Revise Reflect Refine
Q1. What is the probability of getting a head when a coin is tossed once? Also find the probability of getting a tail.
Answer:
When a fair coin is tossed once, there are two equally likely outcomes: Head (H) and Tail (T).
P(Head) = Number of favorable outcomes / Total outcomes = 1/2.
P(Tail) = 1/2.
Q2. A bag contains a red ball, a blue ball and a yellow ball, all the balls being of the same size. Kritika takes out a ball from the bag without looking into it. What is the probability that she takes out the (i) yellow ball? (ii) red ball? (iii) blue ball?
Answer:
Total possible outcomes = 3 (red, blue, yellow). All outcomes equally likely.
(i) P(Yellow) = 1/3
(ii) P(Red) = 1/3
(iii) P(Blue) = 1/3
Q3. Suppose we throw a die once. (i) What is the probability of getting a number greater than 4? (ii) What is the probability of getting a number less than or equal to 4?
Answer:
Total outcomes = 6 (1, 2, 3, 4, 5, 6).
(i) Numbers greater than 4: 5, 6. Number of favorable outcomes = 2. P = 2/6 = 1/3.
(ii) Numbers less than or equal to 4: 1, 2, 3, 4. Number of favorable outcomes = 4. P = 4/6 = 2/3.
Note: The events are complementary; P(E) + P(not E) = 1.
Q4. One card is drawn from a well-shuffled deck of 52 cards. Calculate the probability that the card will (i) be an ace, (ii) not be an ace.
Answer:
Total cards = 52.
(i) There are 4 aces. P(ace) = 4/52 = 1/13.
(ii) Number of non-ace cards = 52 - 4 = 48. P(not ace) = 48/52 = 12/13. Alternatively, using complement: P(not ace) = 1 - P(ace) = 1 - 1/13 = 12/13.
Q5. Two players, Sangeeta and Reshma, play a tennis match. It is known that the probability of Sangeeta winning the match is 0.62. What is the probability of Reshma winning the match?
Answer:
Let S be the event that Sangeeta wins, R be the event that Reshma wins. Since only one can win, they are complementary events. P(Reshma wins) = 1 - P(Sangeeta wins) = 1 - 0.62 = 0.38.
Q6. Savita and Hamida are friends. What is the probability that both will have (i) different birthdays? (ii) the same birthday? (ignoring a leap year).
Answer:
Total days in a year = 365. Savita's birthday can be any day. Hamida's birthday can be any of 365 days.
(i) P(different birthdays) = Number of days different from Savita's / 365 = 364/365.
(ii) P(same birthday) = 1 - P(different birthdays) = 1 - 364/365 = 1/365.
Q7. There are 40 students in Class X of a school of whom 25 are girls and 15 are boys. The class teacher has to select one student as a class representative. She writes the name of each student on a separate card, the cards being identical. Then she puts cards in a bag and stirs them thoroughly. She then draws one card from the bag. What is the probability that the name written on the card is the name of (i) a girl? (ii) a boy?
Answer:
Total cards = 40.
(i) Number of girls = 25. P(girl) = 25/40 = 5/8.
(ii) Number of boys = 15. P(boy) = 15/40 = 3/8. Alternatively, P(boy) = 1 - P(girl) = 1 - 5/8 = 3/8.
Q8. A box contains 3 blue, 2 white, and 4 red marbles. If a marble is drawn at random from the box, what is the probability that it will be (i) white? (ii) blue? (iii) red?
Answer:
Total marbles = 3 + 2 + 4 = 9.
(i) P(white) = 2/9.
(ii) P(blue) = 3/9 = 1/3.
(iii) P(red) = 4/9.
Note: Sum of probabilities = 2/9 + 3/9 + 4/9 = 9/9 = 1.
Q9. Harpreet tosses two different coins simultaneously (say, one is of ₹1 and other of ₹2). What is the probability that she gets at least one head?
Answer:
Possible outcomes: (H,H), (H,T), (T,H), (T,T) — 4 equally likely outcomes. 'At least one head' means one or more heads. Favorable outcomes: (H,H), (H,T), (T,H) — 3 outcomes. P(at least one head) = 3/4. Alternatively, P(no head) = P(T,T) = 1/4. P(at least one head) = 1 - 1/4 = 3/4.
