Introduction to Probability - NCERT Solutions Class 9 Ganita Manjari

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The Mathematics of Maybe: Introduction to Probability – NCERT Solutions Class 9 Maths Ganita Manjari includes all the questions with solutions given in NCERT Class 9 Ganita Manjari textbook.

NCERT Solutions Class 9

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Introduction to Probability – NCERT Solutions

Q.1:

Rank the following events on a scale from 0 (Impossible) to 1 (Certain). Label each event: Impossible, less likely, equally likely (even chance), more likely, certain. Give reasons why you gave each event its ranking.

The next Monday will come after Sunday.
It will snow in Mumbai in July.
An elephant will walk through your classroom today.
You will greet at least one friend at school tomorrow.

Solution:

The ranking of these events is based on the probability scale where {tex}{0}{/tex} is impossible and 1 is certain:

Certain (1): This event is a factual certainty as the sequence of days in a week never changes; Monday always follows Sunday.
Impossible (0): Mumbai has a tropical climate, and July is a monsoon month with high temperatures, making snowfall naturally impossible.
Impossible (0): Under normal circumstances, an elephant entering a classroom is not a realistic occurrence due to safety and physical constraints.
More Likely (0.75): Since you go to school to interact, the chance of greeting a friend is high, though not {tex}100 \%{/tex} guaranteed.

These rankings help quantify how confident we are about the occurrence of different real-world events using the basic definitions of probability.

Q.2:

If you roll a standard 6-sided die, what is the theoretical probability of getting a 4?

Solution:

Number of favourable outcomes = 1 (only the number 4)
Number of all possible outcomes {tex}=6{/tex} (the numbers 1 through 6)
{tex}{P}({/tex}rolling a 4{tex})=\frac{1}{6}=0.1666 \ldots \approx 0.167{/tex} or {tex}16.7 \%{/tex}.

Q.3:

A letter is picked at random from the word ‘PROBABILITY’. What is the probability of picking the letter B?

Solution:

{tex}\text { Number of favourable outcomes }={/tex} {tex}2 \text { (there are } 2 {Bs} \text { in the word }{/tex}{tex}\text { PROBABILITY) }{/tex}
{tex}\text { Number of all possible outcomes }={/tex} {tex}11 \text { (The number of letters in the }{/tex}{tex}\text { word PROBABILITY) }{/tex}
{tex}{P} \text { (Picking the letter B) }=\frac{2}{11}= {/tex} {tex}0.1818 \ldots \approx 0.182 \text { or } 18.2 \% .{/tex}

Q.4:

A teacher mixes a large bag of sweets of different colours and randomly selects a sample of 30 sweets. She counts the number of sweets of each colour:
10 red sweets | 8 green sweets | 7 yellow sweets | 5 blue sweets

Calculate the probability that a randomly picked sweet from the sample is green.
If there are 600 sweets in total in the large bag, estimate how many are likely to be yellow, based on the sample results.

Solution:

Given sample (total = 30 sweets):

Red {tex}=10{/tex}
Green {tex}=8{/tex}
Yellow = 7
Blue {tex}=5{/tex}

Probability of picking a green sweet {tex}P(\text {green})=\frac{\text { Number of green sweets }}{\text { Total sweets }}=\frac{8}{30}{/tex}
{tex}P(\text {green})=\frac{4}{15}{/tex}
Estimate number of yellow sweets in 600 sweets From the sample: {tex}P(\text {yellow})=\frac{7}{30}{/tex} Now estimate in 600 sweets: {tex}\text { Estimated yellow sweets }=\frac{7}{30} \times 600{/tex}
{tex}=7 \times 20=140{/tex}

Q.5:

A survey is conducted at a school where a random sample of 40 students is asked about their favourite club. The responses are:

14 students: Science Club | 11 students: Arts Club | 9 students: Sports Club | 6 students: Debate Club
Assume there are 800 students in the whole school.

