><\/p>\n

Ans.<\/strong> For one coin, S = {H, T}\u00a0= 2<\/p>\n

Let A represents head\u00a0\u00a0= 1<\/p>\n

\u00a0= \u00a0and <\/p>\n

<\/p>\n

P (X = 0) = \u00a0= <\/p>\n

P (X = 1) = 6 P (X) \u00a0= <\/p>\n

P (X = 2) = 15 [P (X)]2<\/sup> \u00a0= <\/p>\n

P (X = 3) = 20 [P (X)]3<\/sup> \u00a0= <\/p>\n

P (X = 4) = 15 [P (X)]4<\/sup> \u00a0= <\/p>\n

P (X = 5) = 6 [P (X)]5<\/sup> \u00a0= <\/p>\n

P (X = 6) = \u00a0= <\/p>\n

\n

4. Find the probability distribution of:<\/strong><\/p>\n

(i) Number of heads in two tosses of a coin.<\/strong><\/p>\n

(ii) Number of tails in the simultaneous tosses of three coins.<\/strong><\/p>\n

(iii) Number of heads in four tosses of a coin.<\/strong><\/p>\n

Ans. (i) <\/strong>S = {T, H}\u00a0= 2<\/p>\n

Let A represents head\u00a0\u00a0= 1<\/p>\n

\u00a0= \u00a0and <\/p>\n

\u00a0<\/p>\n

P (X = 0) = \u00a0= <\/p>\n

P (X = 1) = 2 P (A) \u00a0= <\/p>\n

P (X = 2) = P (A). P (A) = <\/p>\n

Probability distribution<\/p>\n\n\n\n\n
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<\/p>\n<\/td>\n

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0<\/p>\n<\/td>\n

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1<\/p>\n<\/td>\n

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2<\/p>\n<\/td>\n<\/tr>\n

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<\/p>\n<\/td>\n

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<\/p>\n<\/td>\n

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<\/p>\n<\/td>\n

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<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

(ii)<\/strong> Three coins tossed once = one coin tossed three times<\/p>\n

S = {T, H}\u00a0= 2<\/p>\n

\u00a0= \u00a0and <\/p>\n

<\/p>\n

P (X = 0) = \u00a0= <\/p>\n

P (X = 1) = 3 P (A) \u00a0= <\/p>\n

P (X = 2) = 3 P (A). P (A). \u00a0= <\/p>\n

P (X = 3) = P (A). P (A). P (A) = <\/p>\n

Probability distribution<\/p>\n\n\n\n\n
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<\/p>\n<\/td>\n

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0<\/p>\n<\/td>\n

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1<\/p>\n<\/td>\n

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2<\/p>\n<\/td>\n

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3<\/p>\n<\/td>\n<\/tr>\n

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<\/p>\n<\/td>\n

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<\/p>\n<\/td>\n

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<\/p>\n<\/td>\n

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<\/p>\n<\/td>\n

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<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

(iii)<\/strong> S = {H, T}\u00a0= 2<\/p>\n

Let A represents head\u00a0\u00a0= 1<\/p>\n

\u00a0= \u00a0and <\/p>\n

\u00a0<\/p>\n

P (X = 0) = \u00a0= <\/p>\n

P (X = 1) = 4 P (A) \u00a0= <\/p>\n

P (X = 2) = 6 P (A). P (A). \u00a0= <\/p>\n

P (X = 3) = 4 P (A). P (A). P (A). \u00a0= <\/p>\n

P (X = 4) = P (A). P (A). P (A). P (A) = <\/p>\n

Probability distribution<\/p>\n\n\n\n\n
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<\/p>\n<\/td>\n

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0<\/p>\n<\/td>\n

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1<\/p>\n<\/td>\n

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2<\/p>\n<\/td>\n

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3<\/p>\n<\/td>\n

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4<\/p>\n<\/td>\n<\/tr>\n

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<\/p>\n<\/td>\n

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<\/p>\n<\/td>\n

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<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\n

