{"id":14201,"date":"2021-07-30T10:00:10","date_gmt":"2021-07-30T04:30:10","guid":{"rendered":"https:\/\/apboardsolutions.com\/?p=14201"},"modified":"2023-03-29T17:56:59","modified_gmt":"2023-03-29T12:26:59","slug":"ap-10th-class-maths-important-questions-chapter-13","status":"publish","type":"post","link":"https:\/\/apboardsolutions.com\/ap-10th-class-maths-important-questions-chapter-13\/","title":{"rendered":"AP 10th Class Maths Important Questions Chapter 13 Probability"},"content":{"rendered":"

These AP 10th Class Maths Chapter Wise Important Questions<\/a> Chapter 13 Probability will help students prepare well for the exams.<\/p>\n

AP State Syllabus 10th Class Maths 13th Lesson Important Questions and Answers Probability<\/h2>\n

Question 1.
\nIf P(E) = 3\/4, what is the probability of \u201cnot E\u201d?
\nSolution:
\nProbability P(E) = 3\/4.
\nPrabability of \u201cnot E\u201d P (\\((\\overline{\\mathrm{E}})\\)) = 1 – P(E)
\n= 1 – \\(\\frac{3}{4}=\\frac{1}{4}\\)<\/p>\n

Question 2.
\nYou are writing a test of 40 objective type questions. Each question carries 1 mark. What is the probability of marks you may get to be in multiple of 5?
\nSolution:
\nTotal Number of questions = 40
\nTotal Number of outcomes = 40
\nNumber of Multiples of 5 upto 40 = 8
\nFavourable outcomes = 8
\nProbability for getting multiples of 5
\nFavourable outcomes for getting = \\(\\frac{\\text { multiples of } 5}{\\text { Total No. of possible outcomes }}\\)
\n= \\(\\frac{8}{40}=\\frac{1}{5}\\)<\/p>\n

Question 3.
\nA page is opened at random from a book containing 100 pages. Find the probability that the page number is a perfect square.
\nSolution:
\nNumber of pages in given book = 100
\nThe page numbers that will be perfectly a square number (If randomly that are selected) are 1,4,9, 16, 25, 36, 49, 64, 81 and 100.
\n\u2234 Number of favourable outcomes = 100
\nNumber of all possible outcomes = its number of pages = 100
\n\u2234 Probability of getting a perfect number = \\(\\frac{10}{100}\\) = 0.1<\/p>\n

Question 4.
\nIf P(E) = 0.546, what is the probability of ‘not E’ ?
\nSolution:
\nP(E) = 0.546
\nP(E) = 1 – P(E)
\nProbability of “not E” = 1 – 0.546
\n= 0.454<\/p>\n

Question 5.
\nA box contains 3 blue and 4 red balls. What is the probability that the ball taken out randomly will be red ?
\nSolution:
\nTotal number of balls in the box = 3 + 4 = 7
\nNo. of favourable outcomes for picking a red ball = 4
\n\u2234 Probability of red ball
\n= \\(\\frac{\\text { No. of favourable outcomes }}{\\text { Total outcomes }}\\)
\n\u2234 P(E) = \\(\\frac{4}{7}\\)<\/p>\n

Question 6.
\nA three digit number is formed by the digits 2, 3 and 5 without repetition. What is the probability that the number is divisible by 5?
\nSolution:
\nLet ‘E’ be the event of choosing a three digit number divisible by 5.
\nAll possible three digit numbers (without repetition) 235, 253, 325, 352, 523,532.
\n\u2234 n(S) = 6.
\nE = {235,325}
\nn(E) = 2
\n\u2234 P(E) = \\(\\frac{2}{6}=\\frac{1}{3}\\)<\/p>\n

\"AP<\/p>\n

Question 7.
\nIn a class – room, 32 students out of 60 can take tea. Find the probability of “The tea not taken”.
\nSolution:
\nTotal Number of possible outcomes = 60
\nNo. of students doesn’t take tea (No. of favourable outcomes) = 60 – 32
\n= 28
\nProbability of students not taken tea
\n= \\(\\frac{\\text { No. of favourable outcomes for not taken tea }}{\\text { Total No. of possible outcomes }}\\)
\n= \\(\\frac{28}{60}=\\frac{7}{15}\\)<\/p>\n

