{"id":4543,"date":"2016-05-16T11:49:00","date_gmt":"2016-05-16T06:19:00","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/ncert-solutions-class-12-maths-miscellaneous-10\/"},"modified":"2018-05-08T11:50:39","modified_gmt":"2018-05-08T06:20:39","slug":"ncert-solutions-class-12-maths-miscellaneous-10","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-12-maths-miscellaneous-10\/","title":{"rendered":"NCERT Solutions class 12 Maths Miscellaneous"},"content":{"rendered":"

NCERT Solutions class 12 Maths Miscellaneous 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.A and B are two events such that P (A) <\/strong><\/strong>\u00a00. Find P (B\/A) if:<\/strong><\/p>\n

(i)A is a subset of B<\/strong><\/p>\n

(ii)A <\/strong><\/strong>\u00a0B = <\/strong><\/strong>\u00a0<\/strong><\/p>\n

Ans. <\/strong>A and B are two events such that P (A) \u00a00<\/p>\n

To find: P<\/p>\n

(i)<\/strong> A is a subset of B<\/p>\n

\u00a0A \u00a0B<\/p>\n

<\/p>\n

\u00a0<\/p>\n

(ii) <\/strong>A \u00a0B = <\/p>\n

\u00a0<\/p>\n

\n

2.A couple has two children.<\/strong><\/p>\n

(i)Find the probability that both children are males I it is known that at least one of the children is male.<\/strong><\/p>\n

(ii)Find the probability that both children are females if it is known that the elder child is a female.<\/strong><\/p>\n

Ans. (i)<\/strong> Let B1<\/sub> and G1<\/sub> stand for male and female respectively.<\/p>\n

Now the sample space is (S) = {B1<\/sub>B2<\/sub>, B1<\/sub>G2<\/sub>, B2<\/sub>G1<\/sub>, G1<\/sub>G2<\/sub>}<\/p>\n

Let us consider the following events,<\/p>\n

A = both are males<\/p>\n

B = at least one is a male<\/p>\n

\u00a0A = {B1<\/sub>B2<\/sub>} and B = {B1<\/sub>B2<\/sub>, B1<\/sub>G2<\/sub>, B2<\/sub>G1<\/sub>}<\/p>\n

P (B) = , A \u00a0B = {B1<\/sub>B2<\/sub>}\u00a0 and <\/p>\n

Required probability = P<\/p>\n

(ii) <\/strong>Let A = both are females, A = {G1<\/sub>G2<\/sub>} and C = the older is a girl<\/p>\n

C = {B1<\/sub>B2<\/sub>, B1<\/sub>G2<\/sub>}P (C) = <\/p>\n

Required probability = P<\/p>\n

\n

3.Suppose that 5% of men and 0.25% of women have grey hair. A grey haired person is selected at random. What is the probability of this person being male? Assume that there are equal number of males and females.<\/strong><\/p>\n

Ans. <\/strong>Men are represented by E1<\/sub> and women are represented by E2<\/sub>.<\/p>\n

P (E1<\/sub>) = \u00a0and P (E2<\/sub>) = <\/p>\n

A represents grey hair persons.<\/p>\n

<\/p>\n

P\u00a0and P<\/p>\n

P<\/p>\n

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

\n

4.Suppose that 90% of people are right-handed. What is the probability that at most of 6 of a random sample of 10 people are right-handed?<\/strong><\/p>\n

Ans. <\/strong>\u00a0and <\/p>\n

<\/p>\n

P (at most 6 successes) = 1 \u2013 [P (X = 7) + P (X = 8) + P (X = 9) + P (X = 10)]\n

= 1 \u2013 <\/p>\n

= <\/p>\n

\n

5.An urn contains 25 balls of which 10 balls bear a mark \u2018X\u2019 and the remaining 15 bear a mark \u2018Y\u2019. A ball is drawn at random from the urn, its mark noted down and it is replaced. If 6 balls are drawn in this way, find the probability that:<\/strong><\/p>\n