What is the probability that a randomly chosen student from the sample prefers the Arts Club?
Using the sample results, estimate how many students in the whole school are likely to prefer the Sports Club.

Solution:

The probability for the Arts Club is found by dividing the eleven interested students by the forty total students in the sample. This calculation results in a probability of 0.275 or {tex}27.5 \%{/tex} which represents the likelihood for any student in that specific group. For the school-wide estimate of the Sports Club, we take the nine students from the sample and divide by forty to find the relative frequency. Multiplying this fraction by the total population of eight hundred students gives an estimated total of one hundred and eighty students who prefer sports. This objective method relies on experimental data to predict trends within a much larger population without needing to survey every single individual.

Probability for Arts Club {tex}=11 / 40=0.275{/tex}
Estimated Sports Club students {tex}=(9 / 40) \times 800=180{/tex}

Q.6:

Toss a coin 20 times and record the result each time (heads or tails).

How many times did you get heads?
How many times did you get tails?
Calculate the experimental probability of getting heads.
If you toss the coin once more, what is the probability of getting tails?

Solution:

Since I am an AI, I simulated twenty tosses resulting in 11 heads and 9 tails to answer your specific experiment. In this case, I got 11 heads and 9 tails, which directly answers the first two points of your question. The experimental probability of getting heads is calculated as eleven divided by twenty, which equals exactly 0.55 or {tex}55 \%{/tex}. For the final point, the probability of getting tails on the next toss is 0.5 because past results do not affect future independent trials. This exercise highlights the difference between actual observed data and the ideal 50% theoretical expectation for a fair coin.

Heads: 11
Tails: 9
Experimental Probability (Heads): {tex}11 / 20=0.55{/tex}
Probability of Tails next: 0.5 (or 1/2)

Q.7:

Toss a paper cup into the air 100 times. After each toss record whether the cup lands on its bottom, upside down on its top or on its side (See Figure). Assign probabilities to the outcomes by using experimental probability.

Solution:

To solve this experiment, we simulate one hundred tosses and record the frequency of each landing position shown in your image. Suppose the cup lands on its side seventy times, on its bottom twenty times, and upside down on its top ten times. The experimental probability is then calculated by dividing these frequencies by the total trials, resulting in {tex}0.7,0.2{/tex}, and 0.1 respectively. These values represent the objective likelihood of each outcome based specifically on the data collected during this particular set of observations. Unlike a fair coin, a paper cup is biased due to its shape, so experimental data is essential for accurate measurement.

Side: {tex}70 / 100=0.7{/tex}
Bottom: {tex}20 / 100=0.2{/tex}
Top: {tex}10 / 100=0.1{/tex}

Q.8:

What is the probability of getting an even number when rolling a fair 6-sided die?

Solution:

To find the probability of rolling an even number on a fair six-sided die, we look at the theoretical outcomes. First, identify the sample space, which is {tex}S=\{1,2,3,4,5,6\}{/tex}, meaning there are six total possible results. Next, we find the favorable outcomes, which are the even numbers: 2, 4 and 6. Since there are three even numbers, the number of favorable outcomes is 3.

The formula for theoretical probability is the number of favorable outcomes divided by the total number of possible outcomes. By plugging in our values, we get {tex}3 / 6{/tex}, which simplifies to {tex}1 / 2{/tex}. In decimal form, this is 0.5, and as a percentage, it represents a {tex}50 \%{/tex} chance. This means that in an ideal, fair situation, you would expect to see an even number half of the time you roll the die.

Q.9:

Suppose you roll a 6-sided die 12 times and get a ‘3’ three times.

What is the experimental probability of rolling a ‘3’?
What is the theoretical probability of rolling a ‘3’?
Why might these probabilities be different? What would you expect to happen if you roll the die 60, 600, or 6000 times?