Question 8.
\nWhat are equally likely events ? Give one example.
\nAnswer:
\nEqually likely events :
\nTwo events are said to be equally likely events if the probability of occurrence of those events in that experiment is equal.<\/p>\n

Question 9.
\nA bag contains 5 red and 8 white balls. If a ball is drawn at random from the bag, what is the probability that it will be
\ni) white ball
\nii) not to be white ball
\nSolution:
\nTotal number of balls present in bag = 5 (red) + 8 (white) = 13
\nProbability for taking out a white ball
\nP(E) = \\(\\frac{\\text { No.of favourable out comes }}{\\text { Total no.of out comes }}\\) = \\(\\frac{8}{13}\\)
\nProbability for not to be a white ball = p(\\(\\overline{\\mathrm{E}}\\))
\nWe know P(E) + p(\\(\\overline{\\mathrm{E}}\\)) = 1
\n\u21d2 p(\\(\\overline{\\mathrm{E}}\\)) = 1 – p(E) = 1 – \\(\\frac{8}{13}=\\frac{5}{13}\\)<\/p>\n

Question 10.
\nThere are 5 cards in a box with numbers 1 to 5 written on them. If 2 cards are picked out from the box, write all the possible outcomes and find the probability of getting both even numbers.
\nSolution:
\nTotal number of possible outcomes when 2 cards are picked out from the box = 10
\n(1, 2), (1, 3), (1, 4), (1, 5) (2, 3), (2, 4), (2, 5) (3, 4), (3, 5), (4, 5)
\n\u2234 Number of favourable outcomes for getting both even numbers = 1, (2, 4)
\n\u2234 Probability of getting both even no.s
\n= \\(\\frac{\\text { Number of favourable outcomes }}{\\text { Total no. of possible outcomes }}\\) = \\(\\frac{1}{10}\\)<\/p>\n

Question 11.
\nSolution:
\nA die is thrown once. Find the probability of getting
\ni) an even number
\nii) an odd prime number.
\nSolution:
\nWhen a die is thrown once total number of possible outcomes = 6
\n1) For an even number, favourable outcomes = 3
\nProbability for an even number
\n= \\(\\frac{\\text { No. of favourable outcomes }}{\\text { Total No. of possible outcomes }}\\)<\/p>\n

2) For an odd prime number, favourable outcomes = 2
\nProbability for an odd prime number = \\(\\frac{2}{6}=\\frac{1}{3}\\)<\/p>\n

Question 12.
\nA box contains 100 red cards, 200 yellow cards and 50 blue cards. If a card is drawn at random from the box, then find the probability that it will be
\ni) a blue card, ii) not a yellow card.
\nSolution:
\nGiven,
\nNumber of red cards =100
\nNumber of yellow cards = 200
\nNumber of blue cards = 50
\nTotal number of cards in a box = 100 + 200 + 50 = 350<\/p>\n

i) Let E1<\/sub> = Event of selecting that a card drawn is blue.
\nProbability (E1<\/sub>)
\n= \\(\\frac{\\text { Number of favourable outcomes }}{\\text { Total Number of outcomes }}\\)
\n= \\(\\frac{50}{350}=\\frac{1}{7}\\)<\/p>\n

ii) Let E2<\/sub> = Event of selecting that a card drawn is not a yellow card. Probability (E2<\/sub>)
\n= \\(\\frac{\\text { Number of favourable outcomes }}{\\text { Total Number of outcomes }}\\)
\n= \\(\\frac{150}{350}=\\frac{15}{35}=\\frac{3}{7}\\)<\/p>\n

\"AP<\/p>\n

Question 13.
\nPrepare any two problems getting Probability using dice.
\nSolution:
\n1) Find the probability of getting an even number when the dice is thrown once?<\/p>\n

2) What is the probability of getting an odd number when a dice is thrown once?<\/p>\n

