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RD Sharma Solutions are the enthusiastically suggested reference material by the majority of the CBSE schools. Without even a trace of an instructor, this book will direct the students the correct way by explaining their doubts and giving detailed replies. The Class 12 RD Sharma Chapter 30 Exercise 30.5 is the best booklet for any student to prepare their exam with. RD Sharma Solutions Students can try many mock tests. Working on self-assessments using the RD Sharma CLass 12th Exercise 30.5 increases the capability and confidence of the students to face the exams.

The RD Sharma class 12 solutions of Probability exercise 30.5, is utilized by thousands of students and educators for useful information on maths. The RD Sharma class 12th exercise 30.5 comprises of 56 inquiries covering every one of the fundamental ideas identified with requests identified with cards, flip a coin, two balls an and b, discovering probability of at any rate, probably, thrown a dice, and so forth and this inquiry depended on Binomial, Autonomous occasions, Association, and crossing point, with substitution and without substitution.

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Probaility Exercise 30.5 Question 1

Answer:Hint: You must know the rules of finding probability

Given: Bag 1 – 3 white & 6 black balls

Bag 2 – 5 black & 4 white balls

Solution:

Probability Exercise 30.5 Question 2

Answer:Hint: You must know the rules of finding probability

Given: Bag 1 3 Red and 5 Black balls

Bag 2 6 Red and 4 Black balls

Solution:

Bag 1(3R+5B) = 8 balls

Bag 2 (6R+4B) =10 balls

Probability Exercise 30.5 Question 3 (i)

Answer:Hint: You must know the rules of finding probability

Given: Box contains 10 black and 8 red balls

Solution: Box = (10 B and 8 R) balls

Total no. of balls = ( 10 +8) =18 balls

Probability Exercise 30.5 Question 3 (iii)

Answer:

Hint: You must know the rules of finding probability

Given:Box contains 10 black and 8 red balls

Solution:Box = (10B and 8R) balls

Total no. of balls =10 + 8 = 18 balls

P (one is black and other is red) = P(first red ball and second black)+P(first black ball and second red )

Probability Exercise 30.5 Question 4

Answer:Hint: You must know the rules of finding probability .

Given: Two cards are drawn from 52 cards without replacement

Solution: P(exactly one ace)=P(first card ace)+P(second not ace)

Probability Exercise 30.5 Question 5

Answer:

Hint: You must know the rules of finding probability

Given: A speaks of truth

B speaks 80% of truth

Solution:P(both narrating different incidents)=P(A speaks truth and B lies)+P(A lies and B speaks truth)

Probability Exercise 30.5 Question 6 (i)

Answer:Hint: You must know the rules of finding probability

Given: Probability of Kamal’s selection

Probability of Monika’s selection

Solution: P( Kamal selected)

P( Monika selected)

Probability Exercise 30.5 Question 6 (ii)

Answer:Hint: You must know the rules of finding probability

Given:Probability of Kamal’s selection

Probability of Monika’s selection

Solution:P( Kamal selected)

P( Monika selected)

Probability Exercise 30.5 Question 6 (iii)

Answer:Hint: You must know the rules of finding probability

Given:Probability of Kamal’s selection

Probability of Monika’s selection

Solution:P( Kamal selected)

P( Monika selected)

Probability Exercise 30.5 Question 6 (iv)

Answer:Hint: You must know the rules of finding probability

Given:Probability of Kamal’s selection

Probability of Monika’s selection

Solution:P( Kamal selected)

P( Monika selected)

Probability Exercise 30.5 Question 7

Answer:Hint: You must know the rules of finding probability.

Balls are drawn without replacement.

Given: Box contains (3White + 4Red + 5Black) balls

Solution: Box=(3W+4R+5B)

Total no. of balls = 3+ 4+ 5=12 balls

P(one white and one black)=P(first white, second black)+P(first black, second white)

Probability Exercise 30.5 Question 8

Answer:Hint: You must know the rules of finding probability.

Given :Bag contains 8 red and 6 green balls.

The balls are drawn without replacement.

