PROBABILITY - Q&A
EXERCISE 14.1
1. Complete the following statements:
(i) Probability of an event E + Probability of the event 'not E' = __________.
Answer: 1
Explanation: The sum of the probability of an event happening and the probability of it not happening is always 1.
(ii) The probability of an event that cannot happen is __________. Such an event is called __________.
Answer: 0, impossible event
Explanation: An event that has no chance of occurring has a probability of 0.
(iii) The probability of an event that is certain to happen is __________. Such an event is called __________.
Answer: 1, sure or certain event
Explanation: An event that will definitely occur has a probability of 1.
(iv) The sum of the probabilities of all the elementary events of an experiment is __________.
Answer: 1
Explanation: The sum of probabilities of all possible outcomes (elementary events) of an experiment is 1.
(v) The probability of an event is greater than or equal to __________ and less than or equal to __________.
Answer: 0, 1
Explanation: Probability values always range from 0 to 1, inclusive.
2. 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.
Answer: No, they are not equally likely.
Explanation: The car starting depends on various mechanical factors (fuel, battery, engine condition). It is not a random chance event like tossing a coin.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
Answer: No, they are not equally likely.
Explanation: The outcome depends on the player's skill and ability, not just chance. A good player is more likely to score than miss.
(iii) A trial is made to answer a true-false question. The answer is right or wrong.
Answer: Yes, they are equally likely.
Explanation: Assuming the answer is guessed randomly without knowledge, there are only two possibilities (True or False), and both have an equal chance of being selected.
(iv) A baby is born. It is a boy or a girl.
Answer: Yes, they are equally likely.
Explanation: There are two possible outcomes, boy or girl, and theoretically, the probability of each is 1/2.
3. 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 distinct sides, Head and Tail. Assuming the coin is unbiased, the probability of getting a Head is 1/2 and the probability of getting a Tail is 1/2. Since the outcomes are equally likely, neither team has an advantage, making the decision fair.
4. Which of the following cannot be the probability of an event?
(A) 2/3
(B) -1.5
(C) 15%
(D) 0.7
Answer: (B) -1.5
Explanation: The probability of an event must lie between 0 and 1 (inclusive). It cannot be negative.
5. If P(E) = 0.05, what is the probability of ‘not E’?
Answer:
We know that P(E) + P(not E) = 1.
Given P(E) = 0.05.
P(not E) = 1 - P(E)
P(not E) = 1 - 0.05
P(not E) = 0.95.
6. 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?
Answer: 0
Explanation: The bag contains only lemon flavoured candies. Taking out an orange flavoured candy is an impossible event.
(ii) a lemon flavoured candy?
Answer: 1
Explanation: Since all candies are lemon flavoured, taking out a lemon flavoured candy is a sure (certain) event.
7. 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.
Let 'not E' be the event that 2 students do not have the same birthday.
P(not E) = 0.992.
We know that P(E) + P(not E) = 1.
P(E) = 1 - P(not E)
P(E) = 1 - 0.992
P(E) = 0.008.
The probability that the 2 students have the same birthday is 0.008.
8. 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 (red) + 5 (black) = 8.
Total possible outcomes = 8.
(i) Probability of getting a red ball:
Number of favorable outcomes (red balls) = 3.
P(red) = 3/8.
(ii) Probability of getting not a red ball:
P(not red) = 1 - P(red)
P(not red) = 1 - 3/8 = (8 - 3)/8 = 5/8.
(Alternatively, 'not red' means black, and there are 5 black balls, so 5/8).
9. 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 number of marbles = 5 (red) + 8 (white) + 4 (green) = 17.
Total possible outcomes = 17.
(i) Probability of red:
Number of red marbles = 5.
P(red) = 5/17.
(ii) Probability of white:
Number of white marbles = 8.
P(white) = 8/17.
(iii) Probability of not green:
Number of green marbles = 4.
P(green) = 4/17.
P(not green) = 1 - P(green) = 1 - 4/17 = 13/17.
(Alternatively, 'not green' means red or white: 5 + 8 = 13 marbles. So, 13/17).
10. 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 number of coins = 100 (50p) + 50 (₹1) + 20 (₹2) + 10 (₹5) = 180.
Total possible outcomes = 180.
(i) Probability that the coin will be a 50p coin:
Number of 50p coins = 100.