Solution:

The experimental probability of rolling a three is 0.25, which we find by dividing the three times the number appeared by the twelve total rolls.
The theoretical probability for rolling a three is one divided by six, which equals approximately 0.167, as each of the six faces is equally likely.
These probabilities are different because a small sample of twelve rolls allows for random chance to skew the data away from the expected average. As you increase the number of trials to sixty, six hundred, or six thousand, you would expect the experimental results to get closer to the theoretical value. This phenomenon is known as the law of large numbers, where a larger amount of data provides a more accurate reflection of true probability.

Q.10:

When a single 6-sided die is rolled, what is the total number of possible outcomes in the sample space?

Solution:

When you roll a single standard six-sided die, the sample space represents every possible result that can land face up. In this case, the die has six distinct faces, each marked with a different number of dots from one to six. Therefore, the sample space is written as {tex}S=\{1,2,3,4,5,6\}{/tex}, which includes all potential outcomes of a single roll. Since each face is unique and no other results are possible, the total number of elements, or the sample size {tex}n(S){/tex}, is exactly six. This count forms the denominator for calculating any theoretical probability related to a single die roll.

Sample Space (S): {tex}\{1,2,3,4,5,6\}{/tex}
Total Outcomes {tex}{n}({S}): 6{/tex}

Q.11:

For the following experiments write down the sample space S.

Rolling a die and tossing a coin together.
Choosing a random integer between – 5 and + 5
A box containing 5 green and 7 red balls. One ball is drawn at random.

Solution:

When rolling a die and tossing a coin together, every number from one to six can pair with either heads or tails. This results in a sample space of twelve distinct outcomes: {tex}S=\{1 H, 1 T, 2 H, 2 T, 3 H, 3 T, {/tex}{tex}4 H, 4 T, 5 H, 5 T, 6 H, 6 T\}{/tex}.
Choosing a random integer strictly between -5 and +5 means we list all whole numbers that fall within that specific range. The resulting sample space contains nine elements: {tex}S=\{-4,-3,-2,-1,0,1,2,3,4\}{/tex}.
For a box with five green and seven red balls, the sample space represents the possible colors you can draw. Since there are only two types of colors available, the sample space is simply {tex}S=\{{/tex}Green, {tex}\operatorname{Red}\}{/tex}, though the total number of physical balls is twelve.
- {tex}S=\{1 H, 1 T, 2 H, 2 T, 3 H, 3 T, {/tex}{tex}4 H, 4 T, 5 H, 5 T, 6 H, 6 T\}{/tex}
- {tex}S=\{-4,-3,-2,-1,0,1,2,3,4\}{/tex}
- {tex}S=\{{/tex}Green, Red{tex}\}{/tex}

Q.12:

In a village fair, there are 3 popular snacks available: Samosa, Pakora, and Bhaji. For drinks, villagers can choose either Chai or Lassi.

List the sample space of all possible snack and drink combinations a person could choose at the fair.
List the event ‘Selecting Samosa as a snack.’

Solution:

The sample space represents every unique combination of one snack and one drink available at the fair. Since there are three snack options and two drink options, we multiply them to find a total of six possible pairs. The complete sample space is {tex}S={/tex} {(Samosa, Chai), (Samosa, Lassi), (Pakora, Chai), (Pakora, Lassi), (Bhaji, Chai), (Bhaji, Lassi)}. Each element in this set shows a complete meal choice that a villager could possibly make.
An event is a specific subset of the sample space that meets a particular condition. In this case, the condition is that the snack must be a Samosa, regardless of which drink is chosen. Therefore, the event set {tex}E{/tex} contains only the pairs where Samosa is the first element: {tex}E= \{({/tex}Samosa, Chai{tex}),({/tex}Samosa, Lassi{tex})\}{/tex}. This specific subset allows us to focus only on the outcomes relevant to that particular snack choice.
- Sample Space (S): {tex}\{({/tex}Samosa, Chai), (Samosa, Lassi), (Pakora, Chai), (Pakora, Lassi), (Bhaji, Chai), (Bhaji, Lassi)\}
- Event (E): {tex}\{({/tex}Samosa, Chai), (Samosa, Lassi){tex}\}{/tex}

Q.13:

Can you calculate the probability of getting one head and one tail?

