RD Sharma Class 12 Exercise 30.1 Probability Solutions Maths-Download PDF Online

Edited By Lovekush kumar saini | Updated on Jan 25, 2022 04:01 PM IST

The RD Sharma textbook solutions are famous in the entire country for their simple understanding and elaboration for the best knowledge for students. The Class 12 RD Sharma chapter 30 exercise 30.1 solution covers the content of the chapter Probability. RD Sharma solutions This chapter is thought to be complex for students' understanding, but once the basic knowledge is acquired from the right source, it might seem a lot more interesting to solve. The RD Sharma class 12th exercise 30.1 is useful in clearing the basics of the chapter and gives insights into how easily the questions can be solved.

Latest: JEE Main: high scoring chapters | Past 10 year's papers

Don't Miss: Most scoring concepts for NEET | NEET papers with solutions

New: Aakash iACST Scholarship Test. Up to 90% Scholarship. Register Now

This Story also Contains

RD Sharma Class 12 Solutions Chapter30 Probability - Other Exercise
Probability Excercise: 30.1
RD Sharma Chapter-wise Solutions

RD Sharma Class 12 Solutions Chapter30 Probability - Other Exercise

Probability Excercise: 30.1

Probability exercise 30.1 question 1
Answer:
$\frac{4}{7}$
Hint:
Use Formula
$\text { Probability }=\frac{ \text { No. of events }}{\text { Total no. of events }}$
Given,
ten cards in a box.
So, let
Total no. of cards,
S = { 1,2,3,4,5,6,7,8,9,10 }
Consider the given events,
A = Total even no. appears on the card
B = No. from cards which is always more than 3
Therefore,
A = { 2,4,6,8,10 }
B = { 4,5,6,7,8,9,10 }
Now,
A $\cap$ B = { 4,6,8,10 }
Therefore,
Probability that card drawn in even no.
$\begin{aligned} &P(A\mid B)=\frac{ \text { No. of elements of }A\cap B}{\text { Total no. of cards more than 3 }}\\ &\qquad \qquad=\frac{4}{7} \end{aligned}$

Probability exercise 30.1 question 2

Answer:
$\frac{1}{2}, \frac{1}{3}$
Hint:
Use Formula
$\text { Probability }=\frac{\text { No. of event happened }}{ \text { Total no. of events }}$
As per question,
Let , S = Elements of all certain possibilities
= { (B₁B₂), (B₁G₂), (G₁G₂), (G₁B₂) }
Now, A = Event where both child is girl
= { (G1 , G2) }
B = Event where youngest child is a girl
= { (B₁G₂), (G₁G₂) }
C = At Least one child is a girl
= { (B₁G₂), (G₁G₂), (G₁B₂) }
Now,
A ∩ B = (G₁G₂)
A ∩ C = (G₁G₂)

Now,

Required probability
$\begin{aligned} &=\frac{n(A \cap B)}{n(B)}\\ &=\frac{1}{2} \end{aligned}$
2. Required probability
$\begin{aligned} &=\frac{n(A \cap B)}{n(C)}\\ &=\frac{1}{3} \end{aligned}$

Probability exercise 30.1 question 3

Answer:
$\frac{1}{15}$
Hint:
Use Formula
$\text { Probability }=\frac{\text { Required event }}{ \text { Total no. of event }}$
Given, an event where two numbers appear on throwing two dice differently.
So, as per the question
Let, A = No on dice where two values are different
= { (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,3), (2,4), (2,5), (2,6)
(3,1), (3,2), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,5), (4,6),
(5,1), (5,2), (5,3), (5,4), (5,6) }
Similarly,
B= Event where sum of the no. is 4
= (1,3), (2,2), (3,1)
Now,
For event where no. is different and also its sum is 4
$\begin{aligned} &=A \cap B\\ &=\left \{ \left \{ 1,3 \right \}, \left \{ 3,1 \right \} \right \} \end{aligned}$
Thus,
$\begin{aligned} &\text { Probability }=\frac{n(A\cap B)}{ nA }\\ &=\frac{2}{30}\\ &=\frac{1}{15} \end{aligned}$

Probability exercise 30.1 question 4

Answer:
$\frac{1}{2}$
Hint:
$\text { Probability }=\frac{\text { No. of elements required }}{ \text { Total no. of elements }}$
Given, two events
Let, B = Event of getting head on toss third
= { HHH, HTH, THH, TTH }
Similarly, A = Event of getting head on first two tosses
= { HHH, HHT }
Now,
A∩B = { HHH }
Therefore,
Required probability
$\begin{aligned} &=\frac{n(A\cap B)}{n(A)}\\ &=\frac{1}{2} \end{aligned}$

Probability exercise 30.1 question 5

Answer:
$\frac{1}{6}$
Hint:
Find all the possibilities of getting 4 on third toss.
Given, that 6 and 5 appear respectively on the first two tosses.
So, let
A = Event for getting 4 on third toss
= { (1,1,4), (1,2,4), (1,3,4), (1,4,4), (1,5,4), (1,6,4),
(2,1,4), (2,2,4), (2,3,4), (2,4,4), (2,5,4), (2,6,4),
(3,1,4), (3,2,4), (3,3,4), (3,4,4), (3,5,4), (3,6,4),
(4,1,4), (4,2,4), (4,3,4), (4,4,4), (4,5,4), (4,6,4),
(5,1,4), (5,2,4), (5,3,4), (5,4,4), (5,5,4), (5,6,4),
(6,1,4), (6,2,4), (6,3,4), (6,4,4), (6,5,4), (6,6,4) }
Now, let
B = Event of getting 6 on first and 5 on second
= (6,5,1), (6,5,2), (6,5,3), (6,5,4), (6,5,5), (6,5,6),
Now,
A ∩ B = (6,5,4)
Therefore,
$\begin{aligned} &\text { Probability }=\frac{n(A\cap B)}{ n(B)}\\ &=\frac{1}{6} \end{aligned}$

