{"id":3043,"date":"2021-02-15T17:49:30","date_gmt":"2021-02-15T12:19:30","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=3043"},"modified":"2021-09-15T12:09:22","modified_gmt":"2021-09-15T06:39:22","slug":"ap-ssc-10th-class-maths-solutions-chapter-13-ex-13-1","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-ssc-10th-class-maths-solutions-chapter-13-ex-13-1\/","title":{"rendered":"AP SSC 10th Class Maths Solutions Chapter 13 Probability Ex 13.1"},"content":{"rendered":"

AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions<\/a> Chapter 13 Probability Ex 13.1 Textbook Questions and Answers.<\/p>\n

AP State Syllabus SSC 10th Class Maths Solutions 13th Lesson Probability Exercise 13.1<\/h2>\n

10th Class Maths 13th Lesson Probability Ex 13.1 Textbook Questions and Answers<\/h3>\n

Question 1.
\nComplete the following statements:
\ni) Probability of an event E + Probability of the event ‘not E’ = 1\u00a0<\/span>.
\nii) The probability of an event that cannot happen is zero<\/span>.
\nSuch an event is called an impossible event<\/span>.
\niii) The probability of an event that is certain to happen is\u00a0 1\u00a0<\/span> such an event is called sure or certain event<\/span>.
\niv) The sum of the probabilities of all the elementary events of an experiment is 1\u00a0<\/span>.
\nv) The probability of an event is greater than or equal to zero<\/span> and less than or equal to 1\u00a0<\/span>.<\/p>\n

\"AP<\/p>\n

Question 2.
\nWhich of the following experiments have equally likely outcomes? Explain.
\ni) A driver attempts to start a car. The car starts or does not start.
\nAnswer:
\nEqually likely. Since both have the same probability \\(\\frac{1}{2}\\).<\/p>\n

ii) A player attempts to shoot a basket-ball. She\/he shoots or misses the shot.
\nAnswer:
\nEqually likely. Since both have the same probability \\(\\frac{1}{2}\\).<\/p>\n

iii) A trial is made to answer a true-false question. The answer is right or wrong.
\nEqually likely. Since both have the same probability \\(\\frac{1}{2}\\).<\/p>\n

iv) A baby is born. It is a boy or a girl.
\nEqually likely. Since both the events have the same probability \\(\\frac{1}{2}\\).<\/p>\n

Question 3.
\nIf P(E) = 0.05, what is the probability of not E?
\nAnswer:
\nGiven: P(E) = 0.05
\nHence, P(E) + P(\\(\\overline{\\mathrm{E}}\\)) = 1, where P(\\(\\overline{\\mathrm{E}}\\)) is the probability of \u2018not E\u2019
\n0.05 + P(\\(\\overline{\\mathrm{E}}\\)) = 1
\n\u2234 P(\\(\\overline{\\mathrm{E}}\\)) = 1 -0.05 = 0.95.<\/p>\n

\"AP<\/p>\n

Question 4.
\nA bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out
\ni) an orange flavoured candy?
\nii) a lemon flavoured candy?
\nAnswer:
\nBag contains only lemon flavoured candies.
\ni) Taking an orange flavoured candy is an impossible event and hence the probability is zero.
\nii) Also taking a lemon flavoured candy is a sure event and hence its probability is 1.<\/p>\n

Question 5.
\nRahim removes all the hearts from the cards. What is the probability of
\ni. Picking out an ace from the remaining pack.
\nii. Picking out a diamond.
\niii. Picking out a card that is not a heart.
\niv. Picking out the Ace of hearts.
\nAnswer:
\nTotal number of cards in the deck = 52.
\nTotal number of hearts in the deck of cards =13.
\nWhen Hearts are removed, remaining cards = 52 – 13 = 39.
\ni)Picking out an Ace:
\nNumber of outcomes favourable to Ace = 3 [\u2235 \u2666 A, \u2665 A, \u2660 A, \u2663 A]
\nTotal number of possible outcomes from the remaining cards = 39
\n– after removing Hearts.
\nProbability = P(A)
\n= \\(\\frac{\\text { No. of favourable outcomes }}{\\text { Total no. of outcomes }}\\)
\n= \\(\\frac{3}{39}\\) = \\(\\frac{1}{13}\\)<\/p>\n