Solution:

To find the probability of getting exactly one head and one tail when tossing two coins, we first look at the total sample space. A tree diagram shows that the possible outcomes are Head-Head, Head-Tail, Tail-Head, and Tail-Tail, making the total number of outcomes four.

There are two favorable outcomes where we get exactly one of each: Head-Tail (HT) and Tail-Head (TH). By using the theoretical probability formula, we divide these two favorable outcomes by the four total outcomes in the sample space. This calculation results in {tex}2 / 4{/tex}, which simplifies to {tex}1 / 2{/tex} or a {tex}50 \%{/tex} chance. This means that if you toss two coins, you have an even chance of getting one head and one tail.

Conditional Probability Formula

{tex} P(A / B)=\frac{P(A \cap B)}{P(B)}{/tex}

Favorable outcomes: {tex}\{H T, T H\}{/tex}
Total outcomes: 4
Probability: {tex}2 / 4=0.5{/tex} (or {tex}50 \%{/tex})

Q.14:

There are two fruit baskets A and B. Basket A has one apple and two oranges. Basket B has one banana and one mango. You randomly pick one fruit from each basket.

Draw a tree diagram showing all possible pairs of fruits.
List the sample space.
What is the probability of picking one apple and one banana?

Solution:

The experiment involves picking one fruit from Basket A, followed by picking one fruit from Basket B. The tree diagram starts with three branches for the possible choices from Basket A: one branch for the apple ({tex}A{/tex}) and two branches for the two separate oranges ({tex}O_1, O_2{/tex}). Since picking a fruit from Basket A is a single event and we are treating the oranges separately to maintain equally likely outcomes, the probability for each branch from the starting point is {tex}1 / 3{/tex}. From each of these three branches, we draw two new branches to represent the choices from Basket B, which has one banana ({tex}B{/tex}) and one mango ({tex}M{/tex}). Since there are two outcomes, the probability for each of these branches is {tex}1 / 2{/tex}. A complete outcome is a path from the start of the tree to one of the final leaves. There are six final leaves on the tree, which correspond to the six possible pairs of fruits. For example, following the path “Apple” then “Banana” gives the outcome {tex}(A, B){/tex}. Here is the tree diagram visualizing all possible outcomes and their probabilities:
The sample space {tex}S{/tex} is the list of all final pairs of fruits seen at the ends of the branches. Since we have one apple {tex}(A){/tex}, two distinct oranges {tex}\left(O_1, O_2\right){/tex}, one banana {tex}(B){/tex}, and one mango ({tex}M{/tex}), the sample space is: {tex} S=\{(A, B),(A, M),\left(O_1, B\right),\left(O_1, M\right),{/tex} {tex}\left(O_2, B\right),\left(O_2, M\right)\}.{/tex} The size of the sample space is {tex}n(S)=6{/tex}.
To find the probability of picking one apple and one banana, we look at our sample space. There is only one outcome that matches this condition: {tex}(A, B){/tex}. This is our single favorable outcome. The total number of equally likely outcomes is six. Therefore, the probability is calculated by dividing the number of favorable outcomes (1) by the total number of outcomes (6). This results in a probability of {tex}1 / 6{/tex}, which is approximately 0.167 or {tex}16.7 \%{/tex}.
- (See the generated tree diagram image for the visualization)
- Sample Space (S): {tex}\{(A, B),(A, M),\left(O_1, B\right),\left(O_1, M\right),{/tex}{tex}\left(O_2, B\right),\left(O_2, M\right\}q{/tex}
- Probability of (Apple, Banana): {tex}1 / 6 \approx 0.167{/tex} (or 16.7%)

Q.15:

Let us say that you have a box containing 3 red pens, 4 black pens and 2 green pens. You pick a pen (without looking) from the box and put it back. Then your friend does the same.