Q10. In a musical chair game, the person playing the music has been advised to stop playing the music at any time within a 2-minute period. What is the probability that the music will stop within the first half minute after starting?
Answer:
Assuming the stopping time is uniformly distributed along the 2-minute (120 seconds) interval, the favorable time is the first 30 seconds. Probability = (length of favorable interval) / (total length of interval) = 30 seconds / 120 seconds = 1/4.
EXERCISE 14.1
Q1. Complete the following statements: (i) Probability of an event E + Probability of the event ‘not E’ = ______ . (ii) The probability of an event that cannot happen is ______ . Such an event is called ______ . (iii) The probability of an event that is certain to happen is ______ . Such an event is called ______ . (iv) The sum of the probabilities of all the elementary events of an experiment is ______ . (v) The probability of an event is greater than or equal to ______ and less than or equal to ______ .
Answer:
(i) 1
(ii) 0, impossible event
(iii) 1, sure event or certain event
(iv) 1
(v) 0, 1
Q2. Which of the following experiments have equally likely outcomes? Explain. (i) A driver attempts to start a car. The car starts or does not start. (ii) A player attempts to shoot a basketball. She/he shoots or misses the shot. (iii) A trial is made to answer a true-false question. The answer is right or wrong. (iv) A baby is born. It is a boy or a girl.
Answer:
Equally likely outcomes are those where each outcome has the same chance of occurring.
(i) Not equally likely, because the car starting depends on many factors; usually the chance of starting is higher.
(ii) Not equally likely, because the player's skill affects the outcome; the chance of shooting a basket is not equal to missing.
(iii) Equally likely, because if the answer is guessed randomly, right and wrong are equally likely. However, if the student knows, it may not be equally likely. In a fair trial, it is considered equally likely.
(iv) Equally likely, as the probability of a baby being a boy or a girl is approximately 1/2 (ignoring biological variations).
Q3. Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?
Answer:
Tossing a coin is considered fair because the coin has two sides (head and tail) and both outcomes are equally likely. The probability of getting a head is 1/2 and the probability of getting a tail is 1/2, so no team has an advantage.
Q4. Which of the following cannot be the probability of an event? (A) 2/3 (B) –1.5 (C) 15% (D) 0.7
Answer:
Probability of an event lies between 0 and 1 (inclusive).
(A) 2/3 = 0.666... (valid)
(B) –1.5 is negative, so it cannot be a probability.
(C) 15% = 0.15 (valid)
(D) 0.7 (valid)
Therefore, (B) –1.5 cannot be the probability of an event.
Q5. If P(E) = 0.05, what is the probability of ‘not E’?
Answer:
Using the complement rule: P(not E) = 1 – P(E) = 1 – 0.05 = 0.95.
Q6. A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out (i) an orange flavoured candy? (ii) a lemon flavoured candy?
Answer:
Since the bag contains only lemon flavoured candies:
(i) The probability of taking out an orange flavoured candy is 0 (impossible event).
(ii) The probability of taking out a lemon flavoured candy is 1 (certain event).
Q7. It 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?
Answer:
Let E be the event that 2 students have the same birthday. Then not E is the event that they do not have the same birthday. Given P(not E) = 0.992. Therefore, P(E) = 1 – P(not E) = 1 – 0.992 = 0.008.
Q8. A 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 (i) red ? (ii) not red?
Answer:
Total number of balls = 3 + 5 = 8.
(i) Number of red balls = 3. So, P(red) = 3/8.
(ii) P(not red) = 1 – P(red) = 1 – 3/8 = 5/8. Alternatively, number of black balls = 5, so P(not red) = 5/8.
Q9. A 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 (i) red ? (ii) white ? (iii) not green?
Answer:
Total marbles = 5 + 8 + 4 = 17.
(i) P(red) = 5/17.
(ii) P(white) = 8/17.
(iii) Number of marbles that are not green = 5 + 8 = 13. So, P(not green) = 13/17.
Q10. A piggy bank contains hundred 50p coins, fifty ` 1 coins, twenty ` 2 coins and ten ` 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 (i) will be a 50 p coin ? (ii) will not be a ` 5 coin?
Answer:
Total coins = 100 + 50 + 20 + 10 = 180.
