{"id":29189,"date":"2021-07-16T17:28:20","date_gmt":"2021-07-16T11:58:20","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=29189"},"modified":"2021-07-16T17:28:20","modified_gmt":"2021-07-16T11:58:20","slug":"ncert-solutions-for-class-10-maths-chapter-15-ex-15-1","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-10-maths-chapter-15-ex-15-1\/","title":{"rendered":"NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.1"},"content":{"rendered":"

These NCERT Solutions for Class 10 Maths<\/a> Chapter 15 Probability Ex 15.1 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 15 Probability Exercise 15.1<\/h2>\n

\"NCERT<\/p>\n

Question 1.
\nComplete the following statements:
\n(i) Probability of an event E + Probability of the event \u2018not E\u2019 = ………
\n(ii) The probability of an event that cannot happen is ……… Such an event is called ………
\n(iii) The probability of an event that is certain to happen is ………. Such an event is called ………
\n(iv) The sum of the probabilities of all the elementary events of an experiment is ………..
\n(v) The probability of an event is greater than or equal to …………. and less than or equal to ………..
\nSolution:
\n(i) Probability of an event E + Probability of the event \u2018not E\u2019 = 1.
\n(ii) The probability of an event that cannot happen is 0. Such an event is called impossible event.
\n(iii) The probability of an event that is certain to happen is 1. Such an event is called sure event.
\n(iv) The sum of the probabilities of all the elementary events of an experiment is 1.
\n(v) The probability of an event is greater than or equal to 0 and less than or equal to 1.<\/p>\n

Question 2.
\nWhich of the following experiments have equally likely outcomes? Explain.
\n(i) A driver attempts to start a car. The car starts or does not start.
\n(ii) A player attempts to shoot a basketball. She\/he shoots or misses the shot.
\n(iii) A trial is made to answer a true-false question. The answer is right or wrong.
\n(iv) A baby is born. It is a boy or a girl.
\nSolution:
\n(i) The outcome is not equally likely because the car starts normally only when there is some defect, the car does not start.
\n(ii) The outcome is not equally likely because the outcome depends on the training of the player.
\n(iii) The outcome in the trial of true-false question is, either true or false. Hence, the two outcomes are equally likely.
\n(iv) A baby can be either a boy or a girl and both the outcomes have equally likely chances.<\/p>\n

Question 3.
\nWhy is tossing a coin considered to be a fair way of deciding which team should get the bail at the beginning of a football game?
\nSolution:
\nWhen we toss a coin, the outcomes head and tail are equally likely. So, the result of an individual coin toss is completely unpredictable.<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nWhich of the following cannot be the probability of an event?
\n(A) \\(\\frac { 2 }{ 3 }\\)
\n(B) -1.5
\n(C) 15%
\n(D) 0.7
\nSolution:
\nWe know that probability of an event cannot be less than 0 and greater than 1.
\nCorrect option is (B).<\/p>\n

Question 5.
\nIf P (E) = 0.05, what is the probability of \u2018not E\u2019?
\nSolution:
\nWe have, P (E) + P (not E) = 1
\nGiven: P(E) = 0.05
\nP (not E) = 1 – 0.05 = 0.95<\/p>\n

Question 6.
\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
\n(i) an orange flavoured candy?
\n(ii) a lemon flavoured candy?
\nSolution:
\n(i) A bag contains only lemon flavoured candies.
\nP (an orange flavoured candy) = 0
\n(ii) P (a lemon flavoured candy) = 1<\/p>\n

Question 7.
\nIt is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?
\nSolution:
\nWe have, P (E) + P (not E) = 1
\n\u21d2 P (E) + 0.992 = 1
\n\u21d2 P (E) = 1 – 0.992 = 0.008<\/p>\n

\"NCERT<\/p>\n

Question 8.
\nA bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is
\n(i) red?
\n(ii) not red?
\nSolution:
\n(i) Number of red balls = 3
\nNumber of black balls = 5
\nTotal number of balls = 3 + 5 = 8
\n\u2234 Probability (getting red ball) = \\(\\frac { 3 }{ 8 }\\)<\/p>\n

(ii) P (not red) = 1 – P(E)
\n= 1 – \\(\\frac { 3 }{ 8 }\\)
\n= \\(\\frac { 8 – 3 }{ 8 }\\)
\n= \\(\\frac { 5 }{ 8 }\\)<\/p>\n

