Telangana SCERT Solution Class X (10) Maths Chapter 13 Probability Exercise 13.2
Exercise – 13.2
(Q1) A bag contains 3 red balls and 5 black balls. A ball is selected at random from the bag. What is the probability that the ball selected is (i) Red (ii) Not red
=> Solution:
In the bag, no. of red balls = 3
No. Of black balls = 5
Total balls = 3 + 5 = 8
∴ Total possible outcomes = 8
(i) Red: ‘Red ball’ favourable outcomes = 3
P (Red) = 3/8
(ii) Not red:
‘Not red ball’ favourable outcomes = 5
P (Not red) = favourable outcomes/Total balls
= 5/8
(Q2) A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be
(i) Red?
(ii) White?
(iii) Not green?
=> Solution:
Number of red marbles = 5
White marbles = 8
Green marbles = 4
∴ Total marbles in the box = 5 + 8 + 4
= 17
∴ Total number of possible outcomes = 17
(i) Red:
Red marbles = 5
Favourable outcomes = 5
P (Red) = favourable outcomes/Total marbles
= 5/8
(ii) White:
White marbles = 8
Favourable outcomes = 8
P (White marbles) = Favourable outcomes/Total marbles
(iv) Not green:
‘Not green’ marbles = 5+8 = 13
Favourable outcomes = 13
P (not green) = 13/17
(Q3) A kiddy bank contains hundred 50p coins, fifty RS 1 coins, twenty RS 2 coins and ten RS 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (i) will be a 50p coin? (ii) Will not be a RS 5 coin?
=> Solution:
In the kiddy bank,
50p coins = 100
RS 1 coins = 50
RS 2 coins = 20
RS 5 coins = 10
Total coins the bank = 100 + 50 + 20 + 10
∴ Total number of possible outcomes = 180
(i) Will be a 50p coin
50p coins = 100
Favourable outcomes = 100
P (50p coin) = favourable outcome/Total outcome
= 100/180
= 5/9
(ii) Will not be a RS 5 coin
Other than RS 5 coins = 10+50+20
= 170
Favourable out comes = 170
P (not RS 5 coin) = (Favourable outcome/Total outcome)
= 170/180
= 17/18
(Q4) Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish. What is the probability that the fish taken out is a male fish?
=> Solution:
Number of male fish in the tank = 5
Number of female fish in the tank = 8
Total number of fish = 5+8 = 13
Total number of possible outcomes = 13
Favourable outcomes (male fish) = 5
∴ P (Male fish) = 5/13
(Q5) A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, and these are equally likely outcomes. What is the probability that it will point at
(i) 8?
(ii) An odd number
(iii) A number greater than 2?
(iv) A number less than 9?
Solution:
Total number of digits on the spinner = 8
∴ Total number of possible outcomes = 8
(i) 8
Favourable outcomes to ‘8’ = 1
P (8) = 1/8
(ii) An odd number.
Favourable out comes = 4 {1, 3, 5, 7}
P (an odd number) = 4/8
= 1/2
(iii) A number greater than 2.
Greater than ‘2’ number = 3, 4, 5, 6, 7, 8
Favourable outcomes = 6
P (A number greater than 2) = 6/8
= 3/4
(iv) A number less than ‘9’
Less than ‘9’ numbers = 1, 2, 3, 4, 5, 6, 7, 8
Favourable outcomes = 8
P (a number less than 9) = 8/8 = 1
(Q6) One card is selected from a well – shuffled deck of 52 cards. Find the probability of getting
(i) A king of red color
(ii) A face card
(iii) A red face card
(iv) The jack of hearts
(v) A spade
(vi) The queen of diamonds
=> Solution:
Total cards in a deck of cards = 52
∴ Total number of possible outcomes = 52
(i) A king of red color
Red color king cards = 2
Favourable outcomes = 2
P (a king of red color) = 2/52
= 1/26
(ii) A face card
Total face cards = 12
Favourable outcomes = 12
P (a face card) = 12/52
= 3/13
(iii) A red face card
Total face cards = 6
Favourable outcomes = 6
P (a red face card) = 6/52 = 3/26
(iv) The jack of hearts
Total jack of hearts = 1
Favourable outcomes = 1
P (The jack of hearts) = 1/52
(v) A spade
Total spade cards = 13
Favourable outcomes = 13
P (a spade) = 13/52
= 1/4
(vi) The queen of diamonds
‘Total queen of diamond’ cards = 1
Favourable outcomes = 1
P (the queen of diamonds) = 1/52
(Q7) Five cards- the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is selected at random.