Probability exercise 30.1 question 6

Answer:
$=\frac{16}{25}$
Hint:
Just put values and simplify
Given, P (B) = 0.5
$\begin{aligned} &P(A\cap B)=0.32\\ &P(A\mid B)=\frac{P(A\cap B)}{P(B)}\\ &=\frac{0.32}{0.5}\\ &=\frac{32}{50}\\ &=\frac{16}{25} \end{aligned}$

Probability exercise 30.1 question 7

Answer:
$\begin{aligned} &0.2, \frac{2}{3} \end{aligned}$
Hint:
Just simplify to find P (A|B)
Given, P(A) = 0.4
P(B) = 0.3
Now
$\begin{aligned} &P(A\mid B)=\frac{n(A\cap B)}{ n(A)}\\ &0.5=\frac{n(A\cap B)}{ 0.4 }\\ &P(A\cap B)=0.20 \end{aligned}$
Therefore,
$\begin{aligned} &P(A\mid B)=\frac{n(A\cap B)}{ n(B)}\\ &=\frac{0.2}{ 0.3 }\\ &=\frac{2}{3} \end{aligned}$

Probability exercise 30.1 question 8

Answer:
$\frac{5}{6},\frac{1}{2}$
Hint:
Use Formula
$\begin{aligned} &P(A\cup B)=P(A)+P(B)-P(A\cap B) \end{aligned}$ +
Given,
$\begin{aligned} &P(A)=\frac{1}{3}\\ &P(B)=\frac{1}{5}\\ &P(A\cup B)=\frac{11}{30}\\ \end{aligned}$
We know that,
$\begin{aligned} &P(A\cup B)=P(A)+P(B)-P(A\cap B) \end{aligned}$
$\begin{aligned} &\frac{11}{30}=\frac{1}{3}+\frac{1}{5}-P(A\cap B)\\ &P(A\cap B)=\frac{8}{15}-\frac{11}{30}\\ &P(A\cap B)=\frac{16-11}{30}\\ &\qquad \qquad =\frac{5}{30}\\ &P(A\cap B)=\frac{1}{6} \end{aligned}$
Now for,
$\begin{aligned} &P(A\mid B)=\frac{P(A\cap B)}{P(A)}\\ &=\frac{\frac{1}{6}}{\frac{1}{5}}\\ &=\frac{5}{6}\\ &P(B\mid A)=\frac{P(A\cap B)}{P(B)}\\ &=\frac{\frac{1}{6}}{\frac{1}{3}}\\ &=\frac{3}{6}\\ &=\frac{1}{2} \end{aligned}$

Probability exercise 30.1 question 9

Answer:
$\frac{1}{3},\frac{1}{2}$
Hint:
Use Formula
$\text { Probability }=\frac{\text { No. of event happened }}{ \text { Total no. of events }}$
Given, conditions
Let, the whole possibility
S = { (M₁M₂), (M₁F₂), (F₁F₂), (F₁M₂) }
Let , A = Both child are female
= (F₁F₂)
B = Elder child is female
= { (F₁M₂), (F₁F₂) }
C = At Least one child is a male
= { (M₁F₁), (F₁M₂), (M₁M₂) }
D = Both children are male
= (M₁M₂)
D ∩ C = (M₁M₂)
A ∩ B = (F₁F₂)

Required probability

$P(D\mid C)=\frac{n(D\cap C)}{ n(C)}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}$
2. Required probability
$P(A\mid B)=\frac{\frac{1}{4}}{\frac{2}{4}}=\frac{1}{2}$

The RD Sharma class 12 solution of Probability exercise 30.1 consists of a total of 9 questions which are brief enough to make you understand the basics of the chapter and prepare you to face more tough questions in further exercises. The RD Sharma class 12th exercise 30.1 covers all the concepts of this chapter that are mentioned below-

Conditional Probability
Mutually exclusive event
Union and Intersection
Equally Likely events

Listed below are a few benefits of the RD Sharma class 12 solutions chapter 30 exercise 30.1 for reference:-

The RD Sharma class 12th exercise 30.1 is trusted by thousands of students and teachers and is the best preparation material for scoring high marks and acing the maths exam.
The RD Sharma class 12 chapter 30 exercise 30.1 solution is exquisitely prepared and designed by experts in the field of academics in a very lucid manner. It helps all students solve problems in the right way.
The solutions are updated every year so that it corresponds with the syllabus of NCERT and therefore this makes it more useful in the preparation of any public exams.
The RD Sharma class 12th exercise 30.1 is also helpful for completing homework as from these books, students get an idea how to solve the questions. Teachers can also take the reference from these solutions to assign homework and prepare question papers.
The RD Sharma class 12th exercise 30.1 can be accessed through any device online by downloading it from the official website of Career360.
It's free of cost to own the RD Sharma class 12th exercise 30.1, as it doesn't cost a penny to download the solution from the Career360 website.

JEE Main Highest Scoring Chapters & Topics

Just Study 40% Syllabus and Score upto 100%

Download E-book