Exercise 14.1
1. Complete the following statements:
2. Which of the following experiments have equally likely outcomes? Explain.
(i)
(i) A driver attempts to start a car. The car starts or does not start.
Solution:
Given:
A driver attempts to start a car. The car starts or does not start
Explanation:
It is not an equally likely outcome because the probability of it depends on
(ii)
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
Solution:
Given:
A player attempts to shoot a basketball. She/he shoots or misses the shot
Explanation:
It is not an equally likely outcome because the probability depends on the ability of the player.
(iii)
(iii) A trial is made to answer a true-false question. The answer is right or wrong.
Solution:
Given:
A trial is made to answer a true-false question. The answer is right or wrong
Explanation:
It is an equally likely outcome because the probability is
(iv)
(iv) A baby is born. It is a boy or a girl.
Solution:
Given:
A baby is born. It is a boy or a girl
Explanation:
It is an equally likely outcome because the probability is
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?
Solution:
Tossing a coin is considered to be a fair way of deciding which team should get the ball at the beginning of a football game because it is an
Hence, the probability for both the teams are
4. Which of the following cannot be the probability of an event?
(A)
Solution:
We know that the probability of an event E lies in between
Therefore,
5. If P(E) = 0.05, what is the probability of ‘not E’?
Solution:
Probability of (E) =
We know that,
Probability of (E) + Probability of (not E) =
Probability of (not E) = 1
Probability of (not E) =
Therefore, the probability of 'not E' is equal to
6. A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag.
(i)
(i) What is the probability that she takes out an orange flavoured candy?
Solution:
A bag contains only
Therefore, the probability of taking out orange flavoured candy is
(ii)
(ii) What is the probability that she takes out a lemon flavoured candy?
Solution:
As the bag has only
Therefore, the event is sure event and the probability of taking out lemon flavoured candy is
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?
Solution:
We know that the sum of two complementary events is equal to
P(E) + P (not E) =
By putting the given values in the above equation, we can find out the probability of not happening of the event.
The probability of
Probability of 2 students having the same birthday P(E) = 1
Thus, the probability that 2 students have the same birthday is equal to
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?
Solution:
Step (i): First find out the probability of drawing a red ball by using the formula given below.
Probability of an event =
Step (ii): After that, by using the formula for the sum of complementary event, find the probability of not getting a red ball.
P (E) + P (not E) =
(i) Number of red balls in a bag =
Number of black balls in a bag =
Total number of balls =
Probability of drawing red ball P(E) =
Probability of drawing red ball =
(ii) Probability of not getting red ball P(not E) = 1 - P (E) = 1
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?
Solution:
Let's find out the probability of getting red, white, and green marble by using the formula
Probability =
Using the formula of the sum of complementary event, we will find out the probability of not getting the green ball.
P (E) + P (not E) =
Number of red balls in a bag =
Number of white balls in a bag =
Number of green balls in a bag =
Total number of balls =
(i) Probability of drawing red ball =
(ii) Probability of drawing white ball =
(iii) Probability of drawing a green ball =
Let the probability of not getting a green ball be P (not E)
P (not E) = 1 - P (E)
= 1
=
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?
Solution:
We know that,
Probability =
Number of 50 p coins = 100
Number of Rs 1 coins = 50
Number of Rs 2 coins = 20
Number of Rs 5 coins = 10
Total number of coins =
(i) Probability of drawing 50 p coin =
(ii) Probability of getting a Rs 5 coin =
Probability of not getting a 5 rupee coin is 1
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 the below figure). What is the probability that the fish taken out is a male fish?
Solution:
We know that,
Probability =
Number of male fish =
Number of female fish =
Total number of fishes =
The probability that the fish taken out is a male fish =
=
Thus, the probability that the fish that are taken out is a male fish, is
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 below figure), and these are equally likely outcomes. What is the probability that it will point at
(i)
(i) 8 ?
Solution:
We know that,
Probability =
Total possible outcomes =
Probability of getting 8 =
Probability of getting 8 =
(ii)
(ii) an odd number?
Solution:
Total number of odd numbers from 1 to 8 =
Probability of getting odd number =
=
=
(iii)
(iii) a number greater than 2?