\"AP<\/p>\n

ii) Picking out a diamond:
\nNumber of favourable outcomes to diamonds (\u2666) = 13
\nTotal number of possible outcomes = 39
\n\u2234 p(\u2666) = \\(\\frac{13}{39}\\) = \\(\\frac{1}{3}\\)<\/p>\n

iii) Picking out a card that is \u2018not a heart\u2019:
\nAs all hearts are removed, the remain-ing cards are all non-heart cards. So the picked card will be definitely a non-heart card. So this is a sure event.
\nHence its probability is one
\nP(E) = \\(\\frac{39}{39}\\) = 1<\/p>\n

iv) Picking out the Ace of Hearts:
\na) As all the heart cards are removed the left over cards will have three suits (i) spades, (ii), clubs, (iii) dia\u00acmonds of each 13.
\nHence total outcomes = 3 \u00d7 13 = 39 But among them there is no Ace of heart. So number of favourable outcomes for picking Ace of heart = zero.
\n\u2234 Probability P(E) = \\(\\frac{0}{39}\\) = 0
\nSo it is an impossible event.<\/p>\n

b) If picking from the rest of the cards, it is an impossible event and hence probability is zero.
\nIt is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992.<\/p>\n

\"AP<\/p>\n

Question 6.
\nWhat is the probability that the 2 students have the same birthday?
\nAnswer:
\nLet P(E) = The probability that two students not having the same birthday = 0.992
\nThen P(\\(\\overline{\\mathrm{E}}\\)) = The complementary event of E, i.e., two students having the same birthday Also, P(E) + p(\\(\\overline{\\mathrm{E}}\\)) = 1
\n\u2234 The probability that two students have the same birthday P(\\(\\overline{\\mathrm{E}}\\)) = 1 – P(E)
\n= 1 – 0.992 = 0.008<\/p>\n

Question 7.
\nA die is thrown once. Find the probability of getting
\n(i) a prime number;
\n(ii) a number lying between 2 and 6;
\n(iii) an odd number.
\nAnswer:
\ni) When a die is thrown for one time, total number of outcomes = 6
\nNo. of outcomes favourable to a prime number (2, 3, 5) = 3
\n\u2234 Probability of getting a prime = \\(\\frac{\\text { No. of favourable outcomes }}{\\text { Total no. of outcomes }}\\)
\n= \\(\\frac{3}{6}\\) = \\(\\frac{1}{2}\\)<\/p>\n

ii) No. of outcomes favourable to a number lying between 2 and 6 (3, 4, 5) = 3
\n\u2234 Probability of getting a number between 2 and 6
\n= \\(\\frac{\\text { No. of favourable outcomes }}{\\text { Total no. of outcomes }}\\)
\n= \\(\\frac{3}{6}\\) = \\(\\frac{1}{2}\\)<\/p>\n

iii) Number of outcomes favourable to an odd number (1, 3, 5) = 3
\n\u2234 Probability of getting an odd number P(odd)
\n= \\(\\frac{\\text { No. of favourable outcomes }}{\\text { Total no. of outcomes }}\\)
\n= \\(\\frac{3}{6}\\) = \\(\\frac{1}{2}\\)<\/p>\n

\"AP<\/p>\n

Question 8.
\nWhat is the probability of drawing out a red king from a deck of cards?
\nAnswer: Number of favourable outcomes to red king (\u2665 K, \u2666 K) = 2.
\nNumber of total outcomes = 52
\n(\u2235 Number of cards in a deck of cards = 52)
\n\u2234 Probability of getting a red king P (Red king)
\n= \\(\\frac{\\text { No. of favourable outcomes }}{\\text { Total no. of outcomes }}\\)
\n= \\(\\frac{2}{52}\\) = \\(\\frac{1}{26}\\)<\/p>\n

Question 9.
\nMake 5 more problems getting probability using dice, cards or birthdays and discuss with friends and teacher about their solutions.
\nAnswer:
\nClass-room activity.<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions Chapter 13 Probability Ex 13.1 Textbook Questions and Answers. AP State Syllabus SSC 10th Class Maths Solutions 13th Lesson Probability Exercise 13.1 10th Class Maths 13th Lesson Probability Ex 13.1 Textbook Questions and Answers Question 1. Complete the following statements: i) Probability of an event …<\/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.guru\/wp-json\/wp\/v2\/posts\/3043"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=3043"}],"version-history":[{"count":0,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/3043\/revisions"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=3043"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=3043"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=3043"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}