Courses
Courses for Kids
Free study material
Free LIVE classes
More

Maharashtra Board Class 10 Solutions for Maths Chapter 5: Probability

Download Free PDF with Solutions of Probability Class 10, Chapter 5

Last updated date: 17th Mar 2023
Total views: 38.4k
Views today: 0.18k

Probability is an essential chapter for Class 10 students in the Maharashtra Board. The chapter introduces students to the concept of Probability and followed with solved examples and practice questions. In this chapter, students will be able to learn all the methods to solve questions on Probability. In order to understand the chapter in detail, refer to the PDF where solutions can be downloaded for the chapter.

These solutions for Maharashtra Board Class 10 Maths Chapter 5 Probability have been carefully designed by the subject matter experts at Vedantu who are completely well-versed with the essential topics and sub-topics included in the chapter. Downloading the Probability Chapter 5 solutions will help students gain a better understanding of the chapter.

Competitive Exams after 12th Science
Do you need help with your Homework? Are you preparing for Exams?
Study without Internet (Offline)

Access Maharashtra Board solutions for Maths class 10 chapter 5 Probability

Practice set  5.1

1. How many possibilities are there in each of the following?

1. Vanita knows the following sites in Maharashtra. She is planning to visit one of them in her summer vacation. Ajintha, Mahabaleshwar, Lonar Sarovar, Tadoba wild life sanctuary, Amboli, Raigad, Matheran, Anandavan.

2. Any day of a week is to be selected randomly.

3. Select one card from the pack of $52$ cards.

4. One number from $10$ to $20$ is written on each card. Select one card randomly.

Ans:

1. Vanita knows the following sites in Maharashtra. She is planning to visit one of them during her summer vacation. Ajintha, Mahabaleshwar, Lonar Sarovar, Tadoba wildlife sanctuary, Amboli, Raigad, Matheran, Anandavan.

So, given sites are $8$ then the possibilities are $8$.

1. Any day of a week is to be selected randomly, so in a week there are $7$ days thus, there are $7$ possibilities.

2. Select one card from the pack of $52$ cards, hence there are $52$ possibilities are there are $52$ cards available.

3. One number from $10-20$ is written on each card. Select one card randomly so, the possibility is $11$ as there are total $11$ ways to select a card randomly.

Practice set 5.2

1. For each of the following experiments write sample space ‘S’ and number of sample points $n\left( S \right)$.

1. One coin and one die are thrown simultaneously .

2. Two-digit numbers are formed using digits $2,3\And 5$ without repeating a digit.

Ans:

1. One coin and one die are thrown simultaneously

So, sample space will be $\Rightarrow S=\left\{ 1H,1T,2H,2T,3H,3T,4H,4T,5H,5T,6H,6T \right\}$

We calculated the sample space by using the formula $n\left( E_1 \right) \times n\left( E_2 \right)$.

Number of events in sample space $S$ will be $n\left( S \right)=12$.

1. Two-digit numbers are formed using digits $2,3\And 5$ without repeating a digit.

calculate the sample by using the formula $n\left( E_1 \right) \times n\left( E_2 \right)$.

So, sample space will be $S=\left\{ 23,25,32,35,52,53 \right\}$.

Number of events in sample space $S$, $n\left( S \right)=6$.

2. The arrow is rotated and it stops randomly on the disc. Find out on which color it may stop.

(Image will be uploaded soon)

Ans: As per the question it is given that the arrow is rotated and it stops randomly on the disc.

It has six options so it can be stopped at any one.

So, sample space, $\Rightarrow S=\left\{ \operatorname{Re}d,purple,orange,yellow,blue,green \right\}$

Number of events in sample space we can see are $6$thus, $n\left( S \right)=6$.

3. In the month of March $2019$, find the days on which the date is a multiple of $5$.

(Image will be uploaded soon)

Ans: First write all the multiples of $5$so, multiples of $5$are $5,10,15,20,25,30$.

Thus,

• Tuesday: $5-mar-2019$

• Sunday: $10-mar-2019$

• Friday: $15-mar-2019$

• Wednesday: $20-mar-2019$

• Monday: $25-mar-2019$

• Saturday: $30-mar-2019$

4. Form a ‘Road safety committee of two, from $2$ boys $\left( B_1,B_2 \right)$ and $2$ girls $\left( G_1,G_2 \right)$. Complete the following activity to write the sample space.

1. Committee of $2$ boys $=?$

2. Committee of$2$ girls $=?$

3. committee of $1$ boy and $1$ girl is equal to $B_1G_1,?,?,?$

4. Sample space $S=\left\{ ....,....,.....,...,...,..., \right\}$

Ans:

1. As it is clearly seen, the committee of $2$ boys will be $B_1, B_2$.

2. Committee of $2$ girls will be $G_1,G_2$.

3. Committee of $1$ boy and $1$ girl will be $\left( B_1G_1,B_1G_2,B_2G_1,B_2G_2 \right)$ any event can be possible.

4. Sample space $S=\{B_1G_1,B_1G_2,B_2G_1,B_2G_2\}$

Practice set 5.3

1. Write sample space ‘S’ and number of sample points $n\left( S \right)$ for each of the following experiments. Also write events A,B,C in the set form and write $n\left( A \right), n\left( B \right), n\left( C \right).$

1. One die is rolled.

Event A: even number on the upper face.