5. Find the probability distribution of the number of success in two tosses of a die where a success is defined as:<\/strong><\/p>\n

(i) number greater than 4.<\/strong><\/p>\n

(ii) six appears on at least one die.<\/strong><\/p>\n

Ans. <\/strong>S = {1, 2, 3, 4, 5, 6}\u00a0= 6<\/p>\n

(i)<\/strong> Let A be the set of favourable events.\u00a0\u00a0= 1<\/p>\n

\u00a0= \u00a0and <\/p>\n

\u00a0<\/p>\n

P (X = 0) = \u00a0= <\/p>\n

P (X = 1) = 2 P (A) \u00a0= <\/p>\n

P (X = 2) = P (A). P (A) = <\/p>\n

Probability distribution<\/p>\n\n\n\n\n
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<\/p>\n<\/td>\n

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0<\/p>\n<\/td>\n

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1<\/p>\n<\/td>\n

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2<\/p>\n<\/td>\n<\/tr>\n

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<\/p>\n<\/td>\n

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<\/p>\n<\/td>\n

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<\/p>\n<\/td>\n

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<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

(ii)<\/strong> Let A represents that 6 appears on one dieA = {6} \u00a0\u00a0= 1<\/p>\n

\u00a0= \u00a0and <\/p>\n

\u00a0<\/p>\n

P (X = 0) = \u00a0= <\/p>\n

P (X = 1) = 2 P (A) \u00a0= <\/p>\n

P (X = 2) = P (A). P (A) = <\/p>\n

P (at least on six) = <\/p>\n

Probability distribution<\/p>\n\n\n\n\n
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<\/p>\n<\/td>\n

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0<\/p>\n<\/td>\n

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1<\/p>\n<\/td>\n<\/tr>\n

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<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

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6. 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.<\/strong><\/p>\n

Ans.<\/strong> \u00a0= 30, A = {6 defective bulbs}\u00a0= 6<\/p>\n

and <\/p>\n

\u00a0=4 (4 bulbs are drawn with replacement), \u00a0= 0, 1, 2, 3, 4<\/p>\n

P (X = 0) = <\/p>\n

P (X = 1) = <\/p>\n

P (X = 2) = <\/p>\n

P (X = 3) = <\/p>\n

P (X = 4) = <\/p>\n

Probability distribution:<\/p>\n\n\n\n\n
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<\/p>\n<\/td>\n

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0<\/p>\n<\/td>\n

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1<\/p>\n<\/td>\n

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2<\/p>\n<\/td>\n

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3<\/p>\n<\/td>\n

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4<\/p>\n<\/td>\n<\/tr>\n

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<\/p>\n<\/td>\n

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<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

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7. 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.<\/strong><\/p>\n

Ans.<\/strong> Let \u00a0represents the appearance of tail and \u00a0represents the appearance of head.<\/p>\n

Now <\/p>\n

Since \u00a0and <\/p>\n

P (X = 0) = <\/p>\n

P (X = 1) = <\/p>\n

P (X = 2) = <\/p>\n

Probability distribution:<\/p>\n\n\n\n\n
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<\/p>\n<\/td>\n

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0<\/p>\n<\/td>\n

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1<\/p>\n<\/td>\n

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2<\/p>\n<\/td>\n<\/tr>\n

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<\/p>\n<\/td>\n

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<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\n

NCERT Solutions class 12 Maths Exercise 13.4<\/strong><\/p>\n

8. A random variable X has the following probability distribution:<\/strong><\/p>\n\n\n\n\n
\n

X<\/strong><\/p>\n<\/td>\n

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0<\/strong><\/p>\n<\/td>\n

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1<\/strong><\/p>\n<\/td>\n

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2<\/strong><\/p>\n<\/td>\n

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3<\/strong><\/p>\n<\/td>\n

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4<\/strong><\/p>\n<\/td>\n

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5<\/strong><\/p>\n<\/td>\n

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6<\/strong><\/p>\n<\/td>\n

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7<\/strong><\/p>\n<\/td>\n<\/tr>\n

