{"id":11501,"date":"2024-03-09T12:35:48","date_gmt":"2024-03-09T07:05:48","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=11501"},"modified":"2024-03-12T17:49:44","modified_gmt":"2024-03-12T12:19:44","slug":"ts-10th-class-maths-important-questions-chapter-13","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-10th-class-maths-important-questions-chapter-13\/","title":{"rendered":"TS 10th Class Maths Important Questions Chapter 13 Probability"},"content":{"rendered":"

These\u00a0TS 10th Class Maths Chapter Wise Important Questions<\/a> Chapter 13 Probability given here will help you to solve different types of questions.<\/p>\n

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

Previous Exams Questions<\/span><\/p>\n

Question 1.
\n5 red and 8 white balls are present in a bag. If a ball is taken randomly from the bag then find the probability of it to be
\ni) white ball
\nii) not to be white ball (A.P. Mar. ’16)
\nSolution:
\nTotal number of balls present in bag
\n= 5 (red) + 8 (white) = 13
\nProbability for taking out a white ball
\nP(E) = \\(\\frac{\\text { No. of favourable outcomes }}{\\text { Total No. of outcomes }}\\)
\n= \\(\\frac{8}{13}\\)
\nProbability for not to be a white ball
\nP(\\(\\overline{\\mathrm{E}}\\)) = \\(\\frac{8}{13}\\)
\nwe know P(E) + P(\\(\\overline{\\mathrm{E}}\\)) = 1
\n\u21d2 P(\\(\\overline{\\mathrm{E}}\\))
\n= 1 – P(E) = 1 – \\(\\frac{8}{13}\\)
\n= \\(\\frac{5}{13}\\)<\/p>\n

\"TS<\/p>\n

Question 2.
\nWhen die is rolled once unbiased what is the probability of getting a multiple of 3 out of possible out comes ? (T.S. Mar. ’15)
\nSolution:
\nP(E) = \\(\\frac{\\text { favourable outcomes }}{\\text { Total outcomes }}\\) = \\(\\frac{2}{6}\\) = \\(\\frac{1}{3}\\)<\/p>\n

Question 3.
\nThere are 12 red, 18 blue and 6 white balls in a box. When balls is drawn at random from the box, what is the probability of not getting a red ball ? (T.S. Mar. ’15)
\nSolution:
\nTotal Number of balls = 12 + 8 + 6
\n= 36
\nNumber of red balls = 12
\nprobability of getting red ball
\nP(\\(\\overline{\\mathrm{R}}\\)) = \\(\\frac{\\text { favourable outcomes }}{\\text { Total outcomes }}\\)
\n= \\(\\frac{12}{36}\\) = \\(\\frac{1}{3}\\)
\n\u2234 Probability of not getting red ball
\nP(R) = 1 – \\(\\frac{1}{3}\\) = \\(\\frac{2}{3}\\)
\nTotal Number of balls = 12 + 18 + 6 = 36
\nExclude, the red balls, the number of remaining balls = 18 + 6 = 24
\n\u2234 Probability of not getting a
\nRed ball = \\(\\frac{24}{36}\\) = \\(\\frac{2}{3}\\)<\/p>\n

\"TS<\/p>\n

Question 4.
\nThere are 100 flash cards labelled from 1 to 100 in a bag. When a card is drawn from the bag at random, what is the probability of getting …….
\n(i) a card with prime number from possible outcomes ?
\n(ii) a card without prime number from possible outcomes ? (T.S. Mar. ’15)
\nSolution:
\nNumber of prime numbers between 1 and 100 = 25
\nProbability of getting a card with prime numbers = \\(\\frac{25}{100}\\) = \\(\\frac{1}{4}\\) = 0.25
\nProbability of getting a card without prime number = \\(\\frac{75}{100}\\) = 0.75
\n1 – 0.25 = 0.75<\/p>\n