Event B: odd number on the upper face.

Event C: prime number on the upper face.

1. Two dice are rolled.

Event A: The sum of the digits on upper faces is a multiple of $6$.

Event B: The sum of the digits on the upper faces is minimum $10$.

Event C: The same digit on both the upper faces.

1. Three coins are tossed simultaneously.

Condition for event A: To get at least two heads.

Condition for event B: To get no head.

Condition for event C: To get head on the second coin.

1. Two digit numbers are formed using digits $0,1,2,3,4,5$ without repetition of the digits.

Condition for event A: The number formed is even

Condition for event B: The number formed is divisible by $3$.

Condition for event C: The number formed is greater than $50$.

1. From three men and two women, an environment committee of two persons is to be formed.

Condition for event A: There must be at least one woman member.

Condition for event B: One man, one woman committee to be formed.

Condition for event C: There should not be a woman member.

1. One coin and one die are thrown simultaneously.

Condition for event A: To get head and an odd number.

Condition for event B: To get a head or tail and an even number.

Condition for event C: Number on the upper face is greater than $7$ and the tail on the coin.

Ans:

A. One die is rolled.

• Event A: even number on the upper face.

• Event B: odd number on the upper face.

• Event C: prime number on the upper face.

So, sample space $\Rightarrow S=\left\{ 1,2,3,4,5,6 \right\}$, number of sample point will be $n\left( S \right)=6$.

$A=\left\{ 2,4,6 \right\},n\left( A \right)=3$

$B=\left\{ 1,3,5 \right\},n\left( B \right)=3$

$C=\left\{ 2,3,5 \right\},n\left( C \right)=3$

B. Two dice are rolled.

• Event A: The sum of the digits on upper faces is a multiple of $6$.

• Event B: The sum of the digits on the upper faces is minimum $10$.

• Event C: The same digit on both the upper faces.

So, sample space will be

$S=\left\{ \left( 1,1 \right)\left( 1,2 \right)\left( 1,3 \right)\left( 1,4 \right)\left( 1,5 \right)\left( 1,6 \right) \left( 2,1 \right)\left( 2,2 \right)\left( 2,3 \right)\left( 2,4 \right)\left( 2,5 \right)\left( 2,6 \right) \left( 3,1 \right)\left( 3,2 \right)\left( 3,3 \right)\left( 3,4 \right)\left( 3,5 \right)\left( 3,6 \right) \left( 4,1 \right)\left( 4,2 \right)\left( 4,3 \right)\left( 4,4 \right)\left( 4,5 \right)\left( 4,6 \right) \left( 5,1 \right)\left( 5,2 \right)\left( 5,3 \right)\left( 5,4 \right)\left( 5,5 \right)\left( 5,6 \right) \left( 6,1 \right)\left( 6,2 \right)\left( 6,3 \right)\left( 6,4 \right)\left( 6,5 \right)\left( 6,6 \right)\right\}$

Thus, number of sample points $n\left( S \right)=36$

$A=\left\{ \left( 1,5 \right)\left( 2,4 \right)\left( 3,3 \right)\left( 4,2 \right)\left( 5,1 \right)\left( 6,6 \right) \right\},n\left( A \right)=6$

$B=\left\{ \left( 4,6 \right)\left( 5,5 \right)\left( 5,6 \right)\left( 6,4 \right)\left( 6,5 \right)\left( 6,6 \right) \right\},n\left( B \right)=6$

$C=\left\{ \left( 1,1 \right)\left( 2,2 \right)\left( 3,3 \right)\left( 4,4 \right)\left( 5,5 \right)\left( 6,6 \right) \right\},n\left( C \right)=6$

C. Three coins are tossed simultaneously.

• Condition for event A: To get at least two heads.

• Condition for event B: To get no head.

• Condition for event C: To get head on the second coin.

Sample space $S=\left\{ HHH,HHT,HTT,HTH,THT,TTH,THH,TTT \right\}$

Thus, number of sample points $n\left( S \right)=8$.

$A=\left\{ HHH,HTH,HHT,THH \right\},n\left( A \right)=4$

$B=\left\{ TTT \right\},n\left( B \right)=1$

$C=\left\{ HHH,HHT,THH \right\},n\left( C \right)=3$

D. Two digit numbers are formed using digits $0,1,2,3,4,5$ without repetition of the digts.

• Condition for event A: The number formed is even

• Condition for event B: The number formed is divisible by $3$.

• Condition for event C: The number formed is greater than $50$.