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P (X)<\/strong><\/p>\n<\/td>\n

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0<\/strong><\/p>\n<\/td>\n

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\u00a0<\/strong><\/p>\n<\/td>\n

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\u00a0<\/strong><\/p>\n<\/td>\n

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\u00a0<\/strong><\/p>\n<\/td>\n

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\u00a0<\/strong><\/p>\n<\/td>\n

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\u00a0<\/strong><\/p>\n<\/td>\n

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<\/strong><\/p>\n<\/td>\n

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\u00a0<\/strong><\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Determine:<\/strong><\/p>\n

(i) \u00a0<\/strong><\/p>\n

(ii) P (X < 3) <\/strong><\/p>\n

(iii) P (X > 6) <\/strong><\/p>\n

(iv) P (0 < X < 3)<\/strong><\/p>\n

Ans. (i) <\/strong>Since, the sum of all the probabilities of a distribution is 1.<\/p>\n

P (X = 0) + P (X = 1) + \u2026\u2026. + P (X = 7) = 1<\/p>\n

<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0or <\/p>\n

\u00a0\u00a0or <\/p>\n

Since, , therefore \u00a0is not possible.<\/p>\n

<\/p>\n

(ii)<\/strong> P (X < 3) = P (X = 0) + P (X = 1) + P (X = 2)<\/p>\n

\u00a0= <\/p>\n

\u00a0= <\/p>\n

(iii)<\/strong> P (X > 6) = P (X = 7)<\/p>\n

\u00a0= <\/p>\n

(iv)<\/strong> P (0 < X < 3)<\/p>\n

= P (X = 1) + P (X = 2)<\/p>\n

= <\/p>\n

\n

NCERT Solutions class 12 Maths Exercise 13.4<\/strong><\/p>\n

9. The random variable X has a probability distribution <\/strong><\/strong>\u00a0of the following form, where <\/strong><\/strong>\u00a0is some number:<\/strong><\/p>\n

\u00a0<\/strong><\/p>\n

(a) Determine the value of <\/strong><\/strong>\u00a0<\/strong><\/p>\n

(b) Find <\/strong><\/strong>\u00a0<\/strong><\/p>\n

Ans. <\/strong>Probability distribution:<\/p>\n\n\n\n\n
\n

<\/p>\n<\/td>\n

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0<\/p>\n<\/td>\n

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1<\/p>\n<\/td>\n

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2<\/p>\n<\/td>\n<\/tr>\n

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<\/p>\n<\/td>\n

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<\/p>\n<\/td>\n

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<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

(a)<\/strong> P (X = 0) + P (X = 1) + P (X = 2) = 1<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

(b)<\/strong> P (X < 2) = P (X = 0) + P (X = 1)<\/p>\n

\u00a0= <\/p>\n

P (X \u00a02) = P (X = 2) = \u00a0= <\/p>\n

\n

NCERT Solutions class 12 Maths Exercise 13.4<\/strong><\/p>\n

10. Find the mean number of heads in three tosses of fair coin.<\/strong><\/p>\n

Ans.<\/strong> S {H, T}\u00a0=2<\/p>\n

Let A denotes the appearance of head on a toss.<\/p>\n

A = \u00a0\u00a0\u00a0=1<\/p>\n

and <\/p>\n

\u00a0=3, \u00a0= 0, 1, 2, 3<\/p>\n

P (X = 0) = <\/p>\n

P (X = 1) = <\/p>\n

P (X = 2) = <\/p>\n

P (X = 3) = <\/p>\n

Probability distribution:<\/p>\n\n\n\n\n
\n

<\/p>\n<\/td>\n

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0<\/p>\n<\/td>\n

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1<\/p>\n<\/td>\n

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2<\/p>\n<\/td>\n

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3<\/p>\n<\/td>\n<\/tr>\n

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<\/p>\n<\/td>\n

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<\/p>\n<\/td>\n

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<\/p>\n<\/td>\n

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<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Mean = \u00a0= <\/p>\n