P(50p) = 100/180 = 10/18 = 5/9.
(ii) Probability that the coin will not be a ₹ 5 coin:
Number of ₹ 5 coins = 10.
P(₹ 5) = 10/180 = 1/18.
P(not ₹ 5) = 1 - P(₹ 5) = 1 - 1/18 = 17/18.
(Alternatively, count of coins not ₹ 5 = 100 + 50 + 20 = 170. So, 170/180 = 17/18).
11. 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:
Number of male fish = 5.
Number of female fish = 8.
Total number of fish = 5 + 8 = 13.
Total possible outcomes = 13.
Probability of taking out a male fish:
P(male) = (Number of male fish) / (Total number of fish) = 5/13.
12. 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?
Answer:
Total possible outcomes = 8 (numbers 1 to 8).
Number of outcomes pointing at 8 = 1.
P(8) = 1/8.
(ii) an odd number?
Answer:
Odd numbers on the wheel are 1, 3, 5, 7.
Number of favorable outcomes = 4.
P(odd) = 4/8 = 1/2.
(iii) a number greater than 2?
Answer:
Numbers greater than 2 are 3, 4, 5, 6, 7, 8.
Number of favorable outcomes = 6.
P(number > 2) = 6/8 = 3/4.
(iv) a number less than 9?
Answer:
Numbers less than 9 are 1, 2, 3, 4, 5, 6, 7, 8 (all numbers).
Number of favorable outcomes = 8.
P(number < 9) = 8/8 = 1.
13. A die is thrown once. Find the probability of getting
(i) a prime number;
Answer:
Total outcomes on a die = {1, 2, 3, 4, 5, 6}. Total = 6.
Prime numbers are 2, 3, 5.
Number of favorable outcomes = 3.
P(prime number) = 3/6 = 1/2.
(ii) a number lying between 2 and 6;
Answer:
Numbers between 2 and 6 are 3, 4, 5.
Number of favorable outcomes = 3.
P(number between 2 and 6) = 3/6 = 1/2.
(iii) an odd number.
Answer:
Odd numbers are 1, 3, 5.
Number of favorable outcomes = 3.
P(odd number) = 3/6 = 1/2.
14. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
Total number of cards = 52.
(i) a king of red colour
Red kings are King of Hearts and King of Diamonds. Total = 2.
P(red king) = 2/52 = 1/26.
(ii) a face card
Face cards are Kings, Queens, and Jacks. There are 3 per suit and 4 suits (3 × 4 = 12).
P(face card) = 12/52 = 3/13.
(iii) a red face card
Red face cards are in Hearts and Diamonds (2 suits). 3 face cards per suit × 2 suits = 6.
P(red face card) = 6/52 = 3/26.
(iv) the jack of hearts
There is only 1 Jack of Hearts.
P(jack of hearts) = 1/52.
(v) a spade
There are 13 spades in a deck.
P(spade) = 13/52 = 1/4.
(vi) the queen of diamonds
There is only 1 Queen of Diamonds.
P(queen of diamonds) = 1/52.
15. 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?
Answer:
Total number of cards = 5.
Outcomes: Ten, Jack, Queen, King, Ace.
Number of Queens = 1.
P(Queen) = 1/5.
(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:
If the queen is put aside, total cards remaining = 4 (Ten, Jack, King, Ace).
(a) an ace?
Number of Aces = 1.
P(Ace) = 1/4.
(b) a queen?
Number of Queens remaining = 0.
P(Queen) = 0/4 = 0.
16. 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:
Number of defective pens = 12.
Number of good pens = 132.
Total number of pens = 12 + 132 = 144.
Favorable outcomes (good pens) = 132.
P(good pen) = 132/144 = 11/12.
17. (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?
Answer:
Total bulbs = 20.
Defective bulbs = 4.
P(defective) = 4/20 = 1/5.
(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:
If the first bulb drawn is not defective (good), it is removed.
Total bulbs remaining = 20 - 1 = 19.
Total defective bulbs = 4 (since the drawn one was good).
Total good (not defective) bulbs remaining = 16 - 1 = 15.
P(not defective) = 15/19.
18. 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) a two-digit number:
Two-digit numbers are from 10 to 90.
Number of favorable outcomes = 90 - 9 = 81.
P(two-digit number) = 81/90 = 9/10.