(i) Number of 50p coins = 100. So, P(50p coin) = 100/180 = 5/9.
(ii) Number of coins that are not `5 = 180 – 10 = 170. So, P(not `5 coin) = 170/180 = 17/18.
Q11. Gopi 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 Fig. 14.4). What is the probability that the fish taken out is a male fish?
Answer:
Total fish = 5 + 8 = 13. Number of male fish = 5. So, P(male fish) = 5/13.
Q12. A 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 Fig. 14.5), and these are equally likely outcomes. What is the probability that it will point at (i) 8 ? (ii) an odd number? (iii) a number greater than 2? (iv) a number less than 9?
Answer:
Total possible outcomes = 8.
(i) Only one outcome '8'. So, P(8) = 1/8.
(ii) Odd numbers: 1,3,5,7 → 4 outcomes. So, P(odd) = 4/8 = 1/2.
(iii) Numbers greater than 2: 3,4,5,6,7,8 → 6 outcomes. So, P(greater than 2) = 6/8 = 3/4.
(iv) Numbers less than 9: all numbers 1 to 8 → 8 outcomes. So, P(less than 9) = 8/8 = 1.
Q13. A die is thrown once. Find the probability of getting (i) a prime number; (ii) a number lying between 2 and 6; (iii) an odd number.
Answer:
Total outcomes: 1,2,3,4,5,6 → 6 outcomes.
(i) Prime numbers: 2,3,5 → 3 outcomes. So, P(prime) = 3/6 = 1/2.
(ii) Numbers between 2 and 6 (excluding 2 and 6): 3,4,5 → 3 outcomes. So, P(between 2 and 6) = 3/6 = 1/2.
(iii) Odd numbers: 1,3,5 → 3 outcomes. So, P(odd) = 3/6 = 1/2.
Q14. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting (i) a king of red colour (ii) a face card (iii) a red face card (iv) the jack of hearts (v) a spade (vi) the queen of diamonds
Answer:
Total cards = 52.
(i) Red kings: king of hearts and king of diamonds → 2 cards. So, P(red king) = 2/52 = 1/26.
(ii) Face cards: Jack, Queen, King in each suit → 4 suits × 3 = 12 cards. So, P(face card) = 12/52 = 3/13.
(iii) Red face cards: hearts and diamonds, each suit has 3 face cards → 2×3=6 cards. So, P(red face card) = 6/52 = 3/26.
(iv) Jack of hearts: only 1 card. So, P(jack of hearts) = 1/52.
(v) Spade: 13 cards. So, P(spade) = 13/52 = 1/4.
(vi) Queen of diamonds: only 1 card. So, P(queen of diamonds) = 1/52.
Q15. Five 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. (i) What is the probability that the card is the queen? (ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?
Answer:
Initially, 5 cards.
(i) Only one queen. So, P(queen) = 1/5.
(ii) After queen is drawn and put aside, remaining cards = 4.
(a) There is one ace. So, P(ace) = 1/4.
(b) No queen remains. So, P(queen) = 0/4 = 0.
Q16. 12 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.
Answer:
Total pens = 12 + 132 = 144. Good pens = 132. So, P(good pen) = 132/144 = 11/12.
Q17. (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? (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?
Answer:
(i) Total bulbs = 20, defective = 4. So, P(defective) = 4/20 = 1/5.
(ii) After drawing one non-defective bulb, remaining bulbs = 19. Non-defective bulbs originally = 16, one removed, so left = 15. P(not defective) = 15/19.
Q18. A 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 (i) a two-digit number (ii) a perfect square number (iii) a number divisible by 5.
Answer:
Total discs = 90.
(i) Two-digit numbers: 10 to 90 → that's 90 – 9 = 81 numbers. So, P(two-digit) = 81/90 = 9/10.
(ii) Perfect squares from 1 to 90: 1²=1, 2²=4, 3²=9, 4²=16, 5²=25, 6²=36, 7²=49, 8²=64, 9²=81. That's 9 numbers. So, P(perfect square) = 9/90 = 1/10.
(iii) Numbers divisible by 5: 5,10,15,...,90. Count = 90/5 = 18. So, P(divisible by 5) = 18/90 = 1/5.
