{"id":9497,"date":"2024-02-18T10:58:49","date_gmt":"2024-02-18T05:28:49","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=9497"},"modified":"2024-02-21T17:20:18","modified_gmt":"2024-02-21T11:50:18","slug":"ts-10th-class-maths-solutions-chapter-13-intext-questions","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-10th-class-maths-solutions-chapter-13-intext-questions\/","title":{"rendered":"TS 10th Class Maths Solutions Chapter 13 Probability InText Questions"},"content":{"rendered":"

Students can practice\u00a010th Class Maths Textbook SSC Solutions Telangana<\/a> Chapter 13 Probability InText Questions to get the best methods of solving problems.<\/p>\n

TS 10th Class Maths Solutions Chapter 13 Probability InText Questions<\/h2>\n

Do This<\/span><\/p>\n

(a) Outcomes of which of the following experiments cure equally likely. (AS3<\/sub>)(Page No. 307)<\/p>\n

Question 1.
\nGetting a digit 1, 2, 3, 4, 5 or 6 when a die is rolled.
\nSolution:
\nEqually likely<\/p>\n

Question 2.
\nSelecting a different colour ball from a bag of 5 red balls, 4 blue balls and 1 black ball.
\nNote : Picking two different colour balls, i.e., Picking a red, a blue (or) black ball from a ………
\nSolution:
\nNot equally likely<\/p>\n

Question 3.
\nWinning in a game of carrom.
\nSolution:
\nNot equally likely<\/p>\n

Question 4.
\nUnits place of a two digit number selected may be 0, 1, 2, 3, 4, 5, 6, 7, 8 or 9.
\nSolution:
\nequally likely<\/p>\n

Question 5.
\nSelecting a different colour ball from a bag of 10 red balls, 10 blue balls and 10 black balls.
\nSolution:
\nequally likely<\/p>\n

\"TS<\/p>\n

Question 6.
\nRaining on a particular day of July.
\nSolution:
\nequally likely<\/p>\n

(b) Are the outcomes of every experiment equally likely ?
\nSolution:
\nOutcomes of all experiments need not necessarily be equally likely.<\/p>\n

(c) Give examples of 5 experiments that have equally likely outcomes and five more examples that do not have equally likely outcomes.
\nSolution:
\nEqually likely events :
\na) Getting an even or odd number when a die is rolled.
\nb) Getting tail or head when a coin is tossed.
\nc) Getting an even (or) odd number when a card is drawn at random from a pack of cards numbered from 1 to 10.
\nd) Picking a green or black ball from a bag containing 8 green balls and 8 black balls.
\ne) Selecting a boy or girl from a class of 20 boys and 20 girls.
\nf) Selecting a red or black card from a deck of cards.<\/p>\n

Events which are not equally likely :
\na) Getting a prime (or) composite number when a die is thrown.
\nb) Getting an even or odd number when a card is drawn at random from a pack of cards numbered from 1 to 5.
\nc) Getting a number which is a multiple of 3 (or) not a multiple of 3 from numbers 1, 2, …….., 10.
\nd) Getting a number less than 5 or greater than 5.
\ne) Drawing a white ball or green ball from a bag containing 5 green balls and 8 white balls.<\/p>\n

Think of 5 situations with equally likely events and find the sample space.
\na) Tossing a coin : Getting a tail or head when a coin is tossed.
\nSample space = {T, H}
\nb) Getting an even (or) odd number when a die is rolled.
\nSample space = {1, 2, 3, 4, 5, 6}
\nc) Winning a game of shuttle Sample space = {win, loss}
\nd) Picking a black (or) blue ball from a bag containing 3 blue balls and 3 black balls = {blue, black}
\ne) Drawing a red coloured card or black coloured card from a deck of cards = {black, red}<\/p>\n