(ii) a perfect square number:
Perfect squares between 1 and 90 are: 1, 4, 9, 16, 25, 36, 49, 64, 81.
Number of favorable outcomes = 9.
P(perfect square) = 9/90 = 1/10.
(iii) a number divisible by 5:
Numbers divisible by 5 are: 5, 10, 15, ..., 90.
Number of favorable outcomes = 90/5 = 18.
P(number divisible by 5) = 18/90 = 1/5.
19. 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:
Total faces = 6.
(i) getting A:
Letter 'A' appears on 2 faces.
P(A) = 2/6 = 1/3.
(ii) getting D:
Letter 'D' appears on 1 face.
P(D) = 1/6.
20. 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:
Length of rectangle (l) = 3 m.
Breadth of rectangle (b) = 2 m.
Area of rectangle = l × b = 3 × 2 = 6 m2.
Diameter of circle = 1 m, so Radius (r) = 0.5 m.
Area of circle = πr2 = π × (0.5)2 = 0.25π m2.
P(die lands inside circle) = (Area of circle) / (Area of rectangle)
= 0.25π / 6
= π / 24.
21. 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?
Answer:
Total pens = 144.
Defective pens = 20.
Good pens = 144 - 20 = 124.
Nuri buys only if the pen is good.
P(buying) = P(good pen) = 124/144 = 31/36.
(ii) She will not buy it?
Answer:
Nuri does not buy if the pen is defective.
P(not buying) = P(defective pen) = 20/144 = 5/36.
22. 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, ..., 5/36, ..., 1/36
Answer:
Total outcomes when two dice are thrown = 6 × 6 = 36.
Sum 2: (1,1) -> 1 outcome -> 1/36
Sum 3: (1,2), (2,1) -> 2 outcomes -> 2/36
Sum 4: (1,3), (2,2), (3,1) -> 3 outcomes -> 3/36
Sum 5: (1,4), (2,3), (3,2), (4,1) -> 4 outcomes -> 4/36
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
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
Sum 10: (4,6), (5,5), (6,4) -> 3 outcomes -> 3/36
Sum 11: (5,6), (6,5) -> 2 outcomes -> 2/36
Sum 12: (6,6) -> 1 outcome -> 1/36
Completed Table:
Sum: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
Prob: 1/36, 2/36, 3/36, 4/36, 5/36, 6/36, 5/36, 4/36, 3/36, 2/36, 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:
No, I do not agree. The argument is incorrect because the 11 outcomes (sums) are not equally likely. As seen in part (i), different sums have a different number of favourable outcomes (e.g., Sum 2 has 1 outcome, while Sum 7 has 6 outcomes). For probability to be 1/n, the outcomes must be equally likely.
23. 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 possible outcomes = 2 × 2 × 2 = 8.
Outcomes: {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}.
Winning outcomes (same result): {HHH, TTT}. Number of winning outcomes = 2.
Losing outcomes: Total - Winning = 8 - 2 = 6.
P(Hanif will lose) = 6/8 = 3/4.
24. 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 = 6 × 6 = 36.
(i) 5 will not come up either time:
Total outcomes where 5 comes up at least once:
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6) -> 6 outcomes
(1,5), (2,5), (3,5), (4,5), (6,5) -> 5 outcomes (excluding 5,5 which is already counted)
Total favorable for 5 coming up = 11.
Total outcomes where 5 does NOT come up = 36 - 11 = 25.
P(5 will not come up) = 25/36.
(ii) 5 will come up at least once:
As calculated above, there are 11 such outcomes.
P(5 comes up at least once) = 11/36.
25. 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.
Answer: Incorrect.
Explanation: When two coins are tossed, the possible outcomes are HH, HT, TH, TT. There are 4 equally likely outcomes. "One of each" includes both HT and TH. Thus:
P(Two Heads) = 1/4
P(Two Tails) = 1/4
P(One of each) = 2/4 = 1/2.
The probabilities are not 1/3 each.
(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: Correct.
Explanation: On a die, the outcomes are {1, 2, 3, 4, 5, 6}.
Odd numbers: {1, 3, 5} (3 outcomes)
Even numbers: {2, 4, 6} (3 outcomes)
Since the number of odd and even outcomes are equal (3 each out of 6), they are equally likely. P(Odd) = 3/6 = 1/2.