Q19. A child has a die whose six faces show the letters as given below: A B C D E A The die is thrown once. What is the probability of getting (i) A? (ii) D?
Answer:
The die has 6 faces: A appears twice, B appears once, C once, D once, E once.
(i) Number of faces with A = 2. So, P(A) = 2/6 = 1/3.
(ii) Number of faces with D = 1. So, P(D) = 1/6.
Q20. Suppose you drop a die at random on the rectangular region shown in Fig. 14.6. What is the probability that it will land inside the circle with diameter 1m?
Answer:
Description: The rectangular region is 3m by 2m, so area = 3×2 = 6 m². The circle has diameter 1m, so radius = 0.5m, area = π×(0.5)² = π×0.25 = 0.25π m². Assuming the die always lands within the rectangle, the probability = (area of circle)/(area of rectangle) = (0.25π)/6 = π/24.
Q21. A 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 (i) She will buy it ? (ii) She will not buy it ?
Answer:
Total pens = 144. Good pens = 144 – 20 = 124.
(i) P(buy) = P(good) = 124/144 = 31/36.
(ii) P(not buy) = P(defective) = 20/144 = 5/36. Alternatively, 1 – 31/36 = 5/36.
Q22. Refer to Example 13. (i) Complete the following table: Event : ‘Sum on 2 dice’ 2 3 4 5 6 7 8 9 10 11 12 Probability 1/36 ? ? ? ? ? ? ? ? ? 1/36 (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 1/11’. Do you agree with this argument? Justify your answer.
Answer:
(i) From Example 13, total outcomes = 36.
Sum 2: (1,1) → 1 outcome → 1/36
Sum 3: (1,2),(2,1) → 2 outcomes → 2/36 = 1/18
Sum 4: (1,3),(2,2),(3,1) → 3 outcomes → 3/36 = 1/12
Sum 5: (1,4),(2,3),(3,2),(4,1) → 4 outcomes → 4/36 = 1/9
Sum 6: (1,5),(2,4),(3,3),(4,2),(5,1) → 5 outcomes → 5/36
Sum 7: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) → 6 outcomes → 6/36 = 1/6
Sum 8: (2,6),(3,5),(4,4),(5,3),(6,2) → 5 outcomes → 5/36
Sum 9: (3,6),(4,5),(5,4),(6,3) → 4 outcomes → 4/36 = 1/9
Sum 10: (4,6),(5,5),(6,4) → 3 outcomes → 3/36 = 1/12
Sum 11: (5,6),(6,5) → 2 outcomes → 2/36 = 1/18
Sum 12: (6,6) → 1 outcome → 1/36
Table completed.
(ii) No, I do not agree. The outcomes are not equally likely because the number of ways to get each sum is different. For example, sum 7 has 6 outcomes while sum 2 has only 1 outcome. So probability of each sum is not 1/11; it depends on the number of favourable outcomes.
Q23. A 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.
Answer:
Total outcomes when tossing a coin 3 times: 2³ = 8. Winning outcomes: all heads (HHH) or all tails (TTT) → 2 outcomes. Losing outcomes = 8 – 2 = 6. Probability of losing = 6/8 = 3/4.
Q24. A die is thrown twice. What is the probability that (i) 5 will not come up either time? (ii) 5 will come up at least once? [Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]
Answer:
Total outcomes when throwing a die twice: 6×6 = 36.
(i) Outcomes where 5 does not appear at all: first die can be 1,2,3,4,6 (5 possibilities), second die also 5 possibilities → 5×5=25. So, P(no 5) = 25/36.
(ii) P(at least one 5) = 1 – P(no 5) = 1 – 25/36 = 11/36.
Q25. Which of the following arguments are correct and which are not correct? Give reasons for your answer. (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 1/3. (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 1/2.
Answer:
(i) Not correct. The outcomes are not equally likely. When two coins are tossed, the possible outcomes are HH, HT, TH, TT. 'One of each' corresponds to HT and TH, which are two outcomes, not one. So the probabilities are: P(HH)=1/4, P(TT)=1/4, P(one head one tail)=1/2.
(ii) Correct. The possible outcomes are 1,2,3,4,5,6. Odd numbers: 1,3,5 (3 outcomes), even numbers: 2,4,6 (3 outcomes). So P(odd)=3/6=1/2.