\n

NCERT Solutions class 12 Maths Exercise 13.4<\/strong><\/p>\n

11. Two dice are thrown simultaneously. If X denotes the number of sixes, find expectation of X.<\/strong><\/p>\n

Ans.<\/strong> Two dice thrown simultaneously is the same the die thrown 2 times.<\/p>\n

Let S = {1, 2, 3, 4, 5, 6}\u00a0= 6<\/p>\n

Let A denotes the number 6\u00a0A = {6}\u00a0<\/p>\n

P (A) = and <\/p>\n

\u00a0= 2,\u00a0= 0, 1, 2<\/p>\n

P (X = 0) = <\/p>\n

P (X = 1) = 2 P (A).<\/p>\n

P (X = 2) = P (A). P (A) = <\/p>\n

E (X) = <\/p>\n

\n

NCERT Solutions class 12 Maths Exercise 13.4<\/strong><\/p>\n

12. 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).<\/strong><\/p>\n

Ans.<\/strong> S = {(1, 2), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1),<\/p>\n

(1, 3), (2, 3), (3, 2), (4, 2), (5, 2), (6, 2),<\/p>\n

(1, 4), (2, 4), (3, 4), (4, 3), (5, 3), (6, 3),<\/p>\n

(1, 5), (2, 5), (3, 5), (4, 5), (5, 4), (6, 4),<\/p>\n

(1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6)}<\/p>\n

\u00a0= 30<\/p>\n

Let X denotes the larger of the two numbers obtained.<\/p>\n\n\n\n\n\n
\n

<\/p>\n<\/td>\n

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<\/p>\n<\/td>\n

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2<\/p>\n

3<\/p>\n

4<\/p>\n

5<\/p>\n

6<\/p>\n<\/td>\n

\n

2<\/p>\n

4<\/p>\n

6<\/p>\n

8<\/p>\n

10<\/p>\n<\/td>\n

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<\/p>\n

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\n<\/td>\n

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30<\/p>\n<\/td>\n

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\n<\/td>\n

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<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

E (X) = \u00a0= <\/p>\n

\n

NCERT Solutions class 12 Maths Exercise 13.4<\/strong><\/p>\n

13. Let X denotes the sum of the numbers obtained when two fair dice are rolled. Find the variance and standard deviation of X.<\/strong><\/p>\n

Ans.<\/strong> S = {(1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1),<\/p>\n

(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2),<\/p>\n

(1, 3), (2, 3), (3, 3), (4, 3), (5, 3), (6, 3),<\/p>\n

(1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4),<\/p>\n

(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5),<\/p>\n

(1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6)}<\/p>\n

\u00a0= 36<\/p>\n

Let A denotes the sum of the numbers = 2, B denotes the sum of the numbers = 3<\/p>\n

\u00a0C denotes the sum of the numbers = 4, D denotes the sum of the numbers = 5<\/p>\n

\u00a0E denotes the sum of the numbers = 6, F denotes the sum of the numbers = 7<\/p>\n

\u00a0G denotes the sum of the numbers = 8, H denotes the sum of the numbers = 9<\/p>\n

\u00a0I denotes the sum of the numbers = 10, J denotes the sum of the numbers = 11<\/p>\n