Sample space $S=\left\{10,12,13,14,15,20,21,23,24,25,30,31,32,34,35,40,41,42,43,45,50,51,52,53,54 \right\}$

Number of sample points $n\left( S \right)=25$

$A=\left\{ 10,12,14,20,24,30,32,34,40,42,50,52,54 \right\},n\left( A \right)=13$

$B=\left\{ 12,15,21,24,30,42,45,51,54 \right\},n\left( B \right)=9$

$C=\left\{ 51,52,53,54 \right\},n\left( C \right)=4$

2. From three men and two women, an environment committee of two persons is to be formed.

• Condition for event A: There must be at least one woman member.

• Condition for event B: One man, one woman committee to be formed.

• Condition for event C: There should not be a woman member.

So, sample space will be $\Rightarrow S=\left\{{{M}_{1}}{{M}_{2}},{{M}_{1}}{{M}_{3}},{{M}_{1}}{{F}_{1}},{{M}_{1}}{{F}_{2}},{{M}_{2}}{{M}_{3}},{{M}_{2}}{{F}_{1}},{{M}_{2}}{{F}_{2}},{{M}_{3}}{{F}_{1}},{{M}_{3}}{{F}_{2}},{{F}_{1}}{{F}_{2}}\right\}$

Thus, number of sample points is $n\left( S \right)=10$

$A=\left\{ {{M}_{1}}{{F}_{1}},{{M}_{1}}{{F}_{2}},{{M}_{2}}{{F}_{1}},{{M}_{2}}{{F}_{2}},{{M}_{3}}{{F}_{1}},{{M}_{3}}{{F}_{2}},{{F}_{1}}{{F}_{2}} \right\},n\left( A \right)=7$

$B=\left\{ {{M}_{1}}{{F}_{1}},{{M}_{1}}{{F}_{2}},{{M}_{2}}{{F}_{1}},{{M}_{2}}{{F}_{2}},{{M}_{3}}{{F}_{1}},{{M}_{3}}{{F}_{2}} \right\},n\left( B \right)=6$

$C=\left\{ {{M}_{1}}{{M}_{2}},{{M}_{1}}{{M}_{3}},{{M}_{2}}{{M}_{3}} \right\},n\left( C \right)=3$

A. One coin and one die are thrown simultaneously.

• Condition for event A: To get head and an odd number.

• Condition for event B: To get a head or tail and an even number.

• Condition for event C: Number on the upper face is greater than $7$ and tail on the coin.

So, sample space will be $\Rightarrow \left\{ H1,H2,H3,H4,H5,H6,T1,T2,T3,T4,T5,T6 \right\}$

Thus, number of sample points is $n\left( S \right)=12$

$A=\left\{ H1,H3,H5 \right\},n\left( A \right)=3$

$B=\left\{ H2,H4,H6,T2,T4,T6 \right\},n\left( B \right)=6$

As, number on the upper face is greater than $7$ so, we all know that a dice does not have numbers greater than or equal to $7$. So, $n\left( C \right)=0$.

Practice set 5.4

1. If two coins are tossed, find the probability of the following events.

1. Getting at least one head

2. Getting no head

Ans:

1. As per the question given two coins are tossed

So, we all know that the formula for finding probability is

$\Rightarrow P=\dfrac{\text{favorable outcomes}}{\text{total number of outcomes}}$

Thus, sample space will be $S=\left\{HH,HT,TH,TT \right\}$

Favorable outcomes is $3$ which are $\left\{ HH,HT,TH \right\}$

So, the probability of getting at least one head is $p\left( A \right)=\dfrac{3}{4}$.

2. As per the question given, getting no head.

So, we all know that the formula for finding probability is

$\Rightarrow P=\dfrac{\text{favorable outcomes}}{\text{total number of outcomes}}$

Thus, sample space will be $S=\left\{HH,HT,TH,TT \right\}$

Favorable outcome is $1$ which are $\left\{TT\right\}$

So, the probability of getting no heads is $p\left( B \right)=\dfrac{1}{4}$.

2. If two dice are rolled simultaneously, find the probability of the following events.

1. The sum of the digits on the upper face is at least $10$.

2. The sum of the digits on the upper face is $33$.

3. The digit on the first die is greater than the digit on the second die.

Ans:

1. As per the question given the sum of the digits on the upper face is at least $10$

So, probability of getting the sum of the digits on the upper faces is at least $10$

$P(A)=\dfrac{\text{favorable outcomes}}{\text{total number of outcomes}}$

The sample space will be $\Rightarrow S=\{ \left( 1,1 \right)\left( 1,2 \right)\left( 1,3 \right)\left( 1,4 \right)\left( 1,5 \right)\left( 1,6 \right) \left( 2,1 \right)\left( 2,2 \right)\left( 2,3 \right)\left( 2,4 \right)\left( 2,5 \right)\left( 2,6 \right)\left( 3,1 \right)\left( 3,2 \right)\left( 3,3 \right)\left( 3,4 \right)\left( 3,5 \right)\left( 3,6 \right) \left( 4,1 \right)\left( 4,2 \right)\left( 4,3 \right)\left( 4,4 \right)\left( 4,5 \right)\left( 4,6 \right)\left( 5,1 \right)\left( 5,2 \right)\left( 5,3 \right)\left( 5,4 \right)\left( 5,5 \right)\left( 5,6 \right) \}$

favorable outcome will be $6$ which are $\left\{ \left( 4,6 \right)\left( 5,5 \right)\left( 5,6 \right)\left( 6,4 \right)\left( 6,5 \right)\left( 6,6 \right) \right\}$.