Solution: Bag = (8R + 6G) =14 balls

P(atleast 2 balls are green)= 1-P(at most one ball is green)

=1-[P(first green, second and third red)+ P(first red , second green , third red ) +P(first and second red , third green) +P( all red)]

Probability Exercise 30.5 Question 9

Answer:Hint: You must know the rules of finding probability

Given:

Solution:

Probability Exercise 30.5 Question 10

Answer:Hint: You must know the rules of finding probability and G.P series

Given: A and B toss a coin alternately till one of them get a heads and wins thegame

Solution: If A starts the game, then B will win only at even positions

P(winning the game) = P(head at 2

Probability Exercise 30.5 Question 11

Answer:Hint: You must know the rules of finding probability functions.

Given: Two cards are drawn from 52 cards one after another without replacement.

Solution:

P(one red and other black)= P(first red and second black) + P(first black and second red)

Probability Exercise 30.5 Question 12

Answer:Hint: You must know the rules of finding probability functions.

Given: Tickets are number from 1 to 10

Two tickets are drawn one after the other at random.

Solution: We know 5 and 10 are multiples of 5 while 4 and 8 are multiples of 4

Probaility Exercise 30.5 Question 13

Answer:Hint: You must know the rules of finding probability functions.

Given: Husband tells a lie in cases,

Wife tells a lie in cases.

Solution:

Probability Exercise 30.5 Question 14 (i)

Answer:Hint: You must know the rules of finding probability.

Given:

Solution:

Probability Exercise 30.5 Question 14 (ii)

Answer:Hint: You must know the rules of finding probability.

Given:

Solution:

Probability Exercise 30.5 Question 14 (iii)

Hint: You must know the rules of finding probability.

Given:

Solution:

Probability Exercise 30.5 Question 15.

Answer:Hint: You must know the rules of finding probability.

Given: Bag contains 7 white, 5 black, 4 red balls

Four balls are drawn without replacement.

Solution

We have

White=7, Black=5, Red=4

Four balls are drawn without replacement

P(at least 3 black balls are black)

=P(3 black balls and one not black or 4 black balls)

=P(3 black and one not black)+P(4 black balls)

Probability Exercise 30.5 Question 16

Answer:Hint: You must know the rules of finding probability.

Given: A speaks truth three times out of four

B speaks truth four times out of five

C speaks truth five times out of six

Solution:

Probability Exercise 30.5 Question 17 (i)

Answer:Hint: You must know the rules of finding probability.

Given:

Solution:

Total no. of balls = 6 balls in bag (1)

Total no. of balls =8 balls in bag (2)

Probability Exercise 30.5 Question 17 (ii)

Answer:Hint: You must know the rules of finding probability functions.

Given:

Solution:

Total no. of balls = 6 balls in bag (1)

Total no. of balls =8 balls in bag (2)

Probability Exercise 30.5 Question 17 (iii)

Answer:Hint: You must know the rules of finding probability functions.

Given:

Solution:

Total no. of balls = 6 balls in bag (1)

Total no. of balls =8 balls in bag (2)

Probability Exercise 30.5 Question 18

Answer:Hint: You must know the rules of finding probability functions.

Given: Bag contains = (4W+5R+7B) balls

Solution

No. of white balls = 4

No. of black balls = 7

No. of red balls = 5

Total balls = 16

No. of ways in which 4 balls can be drawn from 16 balls

Let A = getting at least two white balls = getting 2, 3, 4 white balls

No. of ways of choosing 2 white balls

No. of ways of choosing 3 white balls

No. of ways of choosing 4 white balls

Probability Exercise 30.5 Question 20 (i)

Answer:Hint: You must know the rules of finding probability functions.

Given:

Solution:

Total no. of balls = 9 balls in bag A

Total no. of balls =10 balls in bag B

P(balls of different colors) =

P (red from bag A and black from bag B) + P (red from bag B + black from bag A)

Probability Exercise 30.5 Question 20 (ii)

Answer:Hint: You must know the rules of finding probability.

Given:

Solution:

Total no. of balls = 9 balls in bag A

Total no. of balls =10 balls in bag B

P (balls of same colors) = P (both red) + P (both black )

Probability Exercise 30.5 Question 22 (i)

Answer:Hint: You must know the rules of finding probability functions.

Given:

Solution:

P(only A passing the examination) = P (A passes) P (B fails)

Probability Exercise 30.5 Question 22 (ii)

Answer:Hint: You must know the rules of finding probability .