Solution:
The numbers that are greater than 2 are (Note: greater than 2 to 8)
=
Probability of getting numbers greater than 2 =
=
=
(iv)
(iv) a number less than 9?
Solution:
The numbers less than 9 are (Note: From 1 to 9)
Probability of getting numbers less than 9 =
=
=
13. A die is thrown once. Find the probability ?
(i)
(i) Probability of getting a prime number.
Solution:
Step 1: Identify the prime numbers on the die.
The prime numbers between 1 and 6 are
Step 2: Find the number of favorable outcomes.
The favorable outcomes (prime numbers) are: 2, 3,
Number of favorable outcomes =
Step 3: Use the formula for probability:
P(prime number)=
=
=
So, the probability of getting a prime number is
(ii)
(ii) Probability of getting a number lying between 2 and 6.
Solution:
Step 1: Identify the numbers between 2 and 6 (not including 2 and 6).The numbers lying strictly between 2 and 6 are
Step 2: Find the number of favorable outcomes.
The favorable outcomes are: 3, 4, 5.
Number of favorable outcomes =
Step 3: Use the formula for probability:
P(number between 2 and 6)=
=
=
(iii)
(iii) Probability of getting an odd number.
Solution:
Step 1: Identify the odd numbers on the die.
The odd numbers between 1 and 6 are
Step 2: Find the number of favorable outcomes.
The favorable outcomes (odd numbers) are: 1, 3, 5.
Number of favorable outcomes =
Step 3: Use the formula for probability:
P(odd number)=
=
=
So, the probability of getting an odd number is
14. One card is drawn from a well-shuffled deck of 52 cards. Find the probability?
(i)
(i) Probability of getting a king of red colour
Solution:
There are
Total number of cards =
Probability =
=
=
(ii)
(ii) Probability of getting a face card
Solution:
Face cards are
There are
Probability =
=
(iii)
(iii) Probability of getting a red face card
Solution:
Red suits are Hearts and Diamonds.
Each red suit has
Therefore, there are 6 red face cards in total.
Probability =
=
(iv)
(iv) Probability of getting the jack of hearts
Solution:
There is only
Probability =
(v)
(v) Probability of getting a spade
Solution:
There are
Probability =
=
(vi)
(vi) Probability of getting the queen of diamonds
Solution:
There is only
Probability =
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)
(i) What is the probability that the card is the queen?
Solution:
Total number of cards =
Number of queen cards =
Probability that the card is the queen =
(ii)
(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?
Solution:
If the queen is drawn and put aside, then four cards are left the ten, jack, king and ace of diamonds
(a) Probability that the card an ace =
(b) Probability that the card is the queen =
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.
Solution:
Total number of pens =
Number of good pens =
We know that,
Probability =
The probability P of drawing a good pen is given by:
P(Good pen) =
Therefore, the probability of drawing a good pen is
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?
(i)
Solution:
Probability that the first bulb drawn is defective:
Total number of bulbs =
Number of defective bulbs =
The probability P of drawing a defective bulb is given by:
P(Defective bulb)=
=
=
(ii)
(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 ?
Solution:
After the first bulb is drawn and found to be not defective, there are now 19 bulbs left.
The number of non-defective bulbs initially was
Since the first bulb was not defective, there are still
The probability P of drawing a non-defective bulb from the remaining 19 is:
P(Non-defective bulb) =
Thus, the probability of drawing a non-defective bulb in the second draw is
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)
(i) a two-digit number
Solution:
Total number of discs =
Probability of getting a two digit number =
=
=
(ii)
(ii) a perfect square number
Solution:
Total number of discs = 90
Total number of perfect square numbers between 1 to 90 are (Note: From
= 1,
Probability of getting a perfect square number =
=
=
(iii)
(iii) a number divisible by 5.
Solution:
Total number of discs = 90
Total numbers that are divisible by 5 are
Probability of getting a number divisible by 5 =
=
=
19. A child has a die whose six faces show the letters as given below:
A BCDEA The die is thrown once. What is the probability of getting (i) A? (ii) D?
(i)
What is the probability of getting A?
Solution:
Probability of getting an A:
There are two faces with the letter A on this die.