Question 14.
\nA bag contains 5 red, 5 green and 5 white balls of the same size. A ball is drawn at random from the bag. Is the probability of picking up a ball of any colour equally likely or not? Justify.
\nSolution:
\nNumber of red balls = 5 = n(R)
\nNo. of green balls = 5 = n(G)
\nNo. of white balls = 5 = n(W)
\nTotal balls = 15 = T(B)
\nNow probability of picking red balls
\n= P (R) = \\(\\frac{150}{350}=\\frac{15}{35}=\\frac{3}{7}\\)
\n= \\(\\frac{5}{15}=\\frac{1}{3}\\)
\nProbability of picking green balls
\n= \\(\\frac{5}{15}=\\frac{1}{3}\\)
\nProbability of picking white balls
\n\\(\\frac{5}{15}=\\frac{1}{3}\\)<\/p>\n

Question 15.
\nFrom a deck of 52 playing cards, King, Ace and 10 of Clubs were removed and remaining cards were well shuffled. If a card is drawn at random from the remaining, find the probability of
\ngetting a card of
\ni) Club
\nii) Ace
\niii) Diamond king
\niv) Club 5.
\nSolution:
\nTotal number of possible outcomes
\n= 52 – 3 = 49
\n(i) Probability of getting a card of club Number of favourable outcomes
\n= \\(\\frac{\\text { for getting a card of club }}{\\text { Total No.of possible outcomes }}\\)
\n= \\(\\frac{10}{49}\\)
\n(ii) Probability of getting a card of ace = \\(\\frac{3}{49}\\)
\n(iii) Probability of getting a card of diamond king = \\(\\frac{1}{49}\\)
\n(iv) Probability of getting a card of club 5 = \\(\\frac{1}{49}\\)<\/p>\n

Question 16.
\nA bag contains 20 discs, which Are numbered from 1 to 20. If one disc is drawn at random from the bag, find the probability that it bears :
\ni) an even number,
\nii) Prime number,
\niii) Multiple of 5,
\niv) Two digit odd number.
\nSolution:
\nTotal number of possible outcomes = 20
\ni) For probability of the disc bears an even number
\nNo. of favourable outcomes = 10 Probability
\n= \\(=\\frac{\\text { No. of favourable outcomes }}{\\text { Total No. of possible outcomes }}\\) = \\(\\frac{10}{20}=\\frac{1}{2}\\)<\/p>\n

ii) For probability of the disc bears a prime number
\nNo.of favourable outcomes = 8
\nProbability = \\(\\frac{8}{20}=\\frac{2}{5}\\)<\/p>\n

iii) For probability of the disc bears a multiple of 5
\nNo. of favourable outcomes = 4
\nProbability = \\(\\frac{4}{20}=\\frac{1}{5}\\)<\/p>\n

iv) For probability of the disc bears a two digit odd number
\nNo. of favourable outcomes = 5
\nProbability = \\(\\frac{5}{20}=\\frac{1}{4}\\)<\/p>\n

Question 17.
\nTwo dice are thrown at the same time. What is the probability that the sum of two numbers appearing on the top of the dice is (a) 10, (b) less than or equal to 12, (c) a prime number, (d) multiple of ‘3’?
\nSolution:
\nTotal number of possible outcomes when rolling two dice at a time 6 \u00d7 6 = 36
\nFavourable outcomes of getting each sum is 10.<\/p>\n

a) Sum be 10 = {(5, 5), (4, 6), (6, 4)}
\nNo. of favourable outcomes = 3
\n\u2234 Required probability
\n= P(E) = \\(\\frac{3}{36}=\\frac{1}{12}\\)<\/p>\n