And the total number of outcomes will be $36$

$p\left( A \right)=\dfrac{6}{36}$

$p\left( A \right)=\dfrac{1}{6}$

Hence, the probability of getting the sum of the digits on the upper face is at least $10$ will be $\dfrac{1}{6}$.

2. As per the question, the sum of the digits on the upper face is $33$.

So, probability of getting the sum of the digits on the upper face is $33$

As we all know that the maximum sum on dice could be $12$, so, the favorable outcome is $0$.

thus the probability will be $p\left( B \right)=0$.

Hence, the probability of getting the sum of the digits on the upper face is $33$ will be $0$.

3. As per the question, the digit on the first die is greater than the digit on the second die.

Thus, the probability of getting the digit on the first die is greater than the digit on the second die.

The sample space will be $\Rightarrow S=\{ \left( 1,1 \right)\left( 1,2 \right)\left( 1,3 \right)\left( 1,4 \right)\left( 1,5 \right)\left( 1,6 \right) \left( 2,1 \right)\left( 2,2 \right)\left( 2,3 \right)\left( 2,4 \right)\left( 2,5 \right)\left( 2,6 \right) \left( 3,1 \right)\left( 3,2 \right)\left( 3,3 \right)\left( 3,4 \right)\left( 3,5 \right)\left( 3,6 \right) \left( 4,1 \right)\left( 4,2 \right)\left( 4,3 \right)\left( 4,4 \right)\left( 4,5 \right)\left( 4,6 \right) \left( 5,1 \right)\left( 5,2 \right)\left( 5,3 \right)\left( 5,4 \right)\left( 5,5 \right)\left( 5,6 \right)\}$

Favorable outcomes will be

$\left( 2,1 \right),\left( 3,1 \right),\left( 3,2 \right),\left( 4,1 \right),\left( 4,2 \right),\left( 4,3 \right),\left( 5,1 \right),\left( 5,2 \right),\left( 5,3 \right),\left( 5,4 \right), \left( 6,1 \right),\left( 6,2 \right),\left( 6,3 \right),\left( 6,4 \right),\left( 6,5 \right)$

Favorable outcomes = 15

So, $P(C)=\dfrac{\text{favorable outcomes}}{\text{total number of outcomes}}$

$p\left( C \right)=\dfrac{15}{36}$

$p\left( C \right)=\dfrac{5}{12}$

Hence, the probability of getting the digit on the first die is greater than the digit on the second die is $\dfrac{5}{12}$.

3. There are $15$ tickets in a box, each bearing one of the numbers from $1-15$One ticket is drawn at random from the box. Find the probability of event that the ticket drawn –

1. Shows an even number

2. Shows a number which is multiple of $5$.

Ans:

1. As per the question it is given that one ticket is drawn at random, even number between $1-15$ are $\left( 2,4,6,8,10,12,14 \right)=7$

And the sample space will be $\Rightarrow S=\left\{ 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15 \right\}=15$

Formula to find probability $p\left( E \right)=\dfrac{\text{favorable outcomes}}{\text{total number of outcomes}}$

So the probability of the ticket showing an even number is $p\left( E \right)=\dfrac{7}{15}$.

2. As per the question it is given that one ticket is drawn at random, so number which are multiple of $5$ between $1-15$ are $\left( 5,10,15 \right)=3$, thus probability of getting a number multiple of $5$ is $\Rightarrow \dfrac{3}{15}=\dfrac{1}{5}$.

4. A two digit number is formed with digits $2,3,5,7,9$ without repetition. What is the probability that the number formed is:

1. An odd number

2. A multiple of $5$

Ans:

1. A two digit number is formed with digits $2,3,5,7,9$ without repetition.

So, the sample space of will be $\Rightarrow S=\left\{ 23,25,27,29,32,35,37,29,52,53,57,59,72,73,75,79,92,93,95,97 \right\}=20$

Favorable outcome is $\Rightarrow \left\{ 23,25,27,29,35,37,39,53,57,59,73,75,79,93,95,97 \right\}=16$

$p\left( E \right)=\dfrac{\text{favorable outcomes}}{\text{total number of outcomes}}$

Hence, the probability is $\Rightarrow \dfrac{16}{20}=\dfrac{4}{5}$.

2. A two digit number is formed with digits $2,3,5,7,9$ without repetition.

Sample space will be $\Rightarrow S=\left\{ 23,25,27,29,32,35,37,29,52,53,57,59,72,73,75,79,92,93,95,97 \right\}=20$

Favorable outcome is $\Rightarrow \left\{ 25,35,75,95 \right\}$

$p\left( E \right)=\dfrac{\text{favorable outcomes}}{\text{total number of outcomes}}$

Hence, the probability is $\Rightarrow \dfrac{4}{20}=\dfrac{1}{5}$.

