Exercise 14.1 Solutions
Probability
1. Complete the following statements:
(i) Probability of an event E + Probability of the event ‘not E’ = 1.
(ii) The probability of an event that cannot happen is 0. Such an event is called an impossible event.
(iii) The probability of an event that is certain to happen is 1. Such an event is called a sure or certain event.
(iv) The sum of the probabilities of all the elementary events of an experiment is 1.
(v) The probability of an event is greater than or equal to 0 and less than or equal to 1.
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.
Not equally likely. The outcome depends on many factors like the car’s condition, fuel, etc. A well-maintained car is much more likely to start.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
Not equally likely. The outcome depends on the player’s skill. A skilled player is more likely to make the shot.
(iii) A trial is made to answer a true-false question. The answer is right or wrong.
Equally likely. If the answer is a pure guess, there are only two possibilities (right or wrong) with an equal chance for each.
(iv) A baby is born. It is a boy or a girl.
Equally likely. Biologically, the chance of being a boy or a girl is considered approximately equal.
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?
Tossing a coin is considered fair because it has two outcomes, Heads and Tails, which are equally likely. The result of a coin toss is unpredictable and not biased towards any particular outcome, giving both teams an equal chance.
4. Which of the following cannot be the probability of an event?
(A) 2/3 (B) –1.5 (C) 15% (D) 0.7
The probability of an event must be between 0 and 1, inclusive.
A) 2/3 ≈ 0.67 (Valid)
B) -1.5 is negative (Invalid)
C) 15% = 0.15 (Valid)
D) 0.7 (Valid)
Answer: (B) –1.5, because probability cannot be negative.
5. If P(E) = 0.05, what is the probability of ‘not E’?
We know that P(E) + P(not E) = 1.
0.05 + P(not E) = 1
P(not E) = 1 – 0.05 = 0.95.
Answer: The probability of ‘not E’ is 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? (ii) a lemon flavoured candy?
(i) an orange flavoured candy?
Since the bag only contains lemon candies, getting an orange candy is an impossible event. The probability is 0.
(ii) a lemon flavoured candy?
Since the bag only contains lemon candies, getting a lemon candy is a sure event. The probability is 1.
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?
Let E be the event that “2 students have the same birthday”.
Then ‘not E’ is the event that “2 students do not have the same birthday”.
We are given P(not E) = 0.992.
P(E) = 1 – P(not E) = 1 – 0.992 = 0.008.
Answer: The probability 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?
Total balls = 3 (red) + 5 (black) = 8.
(i) red?
Favourable outcomes (red) = 3.
P(red) = Favourable / Total = 3/8.
(ii) not red?
P(not red) = 1 – P(red) = 1 – 3/8 = 5/8. (Alternatively, the number of non-red balls is 5, so P = 5/8).
… (Solutions for questions 9 to 25 follow)
9. A box contains 5 red marbles, 8 white marbles and 4 green marbles… What is the probability that the marble taken out will be (i) red? (ii) white? (iii) not green?
Total marbles = 5 + 8 + 4 = 17.
(i) red? P(red) = 5/17.
(ii) white? P(white) = 8/17.
(iii) not green? P(green) = 4/17. So, P(not green) = 1 – 4/17 = 13/17.
10. A piggy bank contains… What is the probability that the coin (i) will be a 50p coin? (ii) will not be a ₹5 coin?
Total coins = 100 (50p) + 50 (₹1) + 20 (₹2) + 10 (₹5) = 180.
(i) will be a 50p coin?
Number of 50p coins = 100. P(50p) = 100/180 = 5/9.
(ii) will not be a ₹5 coin?
Number of ₹5 coins = 10. P(₹5 coin) = 10/180 = 1/18.
P(not ₹5 coin) = 1 – P(₹5 coin) = 1 – 1/18 = 17/18.
11. Gopi buys a fish from a shop… What is the probability that the fish taken out is a male fish?
Total fish = 5 (male) + 8 (female) = 13.
Number of male fish = 5.
P(male fish) = Favourable / Total = 5/13.
Answer: 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… What is the probability that it will point at…
>Total outcomes = 8.
(i) 8? Favourable = 1. P(8) = 1/8.
(ii) an odd number? Odd numbers are 1, 3, 5, 7. Favourable = 4. P(odd) = 4/8 = 1/2.
(iii) a number greater than 2? Numbers are 3, 4, 5, 6, 7, 8. Favourable = 6. P(>2) = 6/8 = 3/4.
(iv) a number less than 9? All numbers are less than 9. Favourable = 8. P(<9) = 8/8 = 1.
13. A die is thrown once. Find the probability of getting…
Total outcomes = {1, 2, 3, 4, 5, 6}. Total = 6.
(i) a prime number; Prime numbers are 2, 3, 5. Favourable = 3. P(prime) = 3/6 = 1/2.
(ii) a number lying between 2 and 6; Numbers are 3, 4, 5. Favourable = 3. P(between 2 and 6) = 3/6 = 1/2.
(iii) an odd number. Odd numbers are 1, 3, 5. Favourable = 3. P(odd) = 3/6 = 1/2.
14. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting…
Total outcomes = 52.
(i) a king of red colour: 2 cards (King of hearts, King of diamonds). P = 2/52 = 1/26.
(ii) a face card: 12 cards (4 Jacks, 4 Queens, 4 Kings). P = 12/52 = 3/13.
(iii) a red face card: 6 cards (J,Q,K of hearts & diamonds). P = 6/52 = 3/26.
(iv) the jack of hearts: 1 card. P = 1/52.
(v) a spade: 13 cards. P = 13/52 = 1/4.
(vi) the queen of diamonds: 1 card. P = 1/52.