(i)all will bear \u2018X\u2019 mark.<\/strong><\/p>\n

(ii)not more than 2 will bear \u2018Y\u2019 mark.<\/strong><\/p>\n

(iii)at least one ball will bear \u2018Y\u2019 mark.<\/strong><\/p>\n

(iv)the number of balls with \u2018X\u2019 mark and \u2018Y\u2019 mark will be equal.<\/strong><\/p>\n

Ans. <\/strong>S = {10 balls with mark X, 15 balls with mark Y}\u00a0\u00a0= 25<\/p>\n

Let a ball drawn with mark X be denoted by A.<\/p>\n

A = {10 balls with mark X}\u00a0= 10<\/p>\n

\u00a0and <\/p>\n

<\/p>\n

(i) <\/strong>All will bear X mark, i.e., \u00a0= 6<\/p>\n

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

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

(ii) <\/strong>Not more than 2 will bear mark Y.<\/p>\n

\u00a0For Y mark, \u00a0= 0, 1, 2and For X mark, \u00a0= 6, 5, 4<\/p>\n

P (not more than 2 will bear mark Y) = P (X = 4) + P (X = 5) + P (X =6)<\/p>\n

= <\/p>\n

= <\/p>\n

= <\/p>\n

= \u00a0= <\/p>\n

(iii) <\/strong>P (at least one ball will bear Y mark) = P (not more than 5 balls will bear mark X)<\/p>\n

= 1 \u2013 P (6)<\/p>\n

= 1 \u2013 <\/p>\n

(iv) <\/strong>P (equal number of balls will bear mark X and Y) = P (3)<\/p>\n

= <\/p>\n

\n

6.In a hurdle race, a player has to cross 10 hurdles. The probability that he will clear each hurdle is <\/strong><\/strong>\u00a0What is the probability that he will knock down fewer than 2 hurdles?<\/strong><\/p>\n

Ans. <\/strong>\u00a0= Probability of knocking down a hurdle = <\/p>\n

\u00a0= Probability of clearing a hurdle = <\/p>\n

<\/p>\n

P (He will knock down fewer than 2 hurdles) = \u00a0= P (X = 0) + P (X = 1)<\/p>\n

= <\/p>\n

= <\/p>\n

\n

7.A die is thrown again and again until three sixes are obtained. Find the probability of obtaining the third six in the sixth throw of the die.<\/strong><\/p>\n

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

A = {6}\u00a0= 1<\/p>\n

\u00a0 and <\/p>\n

Since the 3rd<\/sup> six is obtained in the 6th<\/sup> throw of die, there are two sixes i.e., two successes in the first 5 throws.<\/p>\n

\u00a0 Required probability = P (2 success in the first 5 throws) x P (success in the 6th<\/sup> throw)<\/p>\n

= <\/p>\n

= <\/p>\n

\n

8.If a leap year is selected at random, what is the change that it will contain 52 Tuesday?<\/strong><\/p>\n

Ans. <\/strong>A leap year has 366 days which means 52 complete weeks and 2 days. If any one of these two days in a Tuesday then the year will have 53 Tuesdays.<\/p>\n

Number of total days in a week = 7<\/p>\n

Number of favourable days = 2<\/p>\n

Therefore, P (the year will have 53 Tuesday) = <\/p>\n

\n

9.An experiment succeeds twice as often as it fails. Find the probability that in the next six trails, there will be at least 4 successes.<\/strong><\/p>\n

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

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

P (at least 4 successes) = P (X = 4) + P (X = 5) + P (X = 6)<\/p>\n

= <\/p>\n

= <\/p>\n

= <\/p>\n

= <\/p>\n

= <\/p>\n

= <\/p>\n

\n

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

10.How many times must a man toss a fair coin so that the probability of having at least one head is more than 90%?<\/strong><\/p>\n