5. A card is drawn at random from a pack of well shuffled $52$ playing cards. Find the probability that the card drawn is.

1. An ace

Ans:

1. As per the question given $52$ cards so, in a deck of there are $4$ ace thus, probability of getting an ace is so, favorable outcome is $4$ and total outcome is $52$.

Formula to find probability is $p\left( A \right)=\dfrac{\text{favorable outcomes}}{\text{total number of outcomes}}$

$p\left( A \right)=\dfrac{4}{52}$

$p\left( A \right)=\dfrac{1}{13}$

Hence, the answer is $\dfrac{1}{13}$

2. As per the question given $52$ cards so, in a deck of there are $26$ black cards in which $13$ are spade and $13$ are club cards, favorable outcome is $13$ and total outcome is $52$.

Formula to find probability is $p\left( S \right)=\dfrac{\text{favorable outcomes}}{\text{total number of outcomes}}$.

$p\left( S \right)=\dfrac{13}{52}$ Hence, the probability is $\dfrac{1}{4}$.

Problem set 5

1. Choose the correct alternative answer for each of the following questions.

1. Which number cannot represent a probability?

1. $\dfrac{2}{3}$

2. $1.5$

3. $15\%$

4. $0.7$

Ans: We all know that the probability of any event lies between $0-1$. So, the correct option is (B) that is $1.5$, which is greater than $1$.

2. A die is rolled. What is the probability that the number appearing on the upper face is less than $3$?

1. $\dfrac{1}{6}$

2. $\dfrac{1}{3}$

3. $\dfrac{1}{2}$

4. $0$

Ans: First, write the favorable outcomes that is $\left\{ 1,2 \right\}=2$,

And the total number of outcomes if a single dice is thrown is $\Rightarrow S=\left\{ 1,2,3,4,5,6 \right\}$

To find the probability use $p=\dfrac{\text{favorable outcomes}}{\text{total number of outcomes}}$

So, the probability will be

$p=\dfrac{2}{6}$

$\Rightarrow p=\dfrac{1}{3}$

Hence, the correct option is (B) that is $\dfrac{1}{3}$.

3. What is the probability of the event that a number chosen from $1-100$is a prime number?

1. $\dfrac{1}{5}$

2. $\dfrac{6}{25}$

3. $\dfrac{1}{4}$

4. $\dfrac{13}{50}$

Ans: Firstly, write the favorable outcomes so,

$\{ 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97\}=24$

And total number of outcomes is $100$

Formula to find the probability is $p=\dfrac{\text{favorable outcomes}}{\text{total number of outcomes}}$

$p=\dfrac{24}{100}$

$p =\dfrac{6}{25}$

Hence, the total probability of event that a number chosen from $1-100$ is a prime number is $\dfrac{6}{25}$ and the correct option is (B).

4. There are $40$ cards in a bag. Each bears a number from $1-40$. One card is drawn at random. What is the probability that the card bears a number which is a multiple of $5$?

1. $\dfrac{1}{5}$

2. $\dfrac{3}{5}$

3. $\dfrac{4}{5}$

4. $\dfrac{1}{3}$

Ans: first find out the numbers which are multiple of $5$ so, $\left\{ 5,10,15,20,25,30,35,40 \right\}=8$, thus the favorable outcomes are $8$.

to find probability is $p=\dfrac{\text{favorable outcomes}}{\text{total number of outcomes}}$

Total outcomes are $40$ so probability that the card bears a number which is a multiple of $5$ will be $\Rightarrow \dfrac{8}{40}=\dfrac{1}{5}$.

hence, option (A) is correct.

5. If $n\left( A \right)=2,p\left( A \right)=\dfrac{1}{5}$, then $n\left( S \right)=?$.

1. $10$

2. $\dfrac{5}{2}$

3. $\dfrac{2}{5}$

4. $\dfrac{1}{3}$

Ans: Applying the probability formula

$p\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}$

$\Rightarrow \dfrac{1}{5}=\dfrac{2}{n\left( S \right)}$

$\Rightarrow n\left( S \right)=10$

Hence, the correct option is (A) that is $10$.

1. Basketball players John, Vasim, Akash were practicing the ball drop in the basket. The probabilities of success for John, Vasim and Akash are $\dfrac{4}{5},0.83$ and $58\%$ respectively. Who had the greatest probability of success?

Ans: As per the question, probabilities of success for John, Vasim and Akash are $\dfrac{4}{5},0.83$ and $58\%$ respectively.

So, probability of success for John is $\dfrac{4}{5}=0.8$

Probability of success for Vasim is $0.83$.

Probability of success for Akash is $\Rightarrow \dfrac{58}{100}=0.58$.

Hence, the probability of success is highest for Vasim.

2. In a hockey team there are $6$ defenders, $4$ offenders and $1$ goalie. Out of these, one player is to be selected randomly as a captain. Find the probability of the selection that –

1. The goalee will be selected.

2. The defender will be selected.

Ans:

1. Probability of selection that the goalee will be, favorable outcome is $1$ and total outcome is $11$ using the formula $p\left( G \right)=\dfrac{\text{favorable outcomes}}{\text{total number of outcomes}}$

$\Rightarrow p\left( G \right)=\dfrac{1}{11}$.