\"TS<\/p>\n

d) Is getting a head complementary to getting a tail ? Give reasons.
\nSolution:
\nNumber outcomes favourable to head = 1
\nProbability of getting a head = \\(\\frac{1}{2}\\) [P(E)]
\nNumber of outcomes not favourable to head = 1
\nProbability of not getting a head = \\(\\frac{1}{2}\\) (P (\\(\\overline{\\mathrm{E}}\\)))
\nNow P(E) + P(\\(\\overline{\\mathrm{E}}\\)) = \\(\\frac{1}{2}\\) + \\(\\frac{1}{2}\\) = 1
\n\u2234 Getting a head is complementary to getting a tail.<\/p>\n

e) In case of a die is getting a 1 complementary to events getting 2, 3, 4, 5, 6 ? Give reasons for your answer.
\nSolution:
\nYes, complementary events
\n\u2234 Probability of getting a 1 = \\(\\frac{1}{6}\\) [P(E)]
\nProbability of getting 2, 3, 4, 5, 6
\n= P (\\(\\overline{\\mathrm{E}}\\)) = \\(\\frac{5}{6}\\)
\np(E) + P(\\(\\overline{\\mathrm{E}}\\)) = \\(\\frac{1}{6}\\) + \\(\\frac{5}{6}\\) = \\(\\frac{6}{6}\\) = 1<\/p>\n

f) Write of five new pair of events that are complementary.
\nSolution:
\na) When a die is thrown, getting an even number is complementary to getting an odd number.
\nb) Drawing a red card from a deck of cards is complementary to getting a black card.
\nc) Getting an even number is complementary to getting an odd number from numbers 1, 2, ……….., 8.
\nd) Getting a Sunday is complementary to getting any day other than Sunday in a week.
\ne) Winning a running race is complementary to loosing it.<\/p>\n

Try This<\/span><\/p>\n

Question 1.
\nA child has a die whose six faces show the letters A, B, C, D, E and F. The die is thrown once. What is the probability of getting (i) A ? (ii) D ? (AS4<\/sub>)(Page No. 312)
\nSolution:
\nTotal number of outcomes [A, B, C, D, E and F] = 6
\ni) Number of favourable outcomes to A = 1
\nProbability of getting A = P(A)
\n\"TS
\n= \\(\\frac{1}{6}\\)<\/p>\n

\"TS<\/p>\n

ii) Number of outcomes favourable to D = 1
\nProbability of getting D
\n= P(D)
\n\"TS
\n= \\(\\frac{1}{6}\\)<\/p>\n

Question 2.
\nWhich of the following cannot be the probability of an event ? (AS3<\/sub>)(Page No. 312)
\na) 2.3
\nb) – 1.5
\nc) 15%
\nd) 0.7
\nSolution:
\na) 2.3 – Not possible
\nb) -1.5 – Not possible
\nc) 15% – May be the probability
\nd) 0.7 – May be the probability<\/p>\n

Question 3.
\nYou have a single deck of well shuffled cards. Then,
\ni) What is the probability that the card drawn will be a queen ? (AS4<\/sub>)(Page No. 313)
\nSolution:
\nNumber of all possible outcomes
\n= 4 \u00d7 13 = 1 \u00d7 52 = 52
\nNumber of outcomes favourable to Queen
\n= 4[\u2665Q \u2665Q \u2665Q \u2665Q]
\nProbability P(E)
\n= \\(\\frac{\\text { No. of favourable outcomes }}{\\text { Total no.of outcomes }}\\)
\n= \\(\\frac{4}{52}\\)
\n= \\(\\frac{1}{13}\\)<\/p>\n

ii) What is the probability that it is a face card ? (Page No. 314)
\nSolution:
\nFace cards are J, Q, K.
\n\u2234 Number of outcomes favourable to face cards = 4 \u00d7 3 = 12
\nNo. of all possible outcomes = 52
\nP(E) = \\(\\frac{\\text { No. of favourable outcomes }}{\\text { Total no. of outcomes }}\\)
\n= \\(\\frac{12}{52}\\) = \\(\\frac{3}{13}\\)<\/p>\n