Ans. <\/strong>For one fair coin, probability of head \u00a0= , therefore, <\/p>\n

<\/p>\n

\u00a0<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Therefore, it is clear that <\/p>\n

\n

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

11.In a game a man wins a rupee for a six and looses a rupee for any other number when a fair die is thrown. The man decided to throw a die thrice but to quit as and when he gets a six. Find the expected value of the amounts he wins\/looses.<\/strong><\/p>\n

Ans. <\/strong>When a die is thrown, probability of getting a six \u00a0= <\/p>\n

Therefore, <\/p>\n

(i) <\/strong>If he gets a six in first throw, then,<\/p>\n

Probability of getting a six = <\/p>\n

(ii) <\/strong>If he does not get a six, in first throw, but he gets a six in the second throw, then<\/p>\n

Its probability = <\/p>\n

Probability that he does not get a six in first two throws and he gets a six in third throw<\/p>\n

= <\/p>\n

Probability that he does not get a six in any of the three throws = <\/p>\n

In first throw he gets a six, will receive Re.1<\/p>\n

If he gets a six in second throw, he will receive Rs. (1 \u2013 1) = 0<\/p>\n

If he gets a six in third throw, he will receive Rs. (\u2013 1 \u2013 1 + 1) = Rs. \u2013 1<\/p>\n

= he will loss Rs. 1<\/p>\n

Expected value = <\/p>\n

= \u00a0(loss)<\/p>\n

\n

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

12.Suppose we have four boxes A, B, C and D containing coloured marbles as given below:<\/strong><\/p>\n\n\n\n\n\n
Box<\/strong><\/td>\nMarble colour<\/strong><\/td>\n<\/tr>\n
\u00a0<\/strong><\/td>\nRed<\/strong><\/td>\nWhite<\/strong><\/td>\nBlack<\/strong><\/td>\n<\/tr>\n
A<\/strong><\/p>\n

B<\/strong><\/p>\n

C<\/strong><\/p>\n

D<\/strong><\/td>\n

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

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

8<\/strong><\/p>\n

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

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

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

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

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

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

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

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

4<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A?, box B? box C?<\/strong><\/p>\n

Ans. <\/strong>Let R represents the drawing of red ball and the four boxes are represented by A, B, C and D.<\/p>\n

So\u00a0<\/p>\n

<\/p>\n

Since there are 4 bags.<\/p>\n

Therefore, P (A) = \u00a0P (B) = \u00a0P (C) = \u00a0P (D) = <\/p>\n

<\/p>\n

= <\/p>\n

= <\/p>\n

<\/p>\n

= \u00a0= <\/p>\n

<\/p>\n

= \u00a0= <\/p>\n

\n

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

13.Assume that the chances of a patient having a heart attack is 40%. It is also assumed that a meditation and yoga course reduces the risk of heart attack by 30% and prescription of certain drug reduces its chances by 25%. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga.<\/strong><\/p>\n

Ans. <\/strong>A patient has options to have the treatment of yoga and meditation and that of prescription of drugs.<\/p>\n

Let these events be denoted by E1<\/sub> and E2<\/sub> i.e.,<\/p>\n

E1<\/sub> = Treatment of yoga and meditation<\/p>\n

E2 <\/sub>= Treatment of prescription of certain drugs<\/p>\n

P (E1<\/sub>) = \u00a0and P (E2<\/sub>) = <\/p>\n

Let A denotes that a person has heart attack, then P (A) = 40% = 0.40<\/p>\n

Yoga and meditation reduces heart attack by 30.<\/p>\n

\u00a0Inspite of getting yoga and meditation treatment heart risk is 70% of 0.40<\/p>\n