2. Probability of the selection that the defender will be, favorable outcome is $6$and total outcome is $11$.

Using the formula $p\left( D \right)=\dfrac{\text{favorable outcomes}}{\text{total number of outcomes}}$

$\Rightarrow p\left( D \right)=\dfrac{6}{11}$.

4. Joseph kept $26$cards in a cap, bearing one English alphabet on each card. One card is drawn at random. What is the probability that the card drawn is a vowel card ?

Ans: As it is given that one card is drawn at random.

Sample space will be $\Rightarrow S=\left\{ A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z \right\}=26$

Vowel in English alphabet are $\left\{ A,E,I,O,U \right\}=5$

Using the formula $p=\dfrac{\text{favorable outcomes}}{\text{total number of outcomes}}$

So, the probability of the selection that the card drawn is a vowel card is $p\left( V \right)=\dfrac{5}{26}$.

5. A balloon vendor has $2$ red, $3$ blue and $4$ green balloons. He wants to choose one of them at random to give it to Pranali. What is the probability of the event that Pranali gets,

1. A red balloon

2. A blue balloon

3. A green balloon

Ans: Probability will be find by using $\Rightarrow p=\dfrac{\text{favorable outcomes}}{\text{total number of outcomes}}$

Total number of outcomes will be $\Rightarrow 2+3+4=9$.

a. Probability that Pranali gets a red balloon, favorable outcome will be $2$ so, $p\left( R \right)=\dfrac{2}{9}$

b. Probability that Pranali gets a blue balloon, favorable outcome will be $3$ so,

$p\left( B \right)=\dfrac{3}{9}$

$p\left( B \right) =\dfrac{1}{3}$

c. Probability that Pranali gets a green balloon, favorable outcome will be $4$ so, $p\left( G \right)=\dfrac{4}{9}$.

6. A box contains $5$ red, $8$ blue and $3$ green pens. Rutuja wants to pick a pen at random. What is the probability that the pen is blue?

Ans: To find probability use $p=\dfrac{\text{favorable outcomes}}{\text{total number of outcomes}}$

Total number of outcome will be $5+8+3=16$

Probability that Rutuja picks blue pen, favorable outcome will be $8$ so,

$p\left( B \right)=\dfrac{8}{16} =\dfrac{1}{2}$

Hence, the probability is $\dfrac{1}{2}$.

7.Six faces of a die are as shown below.

(Image will be updated soon)

If the die is rolled once, find the probability of –

1.  ‘A’ appears on upper face.

2. ‘D’ appears on upper face.

Ans: To find probability use $p=\dfrac{\text{favorable outcomes}}{\text{total number of outcomes}}$

Sample space will be $\Rightarrow S=\left\{ A,B,C,D,E,A \right\}=6$

A. We can see that A is appearing two times then the favorable outcome is $2$.

So, probability of getting A on upper face will be

$p\left( A \right)=\dfrac{2}{6} =\dfrac{1}{3}$

B. D is appearing single time so, the probability of getting D on the upper face will be $p\left( D \right)=\dfrac{1}{6}$.

8.A box contains $30$ tickets, bearing only one number from $1$ to $30$ on each. If one ticket is drawn at random, find the probability of an event that the ticket drawn bears.

1. an odd number

2. a complete square number.

Ans:To find probability use $p=\dfrac{\text{favorable outcomes}}{\text{total number of outcomes}}$

Sample space will be $S=\left\{ 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30 \right\}$

n(S) = 30

1. Odd numbers between $1$ to $30$ are $\left( 1,3,5,7,9,11,13,15,17,19,21,23,25,27,29 \right)=15$

favorable outcome will be $15$

So, probability of getting an odd number if one ticket is drawn at random will be

$\Rightarrow p\left( O \right)=\dfrac{15}{30} =\dfrac{1}{2}$

2. Complete squares between $1$ to $30$are $\left\{ 1,4,9,16,25 \right\}=5$.

The favorable outcome will be $5$.

So, probability of getting a complete square if one ticket is drawn at random will be

$\Rightarrow p\left( S \right)=\dfrac{5}{30} =\dfrac{1}{6}$

9. Length and breadth of a rectangular garden are $77m$ and$50m$. There is a circular lake in the garden having diameter$14m$. Due to wind, a towel from a terrace on a nearby building fell into the garden. Then find the probability of the event that it fell in the lake.

(image will be updated soon)

Ans: first find the area of lake by applying $\pi {{r}^{2}}$,

So, the area of lake will be

$A = \dfrac{22}{7}\times \dfrac{14}{2}\times \dfrac{14}{2} = 154{{m}^{2}}$

Now, find the area of the garden so, area of garden= length $\times$ breadth.