(i) What is the probability that the card is the queen?
(ii) If the queen is selected and put aside (without replacement), what is the probability that the second card selected is (a) an ace? (b) a queen?
=> Solution:
Total number of cards = 5 (10, J, Q, K, A)
Total number of possible outcomes = 5
(i) The queen
Number of queen cards = 1
Favourable outcomes = 1
P (queen) = 1/5
(ii) If the selected queen card is put aside, then total number of cards = 4 (10, J, K, A)
Total number of possible outcomes = 4
(ii) (a) an ace
Number of ace cards = 1
Favourable outcomes = 1
P (ace) = 1/4
(ii) (b) a queen
Number of queen cards = 0
Favourable outcomes = 0
P (a queen) = 0/4 = 0
(Q8) 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
=> Solution:
Number of defective pens = 12
Number of good pens = 132
∴ Total number of possible outcomes = 144
Favourable outcomes to good pen = 132
P (good pen) = 132/144 = 11/12
(Q9) A lot of 20 bulbs contain 4 defective ones. One bulb is selected at random from the lot. What is the probability that this bulb is defective? Suppose the bulb selected in previous case is not defective and is not replaced. Now one bulb is selected at random from the rest. What is the probability that this bulb is not defective?
=> Solution:
Total number of bulbs = 20
Number of defective bulbs = 4
∴ Number of good bulbs = 20-4 = 16
∴ Total number of possible outcomes = 20
‘Defective bulb’ favourable outcomes = 4
P (Defective bulb) = 4/20
=- 1/5
* If the bulb selected previously is not defective and it is not replaced then.
Now total number of bulbs = 20-1
= 19
And total number of positive outcomes = 19
Now, total number of good bulbs = 16-1 = 15
∴ Favourable outcomes for ‘not defective’ = 15
P (not defective) = 15/19
(Q10) A box contains 90 discs which are numbered from 1 to 90. If one disc is selected at random from the box, find the probability that it bears
(i) A two-digit number
(ii) A perfect square number
(iii) A number divisible by 5.
=> Solution:
Total number of discs in the box = 90
Total number of possible outcomes = 90
(i) A two digit number.
Number of two digit numbered discs = 81
(They are 10, 11, 12, 13, 14, 15, —–, 89, 90)
Favourable outcomes = 81
P (a two digit number) = 81/90
= 9/10
(ii) A perfect square number
Number of perfect square numbered discs = 9
(They are 1, 4, 9, 16, 25, 36, 49, 64, 81)
Favourable outcomes = 9
P (a perfect square number) = 9/90 = 1/10
(iii) A number divisible by ‘5’
Number of discs which are divisible by ‘5’ = 18
(They are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90)
Favourable outcomes = 18
P (a number divisible by 5) = 18/90
= 1/5
(Q11) Suppose you drop a die at random on the rectangular region shown in figure. What is the probability that it will land inside the circle with diameter 1m?
=> Solution:
Here, the required = Area of the circle/Area of the rectangle ——— (1)
The diameter of the circle,
d = 1m
∴ Radius, r = d/2 = 1/2 m
Area of the circle = πr2
= 22/7 × (1/2)2 ——- (2)
From the figure, area of the rectangle = 3×2
= 6m2 —— (3)
Substituting equation (2) & (3) in equation (1)
Required probability = (22/7 × 1/2 × 1/2)/6
= 22/7 × 1/2 × 1/2 × 1/6
= 11/84
(Q12) A lot consists of 144 ball pens of which 20 are defective and the others are good. The shopkeeper draws one pen at random and gives it to Sudha. What is the probability that
(i) She will buy it?
