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