Total number of faces =
The probability P of getting an A is:
P(A) =
(ii)
What is the probability of getting D?
Solution:
Probability of getting a D:
There is only
Total number of faces =
The probability P of getting a D is:
P(D) =
So, the probability of getting A is
20*. Suppose you drop a die at random on the rectangular region shown in below figure. What is the probability that it will land inside the circle with diameter 1m?
Solution:
We use the concepts of areas of circles and squares and also the basic concepts of probability.
Length of rectangular region =
Breadth of rectangular region =
Area of rectangular region =
=
=
Diameter of circular region =
Radius of circular region =
Area of circular region =
= π ×
=
The probability that it will land inside the circle =
=
=
=
Therefore, the probability that it will land inside the circle is
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 ? (ii) She will not buy it ?
(i)
(i) What is the probability that she will buy it ?
Solution:
We use the basic formula of probability and favourable outcomes.
Total number of ball pens =
Number of defective ball pens =
Number of good ball pens =
Probability that she will buy it =
=
=
(ii)
(ii) What is the probability that she will not buy it?
Solution:
We use the basic formula of probability and favourable outcomes.
Total number of ball pens =
Number of defective ball pens =
Number of good ball pens =
Probability that she will not buy it =
=
=
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 |
(i)
Solution:
We use the basic concepts of probability to solve the problem.
(i) Number of possible outcomes to get the sum as 2
= (1,1) =
Number of possible outcomes to get the sum as 3
= (
Number of possible outcomes to get the sum as 4
= (
Number of possible outcomes to get the sum as 5
= (
Number of possible outcomes to get the sum as 6
= (
Number of possible outcomes to get the sum as 7
= (1,
Number of possible outcomes to get the sum as 8
= (2,
Number of possible outcomes to get the sum as 9
= (3,
Number of possible outcomes to get the sum as 10
= (4,
Number of possible outcomes to get the sum as 11
= (5,
Number of possible outcomes to get the sum as 12
= (6,
Thus, the table is:
| Event: 'Sum on 2 dice' | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|
| Probability |
| Event: 'Sum on 2 dice' | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|
| Probability |
(ii)
(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
Solution:
Probability of each of them is not
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.
Solution:
We use the basic formula of favorable outcomes to solve the problem.
Total possible outcomes when tossing a one rupee coin 3 times are = {
Number of possible outcomes to get three heads or three tails is
The probability that Hanif will win the game =
=
=
The probability that Hanif will lose the game is 1 -
=
Therefore, the probability that Hanif will lose the game is
24. A die is thrown twice.
(i)
(i) What is the probability that 5 will not come up either time?[Hint : Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]
Solution:
Total number of outcomes when die is thrown twice =
(i) Number of possible outcomes when 5 will come up either time
(Hint: die 1: probability of getting 5, die 2: probability of getting 5)
= (5,
Probability that 5 will come up either time =
=
The probability that 5 will not come up either time = 1
=
(ii)
(ii) What is the probability that 5 will come up at least once?[Hint : Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]
Solution:
(ii) Number of possible outcomes when 5 will come up at least once = 11
The probability that 5 comes up at least once =
=
Therefore, the probability that 5 will not come up either time is
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
(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
(i)
Solution:
Given Argument:
"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
This argument is
Reason:
The argument overlooks the fact that the outcomes “one head and one tail” can occur in two different ways: (Head, Tail) or (Tail, Head). These are distinct outcomes when considering the order of the tosses.
The actual sample space for two coin tosses includes four possible outcomes:
So, there are 4 possible outcomes in total, not 3.
Correct probabilities:
P(Two heads) =
P(Two tails) =
P(One of each) =
Thus, the probabilities are
(ii)
Solution:
Given Argument:
"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
This argument is
Reason:
A standard die has six faces with the numbers 1, 2, 3, 4, 5, and 6. Among these, the odd numbers are
The total possible outcomes when throwing a die are
Odd numbers: 1, 3, 5 (3 outcomes)
Even numbers: 2, 4, 6 (3 outcomes)
The probability of getting an odd number is the number of odd numbers divided by the total number of outcomes:
P(Odd number) =
Since the die is fair and the outcomes are equally likely, the probability of getting an odd number is indeed