Given:

Solution:

P(only one of them passing the examination) = P (A passes ,B fails) + P (B passes , A fails)

Probability Exercise 30.5 Question 23

Answer:Hint: You must know the rules of finding probability functions.

Given: Urn A = (4R+3B) = 7 balls

Urn B = (5R+4B) = 9 balls

Urn C = (4R+4B) = 8 balls

Solution

Urn A contain 4 red , 3 black balls

Urn B contain 5 red , 4 black balls

Urn B contain 4 red , 4 black balls

P(3 balls drawn contain 2 red and a black ball)

Probability Exercise 30.5 Question 24 (iii)

Answer:Hint: You must know the rules of finding probability functions.

Given:

Solution:

P(grade A in two subjects) =

P(not grade A in Math )+P(not grade A in Physics)+P(not grade A in Chemistry)

Probability Exercise 30.5 Question 25

Answer:Hint: You must know the rules of finding probability functions.

Given: A and B take turns in throwing two dice, the first to throw 9 being awarded the prize

Solution: Total number of events = 36

P(getting 9)

P(A winning) = P(getting 9 in first throw) + P(getting 9 in third row)+….

P(B winning)=P(getting 9 in second throw)+P(getting 9 in fourth throw)+…

Winning ratio of A to B

Probability Exercise 30.5 Question 26

Answer:Hint: You must know the rules of finding probability functions.

Given: A, B, C are in order. The one to throw a head wins.

Solution: P(A winning) = P(head in first toss)+P(head in 4th toss)+….

P(B winning)= P(head in second toss)+P(head in 5th toss)+….

P (C winning)= P(head in third toss)+P(head in 6th toss)+….

Probability Exercise 30.5 Question 27

Hint: You must know the rules of finding probability functions.

Given: Three persons A,B,C throw a die in succession till one gets a six and wins the game.

Solution:

Probability Exercise 30.5 Question 28

Answer:ProvedHint: You must know the rules of finding probability functions.

Given: A and B take turns in throwing two dice,

The first throw 10 being awarded the prize.

Solution: There are only three possible cases, where the sum of number obtained after throwing

2 dice is 10, i.e.[(4,6) , (5,5) , (6,4)]

Now,

Probability Exercise 30.5 Question 29

Answer:Hint: You must know the rules of finding probability functions.

Given:

Solution:

It is given that bag A contains 3 red and 5 black balls and bag B contains (2R,3B)

Total no. of balls in bag A = 8 balls

Total no. of balls in bag B =5 balls

Probability Exercise 30.5 Question 30 (i)

Answer:Hint: You must know the rules of finding probability functions.

Given:

Solution:

Probability Exercise 30.5 Question 30 (ii)

Answer:Hint: You must know the rules of finding probability functions.

Given:

Solution:

Probability Exercise 30.5 Question 30 (iii)

Answer:Hint: You must know the rules of finding probability functions.

Given:

Solution:

Probability Exercise 30.5 Question 31 (i)

Answer:Hint: You must know the rules of finding probability functions.

Given: Bag contain 3 blue and 5 red marbles

Solution:

Total no. of marbles =8 marbles

Probability Exercise 30.5 Question 31 (ii)

Answer:Hint: You must know the rules of finding probability functions.

Given: Bag contain 3 blue and 5 red marbles

Solution:

Total no. of marbles =8 marbles

Probability Exercise 30.5 Question 31 (iii)

Answer:Hint: You must know the rules of finding probability functions.

Given: Bag contain 3 blue and 5 red marbles

Solution:

Total no. of marbles =8 marbles

Probability Exercise 30.5 Question 32 (i)

Answer:Hint: You must know the rules of finding probability .

Given: Urn contains 7 red and 4 blue balls.

Two balls are drawn with replacement.

Solution:

Probability Exercise 30.5 Question 32 (ii)

Answer:Hint: You must know the rules of finding probability.

Given: Urn contains 7 red and 4 blue balls.

Two balls are drawn with replacement.

Solution:

Probability Exercise 30.5 Question 32 (iii)

Answer:Hint: You must know the rules of finding probability functions.

Given: Urn contains 7 red and 4 blue balls.

Two balls are drawn with replacement

Solution:

Probability Exercise 30.5 Question 33 (i)

Answer:Hint: You must know the rules of finding probability functions.