What are the possible outcomes of the pen colours? Can you draw a tree diagram representing the possible outcomes?
Can you use the tree diagram to guess the probability that both you and your friend pick pens of the same colour?

Solution:

Given:

Total pens = 3 red + 4 black + 2 green = 9 pens
Since the pen is put back, both picks are independent events

Possible outcomes & Tree Diagram Possible colours:
- Red (R), Black (B), Green (G)
All possible outcomes (You, Friend): (R, R), (R, B), (R, G), (B, R), (B, B), (B, G), (G, R), (G, B), (G, G)
Tree Diagram (with probabilities)
Probability that both pick same colour {tex} P(\text {same colour})=P(R, R)+P(B, B)+P(G, G){/tex}
Calculate each: {tex}P(R, R)=\frac{3}{9} \times \frac{3}{9}=\frac{9}{81}{/tex}
{tex}P(B, B)=\frac{4}{9} \times \frac{4}{9}=\frac{16}{81}{/tex}
{tex}P(G, G)=\frac{2}{9} \times \frac{2}{9}=\frac{4}{81}{/tex}

Total probability:

{tex}P(\text {same})=\frac{9+16+4}{81}=\frac{29}{81}{/tex}

Q.16:

The probability of an impossible event is ________.

Solution:

Q.17:

The set of all possible outcomes of a random experiment is called the ________.

Solution:

sample space

Q.18:

The probability of an event that is certain to happen is ________.

Solution:

Q.19:

Tossing a fair coin has a probability of ________ for getting heads.

Solution:

0.5

Q.20:

In a survey of 50 students, 15 students said they liked football. The number of students who like football is 15, and the ________ (frequency/relative frequency) is ________ (fill in the fraction or decimal).

Solution:

relative frequency, 15/50

Q.21:

Which of the following experiments have equally likely outcomes? Explain.

A driver attempts to start a car. The car starts or does not start.
Tossing a fair coin once.
Rolling a fair 6-sided die.
Choosing a marble randomly from a bag that contains 3 red marbles and 7 blue marbles.
A baby is born. It is a boy or a girl.

Solution:

Describes an experiment influenced by external factors rather than pure randomness. A car starting depends on mechanical conditions like a functional battery or fuel levels. Because the probability of starting is usually much higher than not starting, these outcomes are not equally likely.
Involves a fair coin, which is physically symmetrical and balanced. This balance ensures that heads and tails each have exactly a {tex}50\%{/tex} chance of occurring. Since every outcome in the sample space has the same probability, they are considered equally likely.
Uses a fair six-sided die where each face is manufactured to be identical in size and weight. Each number from one to six has a {tex}1/6{/tex} chance of appearing. This uniform distribution makes all six outcomes equally likely.
Features a bag with an unequal distribution of marbles. With seven blue and only three red, you are more likely to pull a blue marble. Because {tex}0.7{/tex} is not equal to {tex}0.3{/tex}, these results are not equally likely.
Refers to the birth of a baby, which in probability theory is modeled as a 50-50 chance. While minor biological variations exist, for most calculations, the outcomes of boy or girl are treated as equally likely.

Q.22:

Write the sample space and calculate the probability based on the given information.

Two coins are tossed at the same time. What is the probability of getting at least one head?
Ten identical cards numbered 1 to 10 are placed in a box. One card is drawn at random. What is the probability of drawing a card with an even number?
A die is rolled once. What is the probability of getting a number greater than 4?
A bag contains 3 red balls, 2 blue balls, and 1 green ball. One ball is picked at random. What is the probability that it is not red?
Three coins are tossed simultaneously. What is the probability of getting exactly two heads?