\u00a0K denotes the sum of the numbers = 12<\/p>\n

A = {1, 1}, \u00a0= 1, P (A) = <\/p>\n

B = {(1, 2), (2, 1)}, \u00a0= 2, P (A) = <\/p>\n

C = {(1, 3), (2, 2), (3, 1)}, \u00a0= 3, P (A) = <\/p>\n

D = {(1, 4), (2, 3), (3, 2), (4, 1)}, \u00a0= 4, P (A) = <\/p>\n

E = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)},<\/p>\n

\u00a0= 5, P (A) = <\/p>\n

F = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)},<\/p>\n

\u00a0\u00a0= 6, P (A) = <\/p>\n

G = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)},<\/p>\n

\u00a0= 5, P (A) = <\/p>\n

H = {(3, 6), (4, 5), (5, 4), (6, 3)},<\/p>\n

\u00a0= 4, P (A) = <\/p>\n

I = {(4, 6), (5, 5), (6, 4)},<\/p>\n

\u00a0= 3, P (A) = <\/p>\n

J = {(5, 6), (6, 5)},<\/p>\n

\u00a0= 2, P (A) = <\/p>\n

K = {6, 6}, \u00a0= 1, P (A) = <\/p>\n\n\n\n\n
\n

<\/p>\n<\/td>\n

\n

<\/p>\n<\/td>\n

\n

<\/p>\n<\/td>\n

\n

<\/p>\n<\/td>\n

\n

<\/p>\n<\/td>\n

\n

<\/p>\n<\/td>\n<\/tr>\n

\n

2<\/p>\n

3<\/p>\n

4<\/p>\n

5<\/p>\n<\/td>\n

\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n<\/td>\n

\n

6<\/p>\n

7<\/p>\n

8<\/p>\n

9<\/p>\n<\/td>\n

\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n<\/td>\n

\n

10<\/p>\n

11<\/p>\n

12<\/p>\n<\/td>\n

\n

<\/p>\n

<\/p>\n

<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Mean = <\/p>\n

\u00a0= <\/p>\n

= <\/p>\n

= <\/p>\n

Now \u00a0= <\/p>\n

\u00a0<\/p>\n

\u00a0= <\/p>\n

\u00a0= <\/p>\n

Variance = \u00a0= \u00a0= 54.83 \u2013 49 = 5.83<\/p>\n

Standard deviation = \u00a0(nearly)<\/p>\n

\n

NCERT Solutions class 12 Maths Exercise 13.4<\/strong><\/p>\n

14. 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.<\/strong><\/p>\n

Ans.<\/strong> \u00a0= 15, P (A) = <\/p>\n\n\n\n\n\n
\n

<\/p>\n<\/td>\n

\n

<\/p>\n<\/td>\n

\n

<\/p>\n<\/td>\n

\n

<\/p>\n<\/td>\n

\n

<\/p>\n<\/td>\n<\/tr>\n

\n

14<\/p>\n

15<\/p>\n

16<\/p>\n

17<\/p>\n

18<\/p>\n

19<\/p>\n

20<\/p>\n

21<\/p>\n

\n<\/td>\n

\n

2<\/p>\n

1<\/p>\n

2<\/p>\n

3<\/p>\n

1<\/p>\n

2<\/p>\n

3<\/p>\n

1<\/p>\n<\/td>\n

\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n<\/td>\n

\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n<\/td>\n

\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n<\/td>\n<\/tr>\n

\n

\n<\/td>\n

\n

15<\/p>\n<\/td>\n

\n

\n<\/td>\n

\n

<\/p>\n<\/td>\n

\n

<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Mean = \u00a0= 17.5<\/p>\n

Variance = \u00a0= <\/p>\n

= 312.20 \u2013 307.4 = 4.78<\/p>\n

Standard deviation = <\/p>\n

\n

NCERT Solutions class 12 Maths Exercise 13.4<\/strong><\/p>\n

15. 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).<\/strong><\/p>\n

Ans. <\/strong><\/p>\n\n\n\n\n\n
\n

<\/p>\n<\/td>\n

\n

<\/p>\n<\/td>\n

\n

<\/p>\n<\/td>\n

\n

<\/p>\n<\/td>\n<\/tr>\n

\n

0<\/p>\n

1<\/p>\n<\/td>\n

\n

<\/p>\n

<\/p>\n<\/td>\n

\n

0<\/p>\n

<\/p>\n<\/td>\n

\n

0<\/p>\n

<\/p>\n<\/td>\n<\/tr>\n

\n

\n<\/td>\n

\n

\n<\/td>\n

\n

<\/p>\n<\/td>\n

\n

<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\u00a0E (X) = Mean = \u00a0= 0.7<\/p>\n