b) The outcomes favourable to the event “Less than or equal to 12” be denotes by’F’are
\n= {(1, 1) (1,2) (1,3) (1,4) (1,5) (1,6)
\n(2, 1) (2, 2) (2,3) (2,4) (2, 5) (2,6)
\n(3, 1) (3, 2) (3, 3) (3,4) (3, 5) (3,6)
\n(4, 1) (4, 2) (4,3) (4,4) (4, 5) (4,6)
\n(5, 1) (5, 2) (5, 3) (5,4) (5, 5) (5,6)
\n(6, 1) (6, 2) (6,3) (6,4) (6, 5) (6,6)}
\nNo. of outcomes favourable to ‘F’ is n(F) = 36
\n\u2234 p(F) = \\(\\frac{\\mathrm{n}(\\mathrm{F})}{\\mathrm{n}(\\mathrm{S})}=\\frac{36}{36}\\) = 36 = 1<\/p>\n

c) The outcomes favourable to the event “Sum of two numbers a prime number” be denoted by ‘G’ are (1, 1) (1, 2) (1,4) (1,6) (2, 1) (2, 3) (2, 5) (3, 2) (3, 4) (4, 1) (4, 3) (5, 2) (5, 6) (6, 1) (6, 5)
\nNo. of outcomes favourable to ‘G’ is n(G) = 15
\n\u2234 p(G) = \\(\\frac{n(G)}{n(S)}=\\frac{15}{36}=\\frac{5}{12}\\)<\/p>\n

d) The outcomes favourable to the event “Sum of two numbers be multiple of 3” be denoted by ‘H’ are (1, 2) (1, 5) (2, 1) (2, 4) (3, 3) (3, 6) (4, 2) (4, 5) (5, 1) (5, 4) (6, 3) (6, 6)
\nNo. of outcomes favourable to ‘H’ is n(H) =12
\n\u2234 p(H) = \\(\\frac{n(H)}{n(S)}=\\frac{12}{36}=\\frac{1}{3}\\)<\/p>\n

Question 18.
\nOne card Is drawn from a well shuffled deck of 52 cards.
\nFind the probability of getting :
\ni) a King of red colour.
\nii) the Jack of black.
\niii) a black face card.
\niv) the Queen of diamonds.
\nSolution:
\nTotal no. of cards = 52
\nTotal no. of possible outcomes = 52
\nA card is drawn randomly from a well shuffled deck of cards<\/p>\n

i) No. of favourable outcomes forgetting a king of red colour = 2 Probability
\n= \\(\\frac{\\text { No. of favourable outcomes }}{\\text { Total No. of possible outcomes }}\\)
\n= \\(\\frac{2}{52}=\\frac{1}{26}\\)<\/p>\n

ii) No. of favourable outcomes forgetting the jack of black = 2
\nProbability = \\(\\frac{2}{52}=\\frac{1}{26}\\)<\/p>\n

iii) No. of favourable outcomes forgetting a black face card = 6
\nProbability = \\(\\frac{6}{52}=\\frac{3}{26}\\)<\/p>\n

iv) No. of favourable outcomes forgetting the Queen of diamonds = 1
\nProbability = \\(\\frac{1}{52}\\)<\/p>\n

\"AP<\/p>\n

Question 19.
\nTwo digit numbers are formed by the digits 0, 1, 2, 3, 4, where the digits are not repeated. Find the probability that
\ni) the number formed is greater than 42.
\nii) the number formed is a multiple of 4.
\nSolution:
\nTwo digit numbers formed by the digits 0, 1,2,3,4
\nwhere the digits are not repeated
\n(10, 12, 13, 14, 20, 21, 23, 24, 30, 31, 32, 34, 40, 41, 42, 43)
\n\u2234 Sample space = (10, 12, 13, 14, 20, 21,23,24,30,31,32,34,40,
\n41, 42, 43)
\n\u2234 n(S) = 16
\ni) Probability of getting the number formed is greater than 42
\n\\(No. of possible outcomes $F$ $=$ Total outcomes\\)
\nhere no. of possible outcomes = 1 (that is 43 only)
\nProbability = (1)
\n\\(\\frac{1}{16}\\) ……………….(1)<\/p>\n

ii) In the sample space multiples of ‘4’ = 12, 20, 24, 32, 40
\n\u2234 No. of multiples of 4 = 5
\nNow probability for forming a multiple of ‘4’ = \\(\\frac{5}{16}\\) …………….. (2)<\/p>\n