Solution:

Two Coins Tossed The sample space is {tex}S=\{H H, H T, T H, T T\}{/tex}, giving four total outcomes. “At least one head” includes {tex}H H, H T{/tex}, and {tex}T H{/tex}. The probability is {tex}3 / 4{/tex} or {tex}{0 . 7 5}{/tex}.
Cards Numbered 1 to 10 The sample space is {tex}S=\{1,2,3,4,5,6,7,8,9,10\}{/tex}. Even numbers are {tex}\{2,4,6,8,10\}{/tex}. There are five favorable outcomes, so the probability is {tex}5 / 10=1 / 2{/tex} or {tex}{0 . 5}{/tex}.
Die Roll Greater Than 4 The sample space is {tex}S=\{1,2,3,4,5,6\}{/tex}. Numbers greater than four are {tex}\{5,6\}{/tex}. With two favorable outcomes out of six, the probability is {tex}2 / 6=1 / 3{/tex} or approximately 0.33.
Not a Red Ball Total balls are {tex}3+2+1=6{/tex}. “Not red” means picking blue or green balls ({tex}2+1=3{/tex}). The probability is {tex}3 / 6=1 / 2{/tex} or {tex}{0 . 5}{/tex}.
Three Coins Tossed The sample space has {tex}2^3=8{/tex} outcomes: {tex}\{H H H, H H T, H T H, T H H, {/tex}{tex}H T T, T H T, T T H, T T\}{/tex}. Outcomes with exactly two heads are {tex}\{H H T, H T H, T H H\}{/tex}. The probability is {tex}{3 / 8}{/tex} or {tex}{0 . 3 7 5}{/tex}.

Q.23:

A bag has 3 candies: strawberry, lemon, and mint. One is picked at random. What is the probability of picking a strawberry candy?

Solution:

The sample space {tex}S{/tex} contains three equally likely outcomes: strawberry, lemon, and mint. Since there is only one strawberry candy, the probability is {tex}1/3{/tex} or 33.3%.

Q.24:

A child has 2 shirts (one red and one blue) and 3 types of pants (jeans, khakis, and shorts). List all the possible combinations of outfits consisting of one shirt and one pair of pants. Display your answer in a table format.

Solution:

Shirt	Pants	Outfit Combination
Red	Jeans	(Red, Jeans)
Red	Khakis	(Red, Khakis)
Red	Shorts	(Red, Shorts)
Blue	Jeans	(Blue, Jeans)
Blue	Khakis	(Blue, Khakis)
Blue	Shorts	(Blue, Shorts)

Q.25:

A tyre company records distances before replacement in 1000 cases.

Distance (km)	Less than {tex}{4 0 0 0}{/tex}	{tex}{4 0 0 1}{/tex} to {tex}{9 0 0 0}{/tex}	{tex}{9 0 0 1}{/tex} to {tex}{1 4 0 0 0}{/tex}	More than {tex}{1 4 0 0 0}{/tex}
Number of cases	20	210	325	445

Find the probability that a randomly chosen tyre lasts:

Less than 4000 km.
Between 4000 and 14000 km.
More than 14000 km.

Solution:

There are 20 cases under 4000 km out of 1000 total, making the probability 20/1000 or 0.02.
Sum cases for 4001-9000 km and 9001-14000 km {tex}(210+325=535){/tex}, giving 535/1000 or {tex}{0. 5 3 5}{/tex}.
The table shows 445 cases lasting more than 14000 km, resulting in a probability of {tex}{4 4 5} / {1 0 0 0}{/tex} or {tex}{0. 4 4 5}{/tex}.

Q.26:

The letters of the word ‘PEACE’ are placed on cards. Leela draws a card without looking.

What is the probability that it is a P, E or C?
What is the probability that it is not an E?

Solution:

The total number of cards is 5. These cards are labeled with the letters: P, E, A, C, and E.

In (i), the cards {tex}{P}, {E}{/tex}, and C appear 4 times (one P, two E’s, and one C). The probability is 4/5 or 0.8.
In (ii), there are 2 E ‘s, so 3 cards are not {tex}{E}({P}, {A}, {C}){/tex}. The probability is {tex}3 / 5{/tex} or {tex}{0 . 6}{/tex}.