Variance (X) = <\/p>\n

\n

NCERT Solutions class 12 Maths Exercise 13.4<\/strong><\/p>\n

Choose the correct answer in each of the following:<\/strong><\/p>\n

16. The mean of the numbers obtained on throwing a die having written 1 on three faces, 2 on two faces and 5 on one face is:<\/strong><\/p>\n

(A) 1 <\/strong><\/p>\n

(B) 2 <\/strong><\/p>\n

(C) 5 <\/strong><\/p>\n

(D) \u00a0<\/strong><\/p>\n

Ans.<\/strong><\/p>\n\n\n\n\n\n
\n

<\/p>\n<\/td>\n

\n

<\/p>\n<\/td>\n

\n

<\/p>\n<\/td>\n<\/tr>\n

\n

1<\/p>\n

2<\/p>\n

5<\/p>\n<\/td>\n

\n

<\/p>\n

<\/p>\n

<\/p>\n<\/td>\n

\n

<\/p>\n

<\/p>\n

<\/p>\n<\/td>\n<\/tr>\n

\n

\n<\/td>\n

\n

\n<\/td>\n

\n

<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Therefore, option (B) is correct.<\/p>\n

\n

NCERT Solutions class 12 Maths Exercise 13.4<\/strong><\/p>\n

17. Suppose that two cards are drawn at random from a deck of cards. Let X be the number of aces obtained. What is the value of E(X)?<\/strong><\/p>\n

(A) <\/strong><\/p>\n

(B) \u00a0<\/strong><\/p>\n

(C) \u00a0<\/strong><\/p>\n

(D) \u00a0<\/strong><\/p>\n

Ans. <\/strong>\u00a0= 52, \u00a0= 4<\/p>\n

P (X = 0) = <\/p>\n\n\n\n\n\n
\n

<\/p>\n<\/td>\n

\n

<\/p>\n<\/td>\n

\n

<\/p>\n<\/td>\n<\/tr>\n

\n

0<\/p>\n

1<\/p>\n

2<\/p>\n<\/td>\n

\n

<\/p>\n

<\/p>\n

<\/p>\n<\/td>\n

\n

<\/p>\n

<\/p>\n

<\/p>\n<\/td>\n<\/tr>\n

\n

\n<\/td>\n

\n

\n<\/td>\n

\n

<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

P (X = 1) = <\/p>\n

P (X = 2) = <\/p>\n

Now E (X) = <\/p>\n

Therefore, option (D) is correct.<\/p>\n

NCERT Solutions class 12 Maths Exercise 13.4<\/h2>\n

NCERT Solutions Class 12 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 12 Maths includes text book solutions from both part 1 and part 2. NCERT Solutions for CBSE Class 12 Maths have total 13 chapters. 12 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 12 solutions PDF and Maths ncert class 12 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide<\/p>\n

CBSE App for Students<\/h2>\n

To download\u00a0NCERT Solutions for class 12 Physics, Chemistry, Biology, History, Political Science, Economics, Geography, Computer Science, Home Science, Accountancy, Business Studies and Home Science; do check myCBSEguide app or website. myCBSEguide provides sample papers with solution, test papers for chapter-wise practice, NCERT solutions, NCERT Exemplar solutions, quick revision notes for ready reference, CBSE guess papers and CBSE important question papers. Sample Paper all are made available through\u00a0the best app for CBSE students<\/strong><\/a>\u00a0and myCBSEguide website.<\/p>\n","protected":false},"excerpt":{"rendered":"

NCERT Solutions class 12 Maths Exercise 13.4 Class 12 Maths book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 12 … Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":-1,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1432,281],"tags":[283,1342,321,216],"class_list":["post-4541","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-mathematics-cbse-class-12","category-ncert-solutions","tag-cbse-study-material","tag-class-12","tag-mathematics","tag-ncert-solutions"],"yoast_head":"\nNCERT Solutions class 12 Maths Exercise 13.4 | myCBSEguide<\/title>\n<meta name=\"description\" content=\"NCERT Solutions class 12 Maths Exercise 13.4 in PDF format for free download. 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