Question 9.
\nA box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be
\n(i) red?
\n(ii) white?
\n(iii) not green?
\nSolution:
\nTotal number of marbles = 5 + 8 + 4 = 17
\n(i) P (red marble) = \\(\\frac { 5 }{ 17 }\\)
\n(ii) P (white marble) = \\(\\frac { 8 }{ 17 }\\)
\n(iii) P (not a green marble) = \\(\\frac { 13 }{ 17 }\\)<\/p>\n

Question 10.
\nA piggy bank contains hundred 50 p coins, fifty \u20b9 1 coins, twenty \u20b9 2 coins and ten \u20b9 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin
\n(i) will be a 50 p coin?
\n(ii) will not be a \u20b9 5 coin?
\nSolution:
\n(i) The total coin contain in piggy bank = 100 + 50 + 20 + 10 = 180
\nThe number of 50 p coin = 100
\nP (getting a 50 p coin) = \\(\\frac { 100 }{ 180 }\\)
\n= \\(\\frac { 10 }{ 18 }\\)
\n= \\(\\frac { 5 }{ 9 }\\)<\/p>\n

(ii) Let F be the event “will not be a \u20b9 coin”.
\nNumber of outcomes favourable to the event F = 180 – 10 = 170 [ \u2235 There are ten \u20b9 5 coins]
\n\u2234 The number of possible outcomes = 180
\n\u2234 P(F) = \\(\\frac { 170 }{ 180 }\\) = \\(\\frac { 17 }{ 18 }\\).<\/p>\n

\"NCERT<\/p>\n

Question 11.
\nGopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish (see figure). What is the probability that the fish taken out is a male fish?
\nSolution:
\nNumber of male fish = 5
\nNumber of female fish = 8
\nTotal number of fish = 5 + 8 = 13
\nP (a male fish) = \\(\\frac { 5 }{ 13 }\\)<\/p>\n

Question 12.
\nA game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see figure.), and these are equally likely outcomes. What is the probability that it will point at
\n(i) 8?
\n(ii) an odd number?
\n(iii) a number greater than 2?
\n(iv) a number less than 9?
\n\"NCERT
\nSolution:
\n(i) P (getting 8) = \\(\\frac { 1 }{ 8 }\\)
\n(ii) P (an odd number) = \\(\\frac { 4 }{ 8 }\\) = \\(\\frac { 1 }{ 2 }\\) ( odd numbers are 1, 3, 5, 7)
\n(iii) P (a number greater than 2) = \\(\\frac { 6 }{ 8 }\\) = \\(\\frac { 3 }{ 4 }\\)
\n(iv) P (a number less than 9) = \\(\\frac { 8 }{ 8 }\\) = 1<\/p>\n

Question 13.
\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(ill) an odd number
\nSolution:
\n(i) Prime numbers on a die = 2, 3, 5
\nP (a prime number) = \\(\\frac { 3 }{ 6 }\\) = \\(\\frac { 1 }{ 2 }\\)<\/p>\n

(ii) Number lying between 2 and 6 = 3, 4, 5
\nP(a number lying between 2 and 6) = \\(\\frac { 3 }{ 6 }\\) = \\(\\frac { 1 }{ 2 }\\)<\/p>\n

(iii) Odd numbers = 1, 3, 5
\nP (an odd number) = \\(\\frac { 3 }{ 6 }\\) = \\(\\frac { 1 }{ 2 }\\)<\/p>\n

\"NCERT<\/p>\n

Question 14.
\nOne card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
\n(i) a king of red colour
\n(ii) a face card
\n(iii) a red face card
\n(iv) the jack of hearts
\n(v) a spade
\n(vi) the queen of diamonds
\nSolution:
\nNumber of cards in a well-shuffled deck = 52.
\n(i) P (a king of red colour) = \\(\\frac { 2 }{ 52 }\\) = \\(\\frac { 1 }{ 26 }\\)
\n(ii) P (a face card) = \\(\\frac { 12 }{ 52 }\\) = \\(\\frac { 3 }{ 13 }\\)
\n(iii) P (a red face card) = \\(\\frac { 6 }{ 52 }\\) = \\(\\frac { 3 }{ 26 }\\)
\n(iv) P (the jack of hearts) = \\(\\frac { 1 }{ 52 }\\)
\n(v) P(a spade) = \\(\\frac { 13 }{ 52 }\\) = \\(\\frac { 1 }{ 4 }\\)
\n(vi) P (the queen of diamonds) = \\(\\frac { 1 }{ 52 }\\)<\/p>\n