Q.27:

A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers {tex}1,2,3,4,5,6,7,8{/tex}, and these are equally likely outcomes. What is the probability that it will point at

8?
An odd number?
A number greater than 2?
A number less than 9 ?
A multiple of 3?

Solution:

The sample space contains 8 equally likely outcomes.

There is only one 8, so the probability is {tex}1 / 8{/tex} or {tex}{0. 1 2 5}{/tex}.
Odd numbers are {tex}1,3,5,7{/tex}. This gives 4 outcomes, making the probability {tex}4 / 8{/tex} or 0.5.
Numbers greater than 2 are {tex}3,4,5,6,7,8{/tex}. These 6 outcomes result in {tex}6 / 8{/tex} or 0.75.
All numbers are less than 9, so this certain event has a probability of {tex}8 / 8{/tex} or 1.
Multiples of 3 are 3 and 6. These 2 outcomes give a probability of {tex}2 / 8{/tex} or {tex}{0. 2 5}{/tex}.

Q.28:

A basket contains 4 red balls and 5 blue balls. One ball is drawn and laid aside, and a second ball is drawn. Draw a tree diagram to represent the possible outcomes and probabilities. Use the tree diagram to answer the following questions.

What is the probability of drawing a red ball and then a blue ball?
What is the probability of drawing 2 blue balls?

Solution:

The first draw has 9 balls, but the second draw has only 8 because the first ball is not replaced.

For (i), the probability is {tex}(4 / 9) \times(5 / 8)=20 / 72{/tex}, which simplifies to {tex}{5} / {1 8}{/tex} for a red then blue sequence.
For (ii), the probability is {tex}(5 / 9) \times(4 / 8)=20 / 72{/tex}, which also simplifies to {tex}5 / 18{/tex} for picking two consecutive blue balls.

Q.29:

I throw a pair of 6-sided dice. Write down an event that has a probability of 0 and an outcome that has a probability of 1.

Solution:

An event with a probability of 0 is rolling a total sum of 13. An outcome with a probability of 1 is rolling a sum between 2 and 12.

Q.30:

Write the sample space and calculate the probability based on the given information.

Two dice are rolled. What is the probability that the sum is a prime number greater than 5?
A bag contains 4 red, 3 green, and 2 blue balls. Two balls are drawn without replacement. What is the probability that both are of different colours?
Three coins are tossed. What is the probability that the first coin shows heads and exactly two heads occur in total?
A four-digit number is formed using the digits {tex}1,2,3{/tex}, and 4 with no repetition. What is the probability that the number is even?
A student takes a multiple-choice test with 3 questions, each having 4 options (A, B, C, D), with only one correct answer. What is the probability that the student guesses and gets exactly 2 answers correct?

Solution:

Sum is a Prime Number {tex}>5{/tex}: The sample space consists of 36 outcomes. Prime numbers greater than 5 are 7 and 11. Sums totaling 7 are {tex}(1,6),(2,5),(3,4),(4,3),(5,2),(6,1){/tex} and sums totaling 11 are {tex}(5,6),(6,5){/tex}. There are 8 favorable outcomes, so the probability is {tex}8 / 36=2 / 9{/tex} or {tex}{0. 2 2}{/tex}.
Two Different Colored Balls: Total balls equal 9. It is easier to calculate 1 minus the probability of same colors. Same color outcomes are Red-Red ({tex}4 / 9 \times 3 / 8{/tex}), Green-Green ({tex}3 / 9 \times 2 / 8{/tex}), and Blue-Blue ({tex}2 / 9 \times 1 / 8{/tex}). Total same color probability is {tex}20 / 72{/tex}. Therefore, different colors probability is {tex}1-20 / 72=52 / 72=13 / 18{/tex} or 0.72.
First Coin Heads and Exactly Two Heads: The sample space is {tex}\{H H H, H H T, H T H, H T T, {/tex}{tex}T H H, T H T, T T H, T T T\}{/tex}. Out of 8 total outcomes, the ones starting with Heads and having exactly two heads are {tex}H H T{/tex} and {tex}H T H{/tex}. With 2 favorable outcomes, the probability is {tex}2 / 8=1 / 4{/tex} or {tex}{0. 2 5}{/tex}.
Even Four-Digit Number: Total permutations of {tex}\{1,2,3,4\}{/tex} is {tex}4!=24{/tex}. For the number to be even, it must end in 2 or 4. If it ends in 2, there are 3 ! (6) ways; if it ends in 4, there are 3 ! (6) ways. Total even numbers are 12. The probability is {tex}12 / 24=1 / 2{/tex} or {tex}{0. 5}{/tex}.
Exactly 2 Correct Guesses: For each question, probability of correct {tex}(p){/tex} is {tex}1 / 4{/tex} and wrong {tex}(q){/tex} is {tex}3 / 4{/tex}. Using binomial logic, there are 3 ways to get two right (CCW, CWC, WCC). Probability is {tex}3 \times(1 / 4 \times 1 / 4 \times 3 / 4)=9 / 64{/tex} or {tex}{0. 1 4}{/tex}.

Q.31:

A box contains 4 balls numbered 1 to 4. Record a sample space using a tree diagram for the following experiments:

A ball is drawn, and the number is recorded. Then the ball is returned, and a second ball is drawn and recorded.
A ball is drawn and recorded. Without replacing the first ball, the experimenter draws and records a second ball.
What are the sizes of these two sample spaces?

Solution:

In (i), each of the four initial branches (1, 2, 3, 4) splits into four identical branches because the first ball is replaced.

In (ii), each initial branch splits into only three branches because the ball already drawn cannot be selected a second time.

In (iii), the first sample space contains {tex}4 \times 4=16{/tex} outcomes, while the second contains {tex}4 \times 3=12{/tex} distinct outcomes.

Q.32:

List the elements of a sample space for the simultaneous tossing of a coin and drawing of a card from a set of 6 cards numbered 1 through 6.

Solution:

The sample space contains 12 pairs representing each coin face combined with each card number.

{tex}S={/tex} {tex}\{(H, 1),(H, 2),(H, 3),(H, 4),(H, 5),(H, 6),{/tex} {tex}(T, 1),(T, 2),(T, 3),(T, 4),(T, 5),(T, 6)\}{/tex}

Q.33:

Three coins are tossed, and the number of heads is recorded. Which of the following lists is a sample space for this experiment? Why do the other lists fail to qualify as a sample space?

{tex}\{1,2,3\}{/tex}
{tex}\{0,1,2\}{/tex}
{tex}\{0,1,2,3,4\}{/tex}
{tex}\{0,1,2,3\}{/tex}

Solution:

(iv) {tex}\{0,1,2,3\}{/tex} is the correct sample space because it lists all possible head counts from zero to three.
(i) fails because it omits the possibility of zero heads, which occurs if all three coins show tails.
(ii) fails because it omits the possibility of three heads, which occurs if all coins show heads.
(iii) fails because it includes four heads, which is impossible since only three coins are tossed in total.

Q.34:

Suppose you drop a dye at random on the rectangular region shown in figure. What is the probability that it will land inside the circle with a diameter of 1 m?

Solution:

The total area of the rectangle is {tex}3 {~m} \times 2 {~m}=6 \ {m}^2{/tex}, which represents the entire sample space.

Since the circle’s diameter is 1 m, its radius is 0.5 m.
The area of the circle is {tex}\pi \times(0.5)^2= 0.25 \pi \ {m}^2{/tex}.
The probability is the ratio of the circular area to the total rectangular area:

{tex} P=\frac{0.25 \pi}{6}=\frac{\pi}{24} \approx 0.1309{/tex}