$Area = 77\times 50 = 3850{{m}^{2}}$

To find probability use $p=\dfrac{\text{favorable outcomes}}{\text{total number of outcomes}}$

So, total number of outcome will be and favorable outcome will be $154$

Hence, the probability of fell in lake will be

$p\left( E \right)=\dfrac{154}{3850} =\dfrac{1}{25}$

10.In a game of chance, a spinning arrow comes to rest at one of the numbers$1,2,3,4,5,6,7,8$.

All these are equally likely outcomes. Find the probability that it will rest at

1. $8$

2. An odd number

3. A number greater then $2$

4. A number less than $9$

Ans: To find probability use $p=\dfrac{\text{favorable outcomes}}{\text{total number of outcomes}}$

So, sample space will be $\Rightarrow S=\left\{ 1,2,3,4,5,6,7,8 \right\}=8$

1. Clearly the probability of rest at $8$ will be $p=\dfrac{1}{8}$.

2. Total odd numbers are $1,3,5,7=4$, so favorable outcomes are $4$. Thus, $\Rightarrow \dfrac{4}{8}=\dfrac{1}{2}$.

3. Numbers greater than $2$ are $3,4,5,6,7,8=6$. Thus, favorable outcome will be $6$ probability of will rest at a number greater than $2$ will be

$\Rightarrow p=\dfrac{6}{8} =\dfrac{3}{4}$

4. Number less than $9$ are $1,2,3,4,5,6,7,8=8$ so, favorable outcome will be $8$, probability of will rest at a number less than $9$will be

$\Rightarrow p=\dfrac{8}{8} =1$

11. There are six cards in a box, each bearing a number from $0$ to $5$. Find the probability of each of the following events, that a card drawn shows.

1. A  natural number

2. A number less than $1$

3. A whole number

4. A number is greater than $5$.

Ans: To find probability use $p=\dfrac{\text{favorable outcomes}}{\text{total number of outcomes}}$

Sample space will be $\Rightarrow S=\left\{ 1,2,3,4,5 \right\}=5$

1. Natural numbers between $0$ to $5$ are $1,2,3,4,5=5$, thus, a favorable outcome will be $5$.

So, the probability of getting a natural number is $p\left( N \right)=\dfrac{5}{6}$.

2. Number less than $1$ is $0$ so, probability of getting a number less than $1$ will be $\Rightarrow p=\dfrac{0}{5} = 0$

3. Whole numbers between $0$ to $5$ are $0,1,2,3,4,5$.

Thus, the favorable outcome will be $6$.

So, probability of getting a whole number is

$p\left( W \right)=\dfrac{6}{6} =1$

4. A number greater than $5$ is none, so the probability of getting a number greater than $5$ will be $0$.

12.A bag contains $3$ red, $3$ white and $3$ green balls. One ball is taken out of the bag at random. What is the probability that the ball drawn is

1. Red

2. Not red

3. Either red or white

Ans: To find probability use $p=\dfrac{\text{favorable outcomes}}{\text{total number of outcomes}}$

sample space will be $\Rightarrow S=\left\{ R,R,R,W,W,W,G,G,G \right\}=9$

1. Probability of getting a red ball, favorable outcome is $3$ thus,

$\Rightarrow p\left( R \right)=\dfrac{3}{9}=\dfrac{1}{3}$

2. Probability of getting not red so, $3$ white and $3$ green are $3+3=6$, favorable outcome is $6$.

so, probability of getting no red is

$\Rightarrow p=\dfrac{6}{9} =\dfrac{2}{3}$

3. Probability of getting either red or white so $3$ red and $3$ white will be $3+3=6$.

Thus, the favorable outcome will be $6$.

So, probability of getting either red or white will be

$\Rightarrow p\left( E \right)=\dfrac{6}{9} =\dfrac{2}{3}$

13.Each card bears one letter from the word ‘mathematics’ The cards are placed on a table upside down. Find the probability that a card drawn bears the letter ‘m’.

Ans: To find probability use, $p=\dfrac{\text{favorable outcomes}}{\text{total number of outcomes}}$

Sample space will be $\Rightarrow S=\left\{ MATHEMATICS \right\}=11$

As ‘m’ appears two times in mathematics, a favourable outcome will be $2$.

Probability that a card drawn bears the letter ‘m’ will be $p\left( M \right)=\dfrac{2}{11}$.

14.Out of $200$ students from a school, $135$ like Kabbaddi and the remaining students do not like the game. If one student is selected at random from all the students, find the probability that the student selected doesn’t like Kabbaddi.

Ans:

To find probability use $p=\dfrac{\text{favorable outcomes}}{\text{total number of outcomes}}$

Total number of outcomes will be $200$.

So, firstly the number of students who don't like kabaddi $\Rightarrow 200-135=65$.

Thus, favorable outcome will be $65$

Hence, the probability of student’s doesn’t like kabbaddi will be

$\Rightarrow p\left( K \right)=\dfrac{65}{200} =\dfrac{13}{40}$

15.A two digit number is to be formed from the digits $0,1,2,3,4$. Repetition of the digits is allowed. Find the probability that the number so formed is a .