Question 15.
\nFive cards – the ten, jack, queen, king and ace of diamonds, are well shuffled with their face downwards. One card is then picked up at random.
\n(i) What is the probability that the card is the queen?
\n(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is
\n(a) an ace?
\n(b) a queen?
\nSolution:
\nOut of 5 cards there is only one queen.
\n(i) P (getting queen) = \\(\\frac { 1 }{ 5 }\\) [when queen is drawn, four cards are left]
\n(ii) (a) P (an ace) = \\(\\frac { 1 }{ 4 }\\)
\n(b) P (a queen) = \\(\\frac { 0 }{ 4 }\\) = 0<\/p>\n

Question 16.
\n12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
\nSolution:
\nNumber of defective pens = 12
\nNumber of good pens = 132
\nTotal number of pens = 12 + 132 = 144
\nP (the pen is good one) = \\(\\frac { 132 }{ 144 }\\) = \\(\\frac { 11 }{ 12 }\\)<\/p>\n

\"NCERT<\/p>\n

Question 17.
\n(i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?
\n(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?
\nSolution:
\n(i) Total number of bulbs = 20
\nNumber of defective bulbs = 4
\nP (bulb drawn is defective) = \\(\\frac { 4 }{ 20 }\\) = \\(\\frac { 1 }{ 5 }\\)
\n(ii) Remaining bulbs = 19
\nP (bulb drawn is not defective) = \\(\\frac { 15 }{ 19 }\\)<\/p>\n

Question 18.
\nA box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears
\n(i) a two digit number.
\n(ii) a perfect square number.
\n(iii) a number divisible by 5.
\nSolution:
\nTotal numbers of discs = 90
\n(i) P (a two digit number) = \\(\\frac { 81 }{ 90 }\\) = \\(\\frac { 9 }{ 10 }\\)<\/p>\n

(ii) Here, perfect square numbers are 1, 4, 9, 16, 25, 36, 49, 64, 81
\nP (getting a perfect square number) = \\(\\frac { 9 }{ 90 }\\) = \\(\\frac { 1 }{ 10 }\\)<\/p>\n

(iii) Numbers divisible by 5 are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90
\nP (getting a number divisible by 5) = \\(\\frac { 18 }{ 90 }\\) = \\(\\frac { 1 }{ 5 }\\)<\/p>\n

\"NCERT<\/p>\n

Question 19.
\nA child has a die whose six faces show the letters as given below:
\n\"NCERT
\nThe die is thrown once. What is the probability of getting
\n(i) A?
\n(ii) D?
\nSolution:
\n(i) P (getting A) = \\(\\frac { 2 }{ 6 }\\) = \\(\\frac { 1 }{ 3 }\\)
\n(ii) P (getting D) = \\(\\frac { 1 }{ 6 }\\)<\/p>\n

Question 20.
\nSuppose you drop a die at random on the rectangular region shown in figure. What is the probability that it will land inside the circle with diameter 1 m?
\n\"NCERT
\nSolution:
\nArea of rectangular region = 1 x b
\n= 3 x 2 = 6m\u00b2
\nArea of circle whose diameter is 1 = \u03c0r\u00b2
\n\"NCERT<\/p>\n

