10th Class Maths 13th Lesson Probability Optional Exercise Textbook Questions and Answers<\/h3>\n
Question 1. Question 2. Question 3. Question 4. AP State Syllabus SSC 10th Class Maths Solutions 13th Lesson Probability Optional Exercise AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions Chapter 13 Probability Optional Exercise Textbook Questions and Answers. 10th Class Maths 13th Lesson Probability Optional Exercise Textbook Questions and Answers Question 1. Two customers Shyam and Ekta are visiting a particular … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/6297"}],"collection":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/comments?post=6297"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/6297\/revisions"}],"predecessor-version":[{"id":40649,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/6297\/revisions\/40649"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=6297"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=6297"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=6297"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\nTwo customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on
\n(i) the same day?
\n(ii) consecutive days?
\n(iii) different days?
\nAnswer:
\nShyam and Ekta can visit the shop in the following combination:
\n![\"AP](\"https:\/\/apboardsolutions.in\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Solutions-Chapter-13-Probability-Optional-Exercise-1.png\")
\n(T Tu) (W, Tu) (Th, Tu) (F, Tu) (S, Tu) (Tu, W) (W, W) (Th, W) (F, W) (S, W) (Tu, Th) (W, Th) (Th, F) (F, Th) (S, Th) (Tu, F) (W, F) (Th, S) (F, F) (S, F) (Tu, S) (W, S) (Th, Th) (F, S) (S, S)
\n\u2234 Number of total outcomes = 5 \u00d7 5 = 52<\/sup> = 25 [also from the above table]
\ni) Number of favourable outcomes to that of visiting on the same day
\n(Tu, Tu), (W, W), (Th, Th), (F, F), (S, S) = 5
\n\u2234 Probability of visiting the shop on the same day = \\(\\frac{\\text { No. of favourable outcomes }}{\\text { No. of total outcomes }}\\) = \\(\\frac{5}{25}\\) = \\(\\frac{1}{5}\\)
\nii) Number of outcomes favourable to consecutive days
\n(Tu, W), (W, Th), (Th, F), (F, S), (W, Tu), (Th, W), (F, Th), (S, F) = 8
\n\u2234 Probability of visiting the shop on consecutive days = \\(\\frac{8}{25}\\)
\niii) If P(E) is the probability of visiting the shop on the same day,
\nthen P(\\(\\overline{\\mathrm{E}}\\)) is the probability of visiting the shop not on the same day.
\ni.e., P(\\(\\overline{\\mathrm{E}}\\)) is the probability of visiting the shop on different days.
\nSuch that P(E) + P(\\(\\overline{\\mathrm{E}}\\)) = 1
\nP(\\(\\overline{\\mathrm{E}}\\)) = 1 – P(E) = 1 – \\(\\frac{1}{5}\\) = \\(\\frac{4}{5}\\)<\/p>\n![\"AP](\"https:\/\/apboardsolutions.in\/wp-content\/uploads\/2020\/12\/AP-Board-Solutions.png\")
<\/p>\n
\nA bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag.
\nAnswer:
\nNumber of red balls in the bag = 5 As the probability of blue balls is double the probability of red balls, we have that number of blue balls is double the number of red balls.
\n\u2234 Blue balls = 5 \u00d7 2 = 10.
\n[!! Let the number of blue balls = x
\nNumber of red balls = 5
\nTotal no. of balls = x + 5
\nTotal outcomes in drawing a ball at random = x + 5
\nNumber of outcomes favourable to red ball = 5
\n\u2234 P(R) = \\(\\frac{5}{x+5}\\)
\nfrom the problem,
\nP(B) = 2 \u00d7 \\(\\frac{5}{x+5}\\) = \\(\\frac{10}{x+5}\\)
\nAlso \\(\\frac{5}{x+5}\\) + \\(\\frac{10}{x+5}\\) = 1
\n[\u2235 P(E) + P(\\(\\overline{\\mathrm{E}}\\)) = 1]
\n\u21d2 \\(\\frac{5+10}{x+5}\\) = 1
\n\u21d2 \\(\\frac{15}{x+5}\\) = 1
\n\u21d2 x + 5 = 15]<\/p>\n
\nA box contains 12 balls out of which x are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball? If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find x.
\nAnswer:
\nNumber of black balls = x
\nTotal number of balls in the box = 12
\nProbability of drawing a black ball = \\(\\frac{\\text { No. of favourable outcomes }}{\\text { No. of total outcomes }}\\) = \\(\\frac{x}{12}\\) …….. (1)
\nWhen 6 more black balls are placed in the box, number of favourable outcomes to black ball becomes = x + 6.
\nTotal number of balls in the box becomes = 12 + 6 = 18.
\nNow the probability of drawing a black ball become = \\(\\frac{x+6}{18}\\) …….. (1)
\nBy Problem,
\n\\(\\frac{x+6}{18}\\) = 2. \\(\\frac{x}{12}\\)
\n\u21d2 \\(\\frac{x+6}{18}\\) = \\(\\frac{x}{6}\\)
\n\u21d2 6(x + 6) = 18(x)
\n\u21d2 6x + 36 = 18x
\n\u21d2 18x – 6x = 36
\n\u21d2 12x = 36
\n\u21d2 x = \\(\\frac{36}{12}\\) = 3
\nCheck:
\n![\"AP](\"https:\/\/apboardsolutions.in\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Solutions-Chapter-13-Probability-Optional-Exercise-2.png\")
\nand hence proved.<\/p>\n<\/p>\n
\nA jar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is 2\/3. Find the number of blue marbles in the jar.
\nAnswer:
\nTotal number of marbles in the jar = 24.
\nLet the number of green marbles = x.
\nThen number of blue marbles = 24 – x.
\nProbability of drawing a green marbles = \\(\\frac{\\text { No. of favourable outcomes }}{\\text { No. of total outcomes }}\\) = \\(\\frac{x}{24}\\)
\nBy problem,
\n\\(\\frac{x}{24}\\) = \\(\\frac{2}{3}\\)
\n\u21d2 3x = 24 \u00d7 2
\nx = \\(\\frac{24 \\times 2}{3}\\) = 16
\n\u2234 Number of green marbles = 16
\nNumber of blue marbles = 24 – 1 = 8
\n\u2234 Probability of picking blue marble = \\(\\frac{8}{24}\\) = \\(\\frac{1}{3}\\)
\n(OR)
\nP(B) = P(E) – P(G) = 1 – \\(\\frac{2}{3}\\) = \\(\\frac{1}{3}\\)
\n[!! P(G) = \\(\\frac{2}{3}\\)
\nP(G) + P(B) = 1
\n\u2234 P(B) = 1 – P(G) = 1 – \\(\\frac{2}{3}\\) = \\(\\frac{1}{3}\\)
\nNumber of blue marbles in the jar = \\(\\frac{1}{3}\\) \u00d7 24 = 8]<\/p>\n","protected":false},"excerpt":{"rendered":"