CBSE Class 10 Mathematics Probability Assignment Set 14

Probability of an Event

Question 1. 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, because \( P(E) + P(\text{not } E) = 1 \).
(ii) 0; such an event is called an impossible event.
(iii) 1; such an event is called a sure (certain) event.
(iv) 1.
(v) 0 and 1, that is, \( 0 \leq P(E) \leq 1 \).
In simple words: These are the basic facts about probability: the values lie from 0 to 1 and complementary events add to 1.

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.

Question 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.
(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:
(i) Not equally likely: the car may or may not start depending on factors like a mechanical fault or fuel.
(ii) Not equally likely: the player may shoot or miss depending on her/his ability and experience.
(iii) Equally likely: the answer is either right or wrong, and each has the same chance.
(iv) Equally likely: the newborn baby can either be a boy or a girl with the same chance.
In simple words: Outcomes are equally likely only when nothing favours one over the other.

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.

Question 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:
A coin toss has only two outcomes, head or tail, and both are equally likely. Hence each team has the same chance (probability \( \frac{1}{2} \)) of winning the toss, so the toss is completely unbiased.
In simple words: Both sides of a fair coin are equally likely.

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.

Question 4. Which of the following cannot be the probability of an event?
(a) \( \frac{2}{3} \)   (b) \( -1.5 \)   (c) 15%   (d) 0.7
Answer:
The probability of any event E satisfies \( 0 \leq P(E) \leq 1 \).
\( \frac{2}{3} \), 15% = 0.15 and 0.7 all lie between 0 and 1, but \( -1.5 \) is negative.
So option (b), -1.5, cannot be a probability.
In simple words: A probability is never negative and never more than 1.

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.

Question 5. If \( P(E) = 0.05 \), what is the probability of 'not E'?
Answer:
\( P(E) + P(\text{not } E) = 1 \)
\( P(\text{not } E) = 1 - 0.05 = 0.95 \).
In simple words: The probabilities of an event and its complement add to 1.

Teacher's Note:
1. Use \( P(\text{not } E) = 1 - P(E) \).
2. Check that your answer lies between 0 and 1.

Question 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?
Answer:
(i) The bag contains only lemon candies, so the number of orange candies is 0. \( P = \frac{0}{n} = 0 \) (an impossible event).
(ii) Every candy is lemon flavoured, so \( P = \frac{n}{n} = 1 \) (a sure event).
In simple words: A candy of a flavour that is not in the bag is impossible; any candy is certainly lemon.

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.

Question 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 the two students have the same birthday. Then \( P(\text{not } E) = 0.992 \).
\( P(E) = 1 - 0.992 = 0.008 \).
In simple words: Subtract the given probability from 1.

Teacher's Note:
1. Identify the event and its complement.
2. Use \( P(E) = 1 - P(\text{not } E) \).

Question 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 + 5 = 8.
(i) \( P(\text{red}) = \frac{3}{8} \).
(ii) \( P(\text{not red}) = 1 - \frac{3}{8} = \frac{5}{8} \).
In simple words: Count the red balls out of 8; 'not red' is the complement.

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.

Question 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 marbles = 5 + 8 + 4 = 17.
(i) \( P(\text{red}) = \frac{5}{17} \).
(ii) \( P(\text{white}) = \frac{8}{17} \).
(iii) \( P(\text{not green}) = 1 - \frac{4}{17} = \frac{13}{17} \).
In simple words: Divide the number of marbles of the wanted kind by 17.

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.

Question 10. A piggy bank contains hundred 50p coins, fifty Rs. 1 coins, twenty Rs. 2 coins and ten Rs. 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 50p coin?
(ii) will not be a Rs. 5 coin?
Answer:
Total number of coins = 100 + 50 + 20 + 10 = 180.
(i) \( P(\text{50p coin}) = \frac{100}{180} = \frac{5}{9} \).
(ii) \( P(\text{Rs. 5 coin}) = \frac{10}{180} = \frac{1}{18} \), so \( P(\text{not a Rs. 5 coin}) = 1 - \frac{1}{18} = \frac{17}{18} \).
In simple words: Count all the coins first, then find the required fraction.

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.

Question 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. What is the probability that the fish taken out is a male fish?
Answer:
Total number of fish = 5 + 8 = 13.
\( P(\text{male fish}) = \frac{5}{13} \).
In simple words: There are 5 male fish out of 13.

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.

Question 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 the figure), 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 number of possible outcomes = 8.
(i) Only one outcome (8): \( P = \frac{1}{8} \).
(ii) Odd numbers are 1, 3, 5, 7: \( P = \frac{4}{8} = \frac{1}{2} \).
(iii) Numbers greater than 2 are 3, 4, 5, 6, 7, 8 (6 numbers): \( P = \frac{6}{8} = \frac{3}{4} \).
(iv) All 8 numbers are less than 9: \( P = \frac{8}{8} = 1 \) (a sure event).
In simple words: Count the favourable numbers out of 8.

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.

Question 13. 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:
Possible outcomes: 1, 2, 3, 4, 5, 6 (total 6).
(i) Prime numbers: 2, 3, 5. \( P = \frac{3}{6} = \frac{1}{2} \).
(ii) Numbers between 2 and 6: 3, 4, 5. \( P = \frac{3}{6} = \frac{1}{2} \).
(iii) Odd numbers: 1, 3, 5. \( P = \frac{3}{6} = \frac{1}{2} \).
In simple words: List the favourable numbers and divide by 6.

Teacher's Note:
1. Write the total number of possible outcomes first, then count the favourable outcomes.
2. Reduce the fraction to its lowest terms and check that the probability lies between 0 and 1.