15. Five cards—the ten, jack, queen, king and ace of diamonds, are well-shuffled… Find the probability…
(i) What is the probability that the card is the queen?
Total cards = 5. Favourable (queen) = 1. P(queen) = 1/5.
(ii) If the queen is drawn and put aside…
Now, total cards remaining = 4. The queen is gone.
(a) an ace? Favourable (ace) = 1. P(ace) = 1/4.
(b) a queen? Favourable (queen) = 0. P(queen) = 0.
16. 12 defective pens are accidentally mixed with 132 good ones… What is the probability that the pen taken out is a good one?
Total pens = 12 (defective) + 132 (good) = 144.
Number of good pens = 132.
P(good pen) = 132/144 = 11/12.
17. (i) A lot of 20 bulbs contain 4 defective ones… What is the probability that this bulb is defective?
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced… What is the probability that this bulb is not defective?
(i) Total bulbs = 20. Defective = 4.
P(defective) = 4/20 = 1/5.
(ii) Now, total bulbs = 19. The bulb drawn was not defective, so good bulbs remaining = 15.
P(not defective) = 15/19.
18. A box contains 90 discs which are numbered from 1 to 90… Find the probability that it bears (i) a two-digit number (ii) a perfect square number (iii) a number divisible by 5.
Total outcomes = 90.
(i) a two-digit number: Numbers are from 10 to 90. Total = 90 – 10 + 1 = 81. P = 81/90 = 9/10.
(ii) a perfect square number: 1, 4, 9, 16, 25, 36, 49, 64, 81. Favourable = 9. P = 9/90 = 1/10.
(iii) a number divisible by 5: 5, 10, …, 90. This is an AP. 90 = 5 + (n-1)5 => n=18. P = 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?
Total outcomes = 6.
(i) A? There are two faces with ‘A’. Favourable = 2. P(A) = 2/6 = 1/3.
(ii) D? There is one face with ‘D’. Favourable = 1. P(D) = 1/6.
20. Suppose you drop a die at random on the rectangular region shown… What is the probability that it will land inside the circle with diameter 1m?
>Area of rectangle = 3m × 2m = 6 m² (Total Area).
Area of circle = πr². Diameter=1m, so r=0.5m. Area = π(0.5)² = 0.25π m² (Favourable Area).
Probability = Favourable Area / Total Area = (0.25π) / 6 = π/24.
21. A lot consists of 144 ball pens… What is the probability that (i) She will buy it ? (ii) She will not buy it ?
Total pens = 144. Defective = 20. Good = 144 – 20 = 124.
(i) She will buy it? (She buys if it is good).
P(good) = 124/144 = 31/36.
(ii) She will not buy it? (She doesn’t buy if it is defective).
P(defective) = 20/144 = 5/36. (or 1 – 31/36).
22. Refer to Example 13. (i) Complete the following table: … (ii) A student argues that ‘there are 11 possible outcomes… Do you agree with this argument? Justify your answer.
(i) Completing the table:
Total outcomes when two dice are thrown = 36.
Sum 2: (1,1) -> P=1/36
Sum 3: (1,2),(2,1) -> P=2/36=1/18
Sum 4: (1,3),(2,2),(3,1) -> P=3/36=1/12
Sum 5: (1,4),(2,3),(3,2),(4,1) -> P=4/36=1/9
Sum 6: (1,5),(2,4),(3,3),(4,2),(5,1) -> P=5/36
Sum 8: (2,6),(3,5),(4,4),(5,3),(6,2) -> P=5/36
Sum 9: (3,6),(4,5),(5,4),(6,3) -> P=4/36=1/9
Sum 10: (4,6),(5,5),(6,4) -> P=3/36=1/12
Sum 11: (5,6),(6,5) -> P=2/36=1/18
Sum 12: (6,6) -> P=1/36
(ii) Do you agree with the argument?
No. The argument is incorrect because the 11 sums (2, 3, …, 12) are not equally likely outcomes. For example, a sum of 7 can be obtained in 6 ways, while a sum of 2 can only be obtained in 1 way. The probability for each sum is different, as calculated above.
23. A game consists of tossing a one rupee coin 3 times… Hanif wins if all the tosses give the same result… and loses otherwise. Calculate the probability that Hanif will lose the game.
Total possible outcomes are {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}. Total = 8.
Hanif wins if the result is HHH or TTT. Favourable outcomes for winning = 2.
P(win) = 2/8 = 1/4.
The probability that Hanif will lose is P(lose) = 1 – P(win) = 1 – 1/4 = 3/4.
Answer: 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?
Total outcomes = 6 × 6 = 36.
(i) 5 will not come up either time?
For each throw, there are 5 outcomes that are not 5 ({1,2,3,4,6}).
Favourable outcomes = 5 × 5 = 25.
P(no 5) = 25/36.
(ii) 5 will come up at least once?
This is the complementary event to ‘no 5 comes up’.
P(at least one 5) = 1 – P(no 5) = 1 – 25/36 = 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…
Not correct. The possible outcomes are {HH, HT, TH, TT}. There are 4 equally likely outcomes.
‘One of each’ corresponds to two outcomes (HT and TH).
P(two heads) = 1/4, P(two tails) = 1/4, P(one of each) = 2/4 = 1/2. The probabilities are not equal to 1/3.
(ii) If a die is thrown…
Not correct. There are 6 equally likely outcomes {1, 2, 3, 4, 5, 6}.
Odd outcomes: {1, 3, 5} (3 outcomes). Even outcomes: {2, 4, 6} (3 outcomes).
P(odd) = 3/6 = 1/2. P(even) = 3/6 = 1/2. The probabilities are equal, and the argument is correct.