Given: Card is drawn from 52 cards, the outcome is noted , then it is replaced and reshuffled,

Another card is drawn.

Solution

We know that there are four suits club (C), spades(S), heart(H), diamond(D), each contain 13 cards.

P(both the cards are of same suit)

Probability Exercise 30.5 Question 33 (ii)

Answer:Hint: You must know the rules of finding probability functions.

Given: Card is drawn from 52 cards, the outcome is noted, the card is replaced and reshuffled,

Another card is drawn.

Solution

We have, four ace and 2 red queens.

P(first card an ace and second card a red queen)

=P(getting an ace)×P(getting a red queen)

Probability Exercise 30.5 Question 34 (i)

Answer:Hint: You must know the rules of finding probability functions.

Given: Out of 100 students, two section of 40 and 60 are formed

Solution: When both enter the same section.

Here are the possibilities of two cases:

Case-(1)enter both are in section A

If both are in section A, 40 students out of 100 can be selected

and (40-2)=38 students out of (100-2)=98

Can be selected

So,

Case-(2)if both are in section B, 60 students out of 100 can be selected

and (60-2)=58 students out of (100-2)=98

Can be selected

So,

Hence, probability that student are either in sector

Probability Exercise 30.5 Question 34 (i)

Answer:Hint: You must know the rules of finding probability functions.

Given: Out of 100 students, two section of 40 and 60 are formed

Solution: When both enter the same section.

Here are the possibilities of two cases:

Case-(1)enter both are in section A

If both are in section A, 40 students out of 100 can be selected

and (40-2)=38 students out of (100-2)=98

Can be selected

So,

Case-(2)if both are in section B, 60 students out of 100 can be selected

and (60-2)=58 students out of (100-2)=98

Can be selected

So,

Hence, probability that student are either in sector

Probability Exercise 30.5 Question 34 (ii)

Answer:Hint: You must know the rules of finding probability functions.

Given: Out of 100 students, two section of 40 and 60 are formed

Solution: We know,

E.g. The probability that both enter different section =

probability that both enter same section.

Probability Exercise 30.5 Question 35

Answer:The decision was fair as the two probabilities are almost equal.

Hint: You must know the rules of finding probability functions.

Given: In a hockey match, both teams scored same number of goals upto end of the game.

Solution:

Probability Exercise 30.5 Question 36

Answer:Hint:

Given: A and B throw a pair of dice alternately.

A wins the game if he gets a total of 7 and B wins if he gets 10

Solution: Total of 7 on dice can be obtained in following ways:

Probabilities of getting 7 in first throw

Probability Exercise 30.5 Question 37

Answer:Hint: You must know the rules of finding probability functions.

cgfyTill one of them gets the sum of numbers as multiples of 6 and wins the game.

Solution: Let S denote the success and F denote the failure (not getting 6)

Thus,

So,

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- Chapter 1 - Relations
- Chapter 2 - Functions
- Chapter 3 - Inverse Trigonometric Functions
- Chapter 4 - Algebra of Matrices
- Chapter 5 - Determinants
- Chapter 6 - Adjoint and Inverse of a Matrix
- Chapter 7 - Solution of Simultaneous Linear Equations
- Chapter 8 - Continuity
- Chapter 9 - Differentiability
- Chapter 10 - Differentiation
- Chapter 11 - Higher Order Derivatives
- Chapter 12 - Derivative as a Rate Measurer
- Chapter 13 - Differentials, Errors and Approximations
- Chapter 14 - Mean Value Theorems
- Chapter 15 - Tangents and Normals
- Chapter 16 - Increasing and Decreasing Functions
- Chapter 17 - Maxima and Minima
- Chapter 18 - Indefinite Integrals
- Chapter 19 - Definite Integrals
- Chapter 20 - Areas of Bounded Regions
- Chapter 21 - Differential Equations
- Chapter 22 - Algebra of Vectors
- Chapter 23 - Scalar Or Dot Product
- Chapter 24 - Vector or Cross Product
- Chapter 25 - Scalar Triple Product
- Chapter 26 - Direction Cosines and Direction Ratios
- Chapter 27 - Straight Line in Space
- Chapter 28 - The Plane
- Chapter 29 - Linear programming
- Chapter 30- Probability
- Chapter 31 - Mean and Variance of a Random Variable

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