Question 20.
\nFrom a pack of 52 playing cards; Jacks, Queens, Kings and Aces of red colour are removed. From the remaining, a card is drawn at random. Find the probability that the card drawn is
\ni) a black queen,
\nii) a face card,
\niii) a blackjack,
\niv) a red card.
\nSolution:
\nTotal number of cards = 52
\nCard removed = 2 + 24-2 + 2 = 8
\n\u2234 Remaining number of cards = 52 – 8 = 44<\/p>\n

i) Number of black queens = 2
\n\u2234 Required probability = \\(\\frac{2}{44}=\\frac{1}{22}\\)<\/p>\n

ii) Number of face cards left = 2 + 2 + 2 = 6
\n\u2234 Required probability = \\(\\frac{6}{44}=\\frac{3}{22}\\)<\/p>\n

iii) Number of blackjacks = 2
\n\u2234 Required probability = \\(\\frac{2}{44}=\\frac{1}{22}\\)<\/p>\n

iv) Number of red cards left = 26 – 8
\n= 18
\n\u2234 Required probability = \\(\\frac{18}{44}=\\frac{9}{22}\\)<\/p>\n

Question 21.
\nThere are 3 red and 4 white balls in a bag. If a ball is taken randomly then calculate the probability of it to be a) red ball b) white ball.
\nSolution:
\nTotal number of balls in the bag = 3 + 4 = 7
\n\u2234 Total outcomes = 7
\na) Number of favourable outcomes for picking a red ball = 3
\n\u2234 Probability of red ball = \\(\\frac{\\text { No. of favourable outcomes }}{\\text { No. of total outcomes }}\\) = \\(\\frac{3}{7}\\)
\noutcomes 7
\nP(E) = \\(\\frac{3}{7}\\)
\nNow number of favourable outcomes for picking a white ball = 4
\n\u2234 Probability for picking a white ball = \\(\\frac{4}{7}\\)<\/p>\n

Question 22.
\nFind the probability of existing 53 Sundays in a common year.
\nSolution:
\nIn a common year, no. of days = 365
\nSo number of weeks = 365 = (52) 7 + (1)
\nThat means there are 52 weeks and 1 more day in a year.
\nSo this 1 day may be any day of the week,
\ni.e., from Sunday to Saturday.
\nSo number of possible outcomes = 7
\nNumber of favourable outcomes for Sunday = 1
\nHence the probability for 53 Sundays = \\(\\frac{1}{7}\\)<\/p>\n

Question 23.
\nIn a skinner numbered from 1-20, find the probability of getting the following
\n(i) prime number
\n(ii) composite number
\n(iii) multiple of three.
\nSolution:
\ni) Numbers in the skinner is 1 to 20.
\nSo the number of all possible outcomes = 20
\nList of prime numbers in the above skinner = 2, 3, 5, 7, 11, 13, 17, 19
\nSo, number of favourable outcomes for a prime number = 8
\n\u2234 Probability of a prime number = \\(\\frac{8}{20}=\\frac{2}{5}\\)<\/p>\n

ii) List of composite numbers in above skinner = 4,6,8,9,10,12,14,15,16,18,20
\n\u2234 Number of favourable outcomes for a composite number = 11
\n\u2234 Probability of getting composite number = \\(\\frac{11}{20}\\)<\/p>\n

iii) List of multiples of 3 in above skinner = 3, 6, 9, 12, 15 and 18
\n\u2234 Number of favourable outcomes for a multiple of 3 = 6
\n\u2234 Probability of getting a multiple of 3 = \\(\\frac{6}{20}=\\frac{3}{10}\\)<\/p>\n

Question 24.
\nWhen a six face die is rolled find the probability of getting the following
\ni) getting less than five
\nii) getting more than five.
\nSolution:
\nFaces on the dice = 6
\n\u2234 number of total possible outcomes = 6<\/p>\n

i) List of possible out comes less than five = 1, 2, 3, 4
\nSo number of favourable outcomes for less than five = 4
\n\u2234 Probability of getting a number less than five = \\(\\frac{4}{6}=\\frac{2}{3}\\)<\/p>\n

ii) List of possible outcomes for getting more than five = (6)
\n\u2234 Number of possible outcomes for getting more than five = 1
\n\u2234 Probability of getting more than five = \\(\\frac{1}{6}\\)<\/p>\n