1. Prime number

2. Multiple of $4$

3. Multiple of $11$

Ans: To find probability use $p=\dfrac{\text{favorable outcomes}}{\text{total number of outcomes}}$

1. Sample space will be $\Rightarrow S=\left\{ 10,11,12,13,14,20,21,22,23,24,30,31,32,33,34,40,41,42,43,44 \right\}=20$

So, the number of sample points will be $20$.

Probability of getting a prime number, $\left\{ 11,13,23,31,41,43 \right\}=6$, favorable outcome is $6$.

$\Rightarrow p\left( P \right)=\dfrac{6}{20} =\dfrac{3}{10}$

2. Probability of getting a multiple of $4$, favorable outcome will be $\left\{ 14,20,24,32,40,44 \right\}=6$

$\Rightarrow p\left( M \right)=\dfrac{6}{20} =\dfrac{3}{10}$

3. Probability of getting a number multiple of $11$, favorable outcome will be $\left\{ 11,22,33,44 \right\}=4$

$\Rightarrow p\left( M \right)=\dfrac{4}{20} =\dfrac{1}{5}$

16.The faces of a die bear numbers $0,1,2,3,4,5$. If the die is rolled twice, then find the probability that the product of digits on the upper face is zero.

Ans: To find probability use $p=\dfrac{\text{favorable outcomes}}{\text{total number of outcomes}}$

Sample space will be $\Rightarrow S=\left\{ \left( 0,0 \right)\left( 0,1 \right)\left( 0,2 \right)\left( 0,3 \right)\left( 0,4 \right)\left( 0,5 \right) \left( 1,0 \right)\left( 1,1 \right)\left( 1,2 \right)\left( 1,3 \right)\left( 1,4 \right)\left( 1,5 \right) \left( 2,0 \right)\left( 2,2 \right)\left( 2,3 \right)\left( 2,4 \right)\left( 2,5 \right) \left( 3,0 \right)\left( 3,1 \right)\left( 3,2 \right)\left( 3,3 \right)\left( 3,4 \right)\left( 3,5 \right) \left( 4,0 \right)\left( 4,1 \right)\left( 4,2 \right)\left( 4,3 \right)\left( 4,4 \right)\left( 4,5 \right) \left( 5,0 \right)\left( 5,1 \right)\left( 5,2 \right)\left( 5,3 \right)\left( 5,4 \right)\left( 5,5 \right)\right\}=36,$

In situation $1$: when the die is rolled twice the face has $0$ twice times.

So, probability of the product of digits on the upper face is zero will be $\Rightarrow \dfrac{1}{6}\times \dfrac{1}{6}=\dfrac{1}{36}$.

In situation $2$: when the die is rolled twice one face is having zero and the second face is having a number between $1$ to $5$.

So, probability of the products of digits on the upper face is zero will be $\Rightarrow \dfrac{1}{6}\times \dfrac{5}{6}=\dfrac{5}{36}$

Thus, total probability of digit on upper face is having zero is

$\Rightarrow \dfrac{1}{36}+\dfrac{5}{36}=\dfrac{6}{36} =\dfrac{1}{6}$

Importance of Maharashtra Board Textbook Solutions for Class 10 Maths Probability

The Chapter 5 of the Class 10 syllabus of Maths for the Maharashtra Board, Probability, is quite crucial. The chapter discusses all the important topics. They also get to learn the formulas and its application in real life.

1. Access to solutions by experts: Getting guidance and learning from experts who have curated the answers that focus on learning standards specific to the age group. It also provides well explained and easy to understand solutions for a student.

2. Beneficial for resolving doubts: Sometimes, questions asked in the chapter become challenging to comprehend, especially for students who find Maths difficult. The textbook solutions can present them with elaborative and comprehensive answers or solutions which will benefit them in case of doubts.

3. Learning through mistakes: When a student is trying to solve a question, it is often observed that they get the correct answers but do not follow the proper method. In such cases, textbook solutions can help them understand the experts' approach, which they can use while finding the answers.

4. Practice with the right approach: Practicing the questions by following correct methods and techniques becomes significant for a student. These practice questions become the building blocks for their further question-solving practice. Hence, it is crucial to solving many questions based on a concept with the right approach.

Benefits of using Vedantus' PDF Solutions for Class 10 Maths Maharashtra Board

• The solutions for Class 10 Chapter 5 for the Maharashtra Board Students have been created by Vedantu experts. With amplified knowledge about the concepts, they have left no detail unexplained in the chapter. Students can use elaborate explanations, examples, etc., to gain further understanding of the chapter.

• Solutions for Probability will help students understand how to solve complex problems related to the chapter. These solutions have been explained in a way that is easy to comprehend for students.

• They can prepare properly for their examinations by solving the problems with the help of the solutions. All details about different ways to solve Probability have been explained. Students can build a strong foundation for the chapter with these solutions.

• During revision, these solutions and practice sets can come in handy to revisit anything when you are having doubts about your answers.

Complete Your Practice With Maharashtra Board Class 10 Maths Chapter 5 Solutions

Download Maharashtra board Class 10 Maths solutions Chapter 5 Probability and start your practice today. With time, you can download more practice sets and solutions to strengthen your problem solving.