Question 5.
\nFind the probability of setting a sum of the numbers on them is 7, when two dice are rolled at a time. (T.S. Mar. ’16)
\nSolution:
\nWhen two dice are rolled at a time the total outcomes are = 62<\/sup> = 36.
\nNumber of outcomes such that their sum of numbers on face is 7 = 6
\n\u2234 Probability of getting sum of numbers on faces to be
\n7 = \\(\\frac{6}{36}\\) = \\(\\frac{1}{6}\\)<\/p>\n

\"TS<\/p>\n

Question 6.
\nA bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of red ball, find the number of blue balls in the bag. (T.S. Mar. ’16)
\nSolution:
\nNumber of red balls present in a bag = 5
\nLet the No. of blue balls = x (say)
\nThen the total No. of balls = 5 + x
\nFrom those (5 + x) balls in the bag the number of favourable outcomes to take a red ball randomly = 5
\nSo, the probability of taking a red ball = \\(\\frac{5}{5+x}\\)
\nNow
\nThe number of favourable outcomes to take a blue ball randomly = x
\nSo, the probability of taking a blue ball = \\(\\frac{x}{5+x}\\)
\nFrom the given problem
\nProbability of blue bell = (Probability of red ball) (2)
\n\\(\\frac{x}{5+x}\\) = \\(\\left[\\frac{5}{5+x}\\right]\\) 2
\n\\(\\frac{x}{5+x}\\) = \\(\\frac{10}{5+x}\\) \u21d2 x = 10
\n\u2234 No. of blue balls in the bag = 10.<\/p>\n

Additional Questions<\/span><\/p>\n

Question 1.
\nKishore buys a fruit from a shop. The shopkeeper have one box. The box contain 18 mangoes, 32 apples so shopkeeper takes out one fruit at random what is the probability that the mango taken out from box.
\nSolution:
\nGiven
\nNumber of mangoes in the box = 18
\nNumber of apples in the box = 32
\nTotal number of fruits in the box = 18 + 32
\nSo total number of outcomes = 50
\nLet E be the even that the mango taken out of the box = 18
\nWe know that,
\nP(E) = \\(\\frac{\\text { No. of outcomes favourable to E}}{\\text { Total No. of all possible outcomes }}\\)
\n= \\(\\frac{18}{50}\\) = \\(\\frac{9}{25}\\)<\/p>\n

Question 2.
\nA room contains 30 green chairs and some white chairs if the probability of drawing a white chair is triple that of green chair determine the number of white chairs in the room.
\nSolution:
\nLet the number of white chairs = x
\nGiven number of green chairs = 30
\nTotal number of chairs in room = x + 30
\nTotal outcomes in drawing a chair at random = x + 30
\nNumber of outcomes favourable to green chair = 30
\n\u2234 P(G) = \\(\\frac{30}{x+30}\\)
\nSo, given in problem P(W) = 3 \u00d7 \\(\\frac{30}{x+30}\\)
\n= \\(\\frac{90}{x+30}\\)
\nWe know that P(G) + P(W) = 1
\n\\(\\frac{30}{x+30}\\) + \\(\\frac{90}{x+30}\\) = 1
\n\\(\\frac{120}{x+30}\\) = 1
\nx + 30 = 120
\nx = 90
\n\u2234 No. of white chairs x = 90.<\/p>\n

