Selina Concise Class 10 Math Chapter 25 Exercise 25A Probability Solutions

Selina Concise Class 10 Math Chapter 25 Exercise 25A Probability Solutions

 

Probability

Exercise: 25A

(Q1) A coin is tossed once find the probability of:

(i) Getting a tail

(ii) Not getting a tail

Solution:

If a coin is tossed once then the simple space is = {H, T}

[H = Head and T – Tail]

∴ n (s) = 2 (n(s) = Number of sample space)

(i) 1 event:

A: Getting a tail

{In a sample space only getting a tail not head}

∴ {T}

∴ n(S) = 2

∴ n(A) = 1

∴ The probability of getting a tail = n(A)/n(S)

= ½

∴ P(A) = ½

(ii) Not getting a tail

B: Not getting a tail

{In sample space only getting a head not tail}

∴ {H}

∴ n(B) = 1

n(S) = 2

Probability, P(B) = n(B)/n(S)

= 1/2

 

(Q2) A bag contains 3 white, 5 black and 2 red balls, all of the same shape and size. A ball is drawn from the bag without looking into it, find the probability that the ball drawn is:

(i) A black ball

(ii) A red ball

(iii) A white ball

(iv) Not a red ball

(v) Not a black ball

Solution:

Given that, A bag contains –

White balls – 3

Black balls – 5

Red balls – 2

∴ Total number of balls = White balls + red balls + black balls

= 3+2+5

= 10 balls

∴ n(s) = 10

(i) A:a black ball

There are 5 black balls

∴ n(A) = 5

∴ Probability of a black balls = n(A)/n(S)

P(A) =5/10 = 1/2

(ii) B: a red ball

There are 2 re balls

∴ n(B)  = 2

∴ Probability of a red balls = n(B)/n(S)

∴ P (Getting a red ball) = n(B)/n(S)

P(B) = 2/10

P(B)  = 1/5

(iii) C:  a white ball

There a 3 white balls

∴ n(C) = 3

Probability of a white ball = n(C)/n(S)

P (getting a white ball) = n(C)/n(S)

P(C) = 3/10

(iv) D: not a red ball

{In a sample space there are getting only a white balls and black balls}

∴ White ball + black balls = 3+5

= 8

∴ n(D) = 8

∴ Probability (Getting a white and black balls) = n(D)/n(S)

P(D) = 8/10 = 4/5

∴ P(D) = 4/5

(v) E: not a black ball

{In sample space there are only white balls and red balls}

∴ White ball + red ball = 3+2

= 5

∴ n(E) = 5

∴ Probability (getting red and white balls) = n(E)/n(S)

P(E) = n(E)/n(S)

P(E) = 5/10

P(E) = ½

 

(Q3) In a single throw of a die find the probability of a getting a number:

(i) Greater than 4

(ii) Less than or equal to 4

(iii) Not greater than 4

Solution:

Given that, a single throw of a die –

Sample space = {1, 2, 3, 4, 5, 6}

n(S) = 6

(i) Event A: greater than 4

{In sample there are greater than 4 numbers only}

{5, 6}

∴ n(A) = 2

Probability (Greater than 4) = n(A)/n(S)

P(A) = 2/6

P(A) = 1/3

(ii) Event B: Less than or equal to 4

Probability (Getting less than or equal to 4) = n(B)/n(S)

{1, 2, 3, 4}

n(B) = 4

P(B) = 4/6

p(B) = 2/3

(iii) Event C: Not greater than 4.

Probability (getting not greater than 4 numbers = n(C)/n(S)

{1, 2, 3, 4}

n(C) = 4

p(C) = 4/6

p(C) = 2/3

 

(Q4) In a single throw of a die, find the probability that the number.

(i) Will be an even number

(ii) Will not be an even number

(iii) Will be an odd number

Solution:

Given that, single throw of a die = {1, 2, 3, 4, 5, 6}

n(S) = 6

(i) Event A: will be an even number = {2, 4, 6}

Probability (getting an even number)

= n(A)/n(S)

{2, 4, 6}

n(A) = 3

p(A) = n(A)/n(S)

= 3/6

p(A) = 1/2

(ii) Event B: will not be an even number = {1, 3, 5}

n(B) = 3

Probability (getting not an even number) = n(B)/n(S)

P(B) = 3/6

P(B) = ½

(iii) Even C: will be an odd number

= {1, 3, 5}

n(C) = 3

Probability (getting an odd number) = n(C)/n(S)

p(C) = 3/6

p(C) = 1/2

 

(Q5) From a well shuffled deck of 52 cards, one card is drawn find the probability that the card drawn will:

(i) Be a black card

(ii) Not be a red card

(iii) Be a red card

(iv) Be a face card

(v) Be a face card of re colour

Solution:

We know that, a well shuffled deck of 52 cards.

n(S) = 52

A black card – 26 (Spade and diamonds)

A red card – (Heart and diamond)

= 26 cards

(i) Event A: be a black cards = 26

∴ n(A) = 26

Probability (Black cards) = n(A)/n(S)

p(A) = 26/52

P(A) = 1/2

(ii) Event B: not be a red = 26

Total number of cards 52,

Red cards – 26 and black cards = 26

Probability (Getting not be a red cards means getting only black cards) = n(B)/n(S)

P(B) = n(B)/n(S)

∴ n(B) = 26 and n(S) = 52

P(B) = 26/52

P(B) = 1/2

(iii) Event C: be a red card = 26

n(C) = 26

Probability (Be a red cards) = n(c)/n(S)

P(C) = 26/52

P(C) = ½

(iv) Event E: be face cards = Total number of cards = 52

In this 52 cards there are 12 face cards (4kings, 4queens and jacks)

∴ n(E) = 12

P(E) = n(E)/n(S)

= 12/52

P(E) = 3/13

(v) Event F: be a face card of red color = 6

Total number of cards 52 in this 52 cards there are 26 red cards and in this 26 red cards there are 6 red face cards.

∴ n(F) = 6

P(F) = n(F)/n(S)

= 6/12

P(F) = 3/26


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