\"TS<\/p>\n

iii) What is the probability that it is a spade ? (Page No. 314)
\nSolution:
\nNumber of spade cards = 13
\nTotal number of cards = 52
\nProbability = \\(\\frac{\\text { Number of outcomes favourable to spade }}{\\text { Number of all outcomes }}\\)
\n= \\(\\frac{13}{52}\\) = \\(\\frac{1}{4}\\)<\/p>\n

iv) What is the probability that is the face cards of spades ? (Page No. 314)
\nSolution:
\nNumber of outcomes favourable to face cards of spades = (K, Q, J) = 3
\nNumber of all outcomes = 52 3
\n\u2234 P(E) = \\(\\frac{3}{52}\\)<\/p>\n

v) What is the probability it is not a face card ? (Page No. 314)
\nSolution:
\nProbability of a face card = \\(\\frac{12}{52}\\)
\n\u2234 Probability that the card is not a face card
\n= 1 – \\(\\frac{12}{52}\\) [P (\\(\\overline{\\mathrm{E}}\\)) = 1 – P(E)]
\n= \\(\\frac{52-12}{52}\\)
\n= \\(\\frac{40}{52}\\) = \\(\\frac{10}{13}\\)<\/p>\n

(Or)<\/p>\n

Number of favourable outcomes = 4 \u00d7 10 = 40
\nNumber of all outcomes = 52
\n\u2234 Probability = \\(\\frac{\\text { No. of favourable outcomes }}{\\text { Total no. of outcomes }}\\)
\n= \\(\\frac{40}{52}\\) = \\(\\frac{10}{13}\\)<\/p>\n

Think – Discuss<\/span><\/p>\n

Question 1.
\nWhy is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of any game ? (Page No. 312)
\nSolution:
\nProbability of getting a head is \\(\\frac{1}{2}\\) and a tail is
\n\\(\\frac{1}{2}\\) = 1
\nHence, tossing a coin is a fair way.<\/p>\n

\"TS<\/p>\n

Question 2.
\nCan \\(\\frac{7}{2}\\) be the probability of an event ? Explain. (AS3<\/sub>) (Page No. 312)
\nSolution:
\n\\(\\frac{7}{2}\\) can’t be the probability of any event. Since probability of any event should lie between 0 and 1.<\/p>\n

Question 3.
\nWhich of the following arguments are correct and which are not correct ? Given reasons. (Page No. 312)<\/p>\n

i) If two coins are tossed simultaneously, there are three possible outcomes – two heads, two tails (or) one of each. Therefore, for each. If these outcomes, the probability is \\(\\frac{1}{3}\\)
\nSolution:
\nFalse
\nReason :
\nAll possible outcomes are 4. They are HH, HT, TH, TT
\nThus, probability of two heads = \\(\\frac{1}{4}\\)
\nProbability of two tails = \\(\\frac{1}{4}\\)
\nProbability of one each = \\(\\frac{2}{4}\\) = \\(\\frac{1}{2}\\)<\/p>\n

\"TS<\/p>\n

ii) If a die is thrown, there are two possible outcomes – an odd number (or) an even number. Therefore, the probability of getting an odd number is \\(\\frac{1}{2}\\).
\nSolution:
\nTrue
\nReason :
\nAll possible outcomes = (1, 2, 3, 4, 5, 6) = 6
\nOutcomes favourable to an odd number = (1, 3, 5) = 3
\nOutcomes favourable to an even number = (2, 4, 6) = 3
\n\u2234 Probability (odd number)
\n= \\(\\frac{\\text { No. of favourable outcomes }}{\\text { Total no. of outcomes }}\\)
\n= \\(\\frac{3}{6}\\) = \\(\\frac{1}{2}\\)<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can practice\u00a010th Class Maths Textbook SSC Solutions Telangana Chapter 13 Probability InText Questions to get the best methods of solving problems. TS 10th Class Maths Solutions Chapter 13 Probability InText Questions Do This (a) Outcomes of which of the following experiments cure equally likely. (AS3)(Page No. 307) Question 1. Getting a digit 1, 2, … Read more<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[16],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/9497"}],"collection":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/comments?post=9497"}],"version-history":[{"count":1,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/9497\/revisions"}],"predecessor-version":[{"id":9500,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/9497\/revisions\/9500"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=9497"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=9497"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=9497"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}