\"TS<\/p>\n

Question 3.
\nThere are 25 cards of same size in a bag on which number 1 to 25 are written one card is taken out of the bag at random. Find the probability that the number on the selected card is not divisible by 5.
\nSolution:
\nGiven total number of cards = 25
\nThe number which are divisible by ‘5’ are 5, 10, 15, 20, 25
\nNo. of all possible outcomes n(5) = 25
\nNumber of out comes favourable to
\nE = n(E) = 5
\n\u2234 P(E) = \\(\\frac{n(E)}{n(S)}\\)
\n= \\(\\frac{5}{25}\\) = \\(\\frac{1}{5}\\)
\nWe know that P(E) + P(\\(\\overline{\\mathrm{E}}\\)) = 1
\nP(\\(\\overline{\\mathrm{E}}\\)) = 1 – (P(E))
\np(\\(\\overline{\\mathrm{E}}\\)) = 1 – \\(\\frac{1}{5}\\)
\n= \\(\\frac{5-1}{5}\\)
\nP(\\(\\overline{\\mathrm{E}}\\)) = \\(\\frac{4}{5}\\)
\n\u2234 probability that the number selected card is not divisible by 5 = \\(\\frac{4}{5}\\)<\/p>\n

Question 4.
\nA jar contains 18 marbles, some are red and other white if a marble is drawn at random from the jar the probability that it is white \\(\\frac{5}{6}\\) . Find the number of white marbles.
\nSolution:
\nTotal number of marbles in the jar = 18
\nLet the number of red marbles = k
\nThe number of white marbles = 18 – k
\nprobability of drawing a red marble = \\(\\frac{\\mathrm{k}}{18}\\)
\nFrom problem = \\(\\frac{\\mathrm{k}}{18}\\) = \\(\\frac{5}{6}\\)
\n\u21d2 k = \\(\\frac{90}{6}\\)
\n\u21d2 k = 15
\nNo. of red marbles = k = 15
\nNo. of white marbles = 18 – 15 = 3.<\/p>\n

\"TS<\/p>\n

Question 5.
\nA game consists of tossing a one rupee coin 2 times and noting its outcome each time. Ravi wins if all the wins give the same result, i.e., two heads or two tails and lose otherwise. Calculate the probability that he will lose the game.
\nSolution:
\nWe know if a coin is tossed for n times, then the total number of outcomes = 2n<\/sup>
\nSo, a coin is tossed for 2 times, then the total number of outcomes 22<\/sup> = 4.
\nsee here
\nT T
\nT H
\nH T
\nH H
\nof the above no. of outcomes with different result = 2.
\nprobability of lossing the game
\n= \\(\\frac{\\text { No. of favourable outcomes of lose }}{\\text { No. of total outcomes }}\\)
\n= \\(\\frac{2}{4}\\) = \\(\\frac{1}{2}\\)<\/p>\n

Question 6.
\nA lot consists of 200 ball pens of which 50 are defective and others are good. The shop keeper draws one pen at random and gives to sindhu. what is the probability that,
\n1) She will buy it ?
\n2) She will not buy it ?
\nSolution:
\ni) Total no. of ball pens = 200
\n\u2234 Number of all possible outcomes = 200
\nNumber of defective ball pens = 50
\nNumber of good ball pens = 200 – 50
\n\u21d2 No. of favourable outcomes = 150
\nprobability that sindhu will buy it
\n= \\(\\frac{\\text { No. of favourable outcomes }}{\\text { Total No. of all possible outcomes }}\\)
\n= \\(\\frac{150}{200}\\) = \\(\\frac{3}{4}\\)<\/p>\n

\"TS<\/p>\n

ii) Probability that sindhu will not buy it
\n= 1 – (probability that sindhu will buy it)
\n= 1 – \\(\\frac{3}{4}\\)
\n= \\(\\frac{4-3}{4}\\)
\n= \\(\\frac{1}{4}\\)<\/p>\n","protected":false},"excerpt":{"rendered":"

These\u00a0TS 10th Class Maths Chapter Wise Important Questions Chapter 13 Probability given here will help you to solve different types of questions. TS 10th Class Maths Important Questions Chapter 13 Probability Previous Exams Questions Question 1. 5 red and 8 white balls are present in a bag. If a ball is taken randomly from the … 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\/11501"}],"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=11501"}],"version-history":[{"count":3,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/11501\/revisions"}],"predecessor-version":[{"id":11882,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/11501\/revisions\/11882"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=11501"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=11501"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=11501"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}