\u00a0\u00a0= 0.40 x 0.70 = 0.28<\/p>\n

Also, Drug prescription reduces the heart attack rick by 25%<\/p>\n

Even after adopting the drug prescription hear rick is 75% of 0.40<\/p>\n

\u00a0\u00a0= 0.40 x 0.75 = 0.30<\/p>\n

<\/p>\n

= <\/p>\n

= <\/p>\n

\n

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

14.If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability <\/strong><\/strong>).<\/strong><\/p>\n

Ans. <\/strong>There are four entries in a determinant of 2 x 2 order. Each entry may be filled up in two ways with 0 or 1.<\/p>\n

\u00a0Number of determinants that can be formed = \u00a0= 16<\/p>\n

The value of determinants is positive in the following cases:<\/p>\n

<\/p>\n

Therefore, the probability that the determinant is positive = <\/p>\n

\n

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

15. An electronic assembly consists of two sub-systems say A and B. From previous testing procedures, the following probabilities are assumed to be known:<\/strong><\/p>\n

P (A fails) = 0.2<\/strong><\/p>\n

P (B fails alone) = 0.15<\/strong><\/p>\n

P (A and B fail) = 0.15<\/strong><\/p>\n

Evaluate the following probabilities.<\/strong><\/p>\n

Ans. <\/strong>Event A fails and B fails denoted by \u00a0and \u00a0respectively.<\/p>\n

\u00a0and P (A and B fails) = 0.15<\/p>\n

\u00a0\u00a0= 0.15<\/p>\n

P(\u00a0above) = <\/p>\n

0.15 = \u00a00.15<\/p>\n

\u00a0= 0.30<\/p>\n

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

(ii) <\/strong>P (A fails alone) = P (\u00a0alone) = \u00a0= 0.20 \u2013 0.15 = 0.05<\/p>\n

\n

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

16.Bag I contains 3 Red and 4 Black balls and B II contains 4 Red and % Black balls. One ball is transferred from Bag I to bag II and then a ball is drawn from bag II. The ball so drawn is found to be Red in colour. Find the probability that the transferred ball is Black.<\/strong><\/p>\n

Ans. <\/strong>Let E1<\/sub> = Ball transferred from Bag I to Bag II is red<\/p>\n

E2<\/sub> = Ball transferred from Bag I to Bag Ii is black<\/p>\n

A = Ball drawn from Bag II is red in colour<\/p>\n

P (E1<\/sub>) = \u00a0and P (E2<\/sub>) = <\/p>\n

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

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

\n

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

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

17.If A and B are two events such that P (A) <\/strong><\/strong>\u00a00 and P<\/strong><\/strong>\u00a0= 1, then:<\/strong><\/p>\n

(A) A \u00a0B<\/strong><\/p>\n

(B) B \u00a0A<\/strong><\/p>\n

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

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

Ans. <\/strong>A and B are two events such that P (A) \u00a00 and P\u00a0= 1<\/p>\n

\u00a0\u00a0<\/p>\n

A \u00a0B<\/p>\n

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

\n

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

18. If P<\/strong><\/strong>\u00a0> P (A), then which of the following is correct:<\/strong><\/p>\n

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

(B) P (A \u00a0B) < P (A) . P (B)<\/strong><\/p>\n

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

(D) P \u00a0= P (B)<\/strong><\/p>\n

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

<\/p>\n

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

\n

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

19.If A and B are any two events such that P (A) + P (B) \u2013 P (A and B) = P (A), then<\/strong><\/p>\n

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

(B) P \u00a0= 1<\/strong><\/p>\n

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

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

Ans. <\/strong>P (A) + P (B) \u2013 P (A and B) = P (A)<\/p>\n

P (B) \u2013 P (A and B)<\/p>\n

<\/p>\n

\u00a0= 1<\/p>\n

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

NCERT Solutions class 12 Maths Miscellaneous<\/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 Miscellaneous 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 … 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-4543","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 Miscellaneous | myCBSEguide<\/title>\n<meta name=\"description\" content=\"NCERT Solutions class 12 Maths Miscellaneous in PDF format for free download. 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