(ii) She will not buy it?
=> Solution:
Total number of pens in the 10+ = 144
Number of defective pens = 20
∴ Number of good pens = 144 – 20
= 124
∴ Total number of possible outcomes = 144
(i) She will buy it
Here, the probability that Sudha will buy the pen is equal to the probability that the pen drawn is a good pen.
Favourite outcomes for good pen = 124
∴ The probability that she will buy it = 124/144 = 31/36
(ii) She will not buy it
Here, the probability that Sudha will not buy the pen is equal to the probability that the pen drawn is a defective pen.
Favourable outcomes for = 20
Defective pen
∴ The probability that she will not buy it = 20/144
= 5/36
(Q13) Two dice are rolled simultaneously and counts are added
(i) Complete the table given below
Event: sum on ‘2’ dice probability
2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
1/36 | 5/36 | 1/36 |
=> Solution:
(i) Here two dice are rolled one then.
Dice | 1 | 2 | 3 | 4 | 5 | 6 |
1 | (1,1) | (1,2) | (1,3) | (1,4) | (1,5) | (1,6) |
2 | (2,1) | (2,2) | (2,3) | (2,4) | (2,5) | (2,6) |
3 | (3,1) | (3,2) | (3,3) | (3,4) | (3,5) | (3,6) |
4 | (4,1) | (4,2) | (4,3) | (4,4) | (4,5) | (4,6) |
5 | (5,1) | (5,2) | (5,3) | (5,4) | (5,5) | (5,6) |
6 | (6,1) | (6,2) | (6,3) | (6,4) | (6,5) | (6,6) |
Total number of possible outcomes = 62 = 6×6 = 36
Sum on two dice:
2 = (1, 1)
3 = (1, 2), (2, 1)
4 = (1, 3), (2, 2), (3, 1)
5 = (1, 4), (2, 3), (3, 2), (4, 1)
6 = (1, 5), (2, 4), (3, 3), (4, 2), (5, 1)
7 = (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)
8 = (2, 6), (3, 5), (4, 4), (5, 3), (6, 2)
9 = (3, 6), (4, 5), (5, 4), (6,3)
10 = (4, 6), (5, 5), (6, 4)
11 = (5, 6), (6, 5)
12 = (6, 6)
Event: Sum of ‘2’ dice | 2 | 3 | 4 | 5 | 6 | |
Probability | 1/36 | 2/36 | 3/36 | 4/36 | 5/36 | |
Event: Sum of ‘2’ dice | 7 | 8 | 9 | 10 | 11 | 12 |
Probability | 6/36 | 5/36 | 4/36 | 3/36 | 2/36 | 1/36 |
(ii) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability 1/11. Do you agree with this argument? Justify your answer.
=> Solution: No, I don’t agree.
Here, two dice are rolled at once. Hence the possible outcomes are not ‘11’ they are 36.
∴ The probability of each outcome is not 1/11
(14) A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Deskhitha wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that she will lose the game.
=> Solution:
Here, a one rupee coin tossed 3 times.
The total number of possible outcomes = 23 = 2×2×2
= 8
They are,
HHH
HHT
HTH
HTT
THH
THT
TTH
TTT
She, loses the game, if all the tosses don’t give, the same result.
∴ Here favourable outcomes that she loses = 6
∴ The probability that she loses the game = 6/8 = 3/4
(Q15) A dice is thrown twice. What is the probability that (i) 5 will not come up either time? (ii) 5 will come up at least once?
=> Solution:
If one dice is thrown twice then. Total number of possible outcomes = 62 = 6×6 = 36
(i) 5 will not come up either time favourable outcomes = 25
{(1, 1), (1, 2), (1, 3), (1, 4), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 6)}
∴ P (5 will not come either time) = 25/36
(ii) 5 will come up at least once.
Favourable outcomes = 11
{1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5), (5, 1), (5, 2), (5, 4), (5, 6)}
P (5 will come up at least once) = 11/36
Here is your solution of Telangana SCERT Class 10 Math Chapter 13 Probability Exercise 13.2
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