Question 21.
\nA lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that
\n(i) she will buy it?
\n(ii) she will not buy it?
\nSolution:
\nTotal number of ball pens = 144
\nNumber of defective pens = 20
\nNumber of good pens = 144 – 20 = 124
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 22.
\nTwo dice, one blue and one grey, are thrown at the same time. Now
\n(i) Complete the following table:
\n\"NCERT
\n(ii) A student argues that-there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability \\(\\frac { 1 }{ 11 }\\). Do you agree with this argument? Justify your answer.
\nSolution:
\n(i) Total number of possible outcomes = 36
\n(1, 2) and (2, 1) are the favourable events of getting the sum 3.
\nP(sum 3) = \\(\\frac { 2 }{ 36 }\\) = \\(\\frac { 1 }{ 18 }\\)
\n(1, 3) , (2, 2) and (3, 1) are the favourable events of getting the sum 4.
\nP(sum 4) = \\(\\frac { 3 }{ 36 }\\) = \\(\\frac { 1 }{ 12 }\\)
\n(1, 4) , (2, 3), (3, 2) and (4, 1) are the favourable events of getting the sum 5.
\nP(sum 5) = \\(\\frac { 4 }{ 36 }\\) = \\(\\frac { 1 }{ 9 }\\)
\n(1, 5) , (2, 4), (3, 3), (4, 2) and (5, 1) are the favourable events of getting the sum 6.
\nP (sum 6) = \\(\\frac { 5 }{ 36 }\\)
\n(1, 6) , (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1) are the favourable events of getting the sum 7.
\nP(sum 7) = \\(\\frac { 6 }{ 36 }\\) = \\(\\frac { 1 }{ 6 }\\)
\n(3, 6) , (4, 5), (5, 4) and (6, 3) are the favourable events of getting the sum 9.
\nP(sum 9) = \\(\\frac { 4 }{ 36 }\\) = \\(\\frac { 1 }{ 9 }\\)
\n(4, 6) , (5, 5) and (6, 4) are the favourable events of getting the sum 10.
\nP(sum 10 = \\(\\frac { 3 }{ 36 }\\) = \\(\\frac { 1 }{ 12 }\\)
\n(5,6) and (6,5) are the favourable events of getting the sum 11.
\nP(sum 11) = \\(\\frac { 2 }{ 36 }\\) = \\(\\frac { 1 }{ 18 }\\)<\/p>\n

(ii) No, because the outcomes as 11 different sum are not equally likely.<\/p>\n

Question 23.
\nA game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result, i.e. three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.
\nSolution:
\nPossible outcomes are
\nHHH, TTT, HHT, HTH, THH, TTH, THT, HTT = 8
\nP (win the game) = \\(\\frac { 2 }{ 8 }\\) = \\(\\frac { 1 }{ 4 }\\)
\nP (lose the game) = 1 – \\(\\frac { 1 }{ 4 }\\) = \\(\\frac { 3 }{ 4 }\\)<\/p>\n

Question 24.
\nA die is thrown twice. What is the probability that
\n(i) 5 will not come up either time?
\n(ii) 5 will come up at least once?
\n[Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment.]
\nSolution:
\nTotal outcomes = 36
\nNumber of outcomes in favour of 5 is (1, 5) (2, 5) (3, 5) (4, 5) (5, 5) (6, 5) (5, 1) (5, 2) (5, 3) (5, 4) (5, 6) = 11
\n(i) P (5 will not come up either time) = \\(\\frac { 25 }{ 36 }\\)
\n(ii) P (5 will come up at least once) = \\(\\frac { 11 }{ 36 }\\)<\/p>\n

\"NCERT<\/p>\n

Question 25.
\nWhich of the following arguments are correct and which are not correct? Give reasons for your answer.
\n(i) If two coins are tossed simultaneously there are three possible outcomes- two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is \\(\\frac { 1 }{ 3 }\\).
\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:
\n(i) Argument is incorrect.
\nThe possible outcomes are (HH), (HT), (TH), (TT)
\nP(HH) = \\(\\frac { 1 }{ 4 }\\)
\nP(TT) = \\(\\frac { 1 }{ 4 }\\)
\nP(HT or TH) = \\(\\frac { 2 }{ 4 }\\) = \\(\\frac { 1 }{ 2 }\\)<\/p>\n

(ii) Argument is correct.
\nPossible outcomes = 1, 2, 3, 4, 5, 6
\nOdd numbers are = 1, 3, 5
\nP (an odd number) = \\(\\frac { 3 }{ 6 }\\) = \\(\\frac { 1 }{ 2 }\\).<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.1 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 10 Maths Chapter 15 Probability Exercise 15.1 Question 1. Complete the following statements: (i) Probability of an event E + Probability of the event \u2018not E\u2019 = ……… …<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.1<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[2],"tags":[],"yoast_head":"\nNCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.1 - NCERT Solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-10-maths-chapter-15-ex-15-1\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.1 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.1 Questions and Answers are prepared by our highly skilled subject experts. 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