\"AP<\/p>\n

Question 25.
\nWhen two dice are rolled.
\nFill the probability of the sum on their faces in the table given below.
\n\"AP
\nDraw the bar graph for the above.
\nSolution:
\n\"AP
\n\"AP
\nNow on a graph paper
\nConsider \u201cSum on faces\u201d on x-axis and respective probabilities on y-axis to draw bar graphs from the above table.
\n\"AP<\/p>\n

Question 26.
\nIn a 50 marks examination, there is 80% possibility to pass in that exam. So find the probability for pass in exam.
\nSolution:
\nPossibility for pass in Exam = 80% = \\(\\frac{80}{100}\\)
\nthat means in out of 100 chances (total possible out comes = 100) 80 chances are favourable to pass (number of favourable out comes = 80)
\n\u2234 Probability for pass = \\(=\\frac{\\text { number of favourable outcomes }}{\\text { number of total possible outcomes }}\\)
\n= \\(\\frac{80}{100}=\\frac{4}{5}\\)
\nthen its probability = \\(\\frac{4}{5}\\)<\/p>\n

Question 27.
\nIn a class 32 students out of 60 take tea. So find the probability of choosing randomly at student who doesn\u2019t take tea.
\nSolution:
\nTotal number of students = 60 = Number of possible outcomes.
\nNumber of students take tea = 32
\nSo number of students who don’t take tea = 60 – 32 = 28
\nNumber of favourable outcomes for choosing a boy who doesn\u2019t take tea = 28
\n\u2234 Its probability = \\(\\frac{28}{60}=\\frac{7}{15}\\)<\/p>\n

\"AP<\/p>\n

Question 28.
\nIn the following table, temperature and rainfall conditions are stated. Then find the possibility of rainfall and no rainfall.
\n\"AP
\nSolution:
\nPosition – I:
\ni) Rainfall possibility is high when temperature is more than 35\u00b0C.<\/p>\n

Position – II :
\nRainfall possibility is low when temperature is below 31\u00b0C.<\/p>\n

Question 9.
\nFind the probability of winning of each team from the given table.
\n\"AP
\nSolution:
\ni) Probability of winning to India when bats first = \\(\\frac{20}{30}\\left(\\frac{\\text { Won }}{\\text { Played }}\\right)=\\frac{2}{3}\\) = 0.66
\nii) Probability of Pakistan ~ 28 = \\(\\frac{12}{28}=\\frac{3}{7}\\) = 0.43
\niii) Probability of Australia = \\(\\frac{22}{44}=\\frac{1}{2}\\) = 0.50
\niv) Probability of Sri Lanka = \\(\\frac{20}{50}=\\frac{2}{5}\\) = 0.40
\nSo highest probability of winning is with INDIA i.e., = \\(\\frac{140}{210}=\\frac{2}{3}\\) = 0.66<\/p>\n","protected":false},"excerpt":{"rendered":"

These AP 10th Class Maths Chapter Wise Important Questions Chapter 13 Probability will help students prepare well for the exams. AP State Syllabus 10th Class Maths 13th Lesson Important Questions and Answers Probability Question 1. If P(E) = 3\/4, what is the probability of \u201cnot E\u201d? Solution: Probability P(E) = 3\/4. Prabability of \u201cnot E\u201d … Read more<\/a><\/p>\n","protected":false},"author":4,"featured_media":35715,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[12],"tags":[],"jetpack_featured_media_url":"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2021\/07\/Chapter-13-Probability.png","_links":{"self":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/14201"}],"collection":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/users\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/comments?post=14201"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/14201\/revisions"}],"predecessor-version":[{"id":35716,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/14201\/revisions\/35716"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media\/35715"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media?parent=14201"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/categories?post=14201"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/tags?post=14201"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}