HBSE 10th Class Maths Important Questions Chapter 15 Probability

Haryana State Board HBSE 10th Class Maths Important Questions Chapter 15 Probability Important Questions and Answers.

Haryana Board 10th Class Maths Important Questions Chapter 15 Probability

Short Answer Type Questions

Question 1.
If the probability of winning a game is 0.07, what is probability of lossing it?
Solution :
P (winning a game) = 0.07 (given)
P (winning a game) + P (lossing a game) = 1
0.07 + P (lossing a game) = 1
P (lossing a game) = 1 – 0.07
= 0.93

Question 2.
If probability of ‘not E’ = 0.95, then find P (E).
Solution :
Probability of “not E’ = 0.95 (given)
P (E) + P (not E) = 1
P (E) + 0.95 = 1
P (E) = 1 – 0.95
P (E) = 0.05

Question 3.
A die is thrown once. Find the probability of getting ‘at most’?
Solution :
A die is thrown once, the possible numbers are: 1, 2, 3, 4, 5, 6
The numbers ‘at most’ are = 1, 2
Let E be event of getting at most 2
Number of outcomes to favourable event E = 2
P (E) = \(\frac{2}{6}=\frac{1}{3}\)

Question 4.
A dice is thrown once, find the probability of getting
(i) composite number
(ii) a prime numbers.
Solution :
A dice is thrown once, the possible numbers are: 1, 2, 3, 4, 5, 6
The numbers of all possible outcomes = 6
(i) The composite numbers are 4, 6
Let E₁ be event of a getting of composite number
Number of outcomes to favourable event
E₁ = 2
P (E₁) = \(\frac{2}{6}=\frac{1}{3}\)

(ii) The prime number are 2, 3, 5
Let E₂ be event of a getting of a prime number
Number of outcomes to favourable to event
E₂= 3
P (E₂) = \(\frac{3}{6}=\frac{1}{2}\)

Question 5.
A child has a die whose six faces show the letters as shown below:

The die is thrown once. What is the probability of getting
(i) A?
(ii) C?
Solution :
In throwing the die any one of the six faces may come upward
The number of all possible outcomes = 6
(i) Since, there are two faces with letter A
Let E₁ be event of getting faces with letter A
Number of outcomes to favourable to event E₁ = 2

(ii) The prime numbers from 1 to 20 are 2, 3, 12, 15, 18 i.e. 6 numbers
Let E₃ be event of drawn card a divisible by 3
Number of outcomes to favourable to event
P(E₃) = \(\frac {3}{10}\)

Question 7.
If a number x is chosen at random from the numbers -3, -2, -1, 0, 1, 2, 3. What is the probability that x² ≤ 4?
Solution :
A number x is chosen at random from the numbers -3, -2, -1, 0, 1, 2, 3
x² = 9, 4, 1, 0, 1, 4, 9
The number all possible outcomes = 7
The numbers x² ≤ 4 are 4, 1, 0, 1, 4, 9 i.e. 5 numbers
Let E be event of chosen at random x² ≤ 4
Number of outcomes to favourable to event E = 5
∴ P (E) = \(\frac {5}{7}\)

Question 8.
A pair of dice is thrown once. What is probability of getting a doublet?
Solution :
A pair of dice is thrown once. The number of all possible outcomes = 6 × 6 = 36

Some number on both dice are:
(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) i.e. 6 faces are doublet.
Let E be event of getting the same number on both dice.
Number of outcomes to favourable to event E = 6
∴ P (E) = \(\frac{6}{36}=\frac{1}{6}\)

Question 9.
A piggy bank contains hundred coins of Rs. 1, twenty coins of Rs. 2, fifteen coins of Rs. 5 and ten coins of Rs. 10. If it is equally likely that one coin will fall, when the bank is turned upside down,
(i) will be a Rs. 2 coin,
(ii) will not be Rs. 5 coin.
Solution :
Total number of coins = 100 + 20 + 15 = 145
The number of all possible outcomes 145
(i) Number of Rs 2 coins = 20
Number of favourble outcomes = 20
∴ P (Rs. 2 coins) = \(\frac{20}{145}=\frac{4}{29}\)

(ii) Number of coins other than Rs. 5 coins = 100 + 20 + 10 = 130
Number of favourable outcomes = 130
∴ P (will not be a Rs 5 coins) = \(\frac{130}{145}=\frac{26}{29}\)

Question 10.
A die is thrown twice. Find the probability that (i) 5 will come up at least once. (ii) 5 will not come up either time.
Solution :
A die is thrown twice. The number of all possible outcomes 36

(i) The number 5 will come up at least once are (1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) i.e. 11 faces
Let E be the event of getting 5 will come up at least once
∴ Number of outcomes to favourable to event E = 11
∴ P (E) = \(\frac {11}{36}\)

(ii) P (5 will not come up either time)
= 1 – P(E)
= 1 – \(\frac {11}{36}\)
= \(\frac {25}{36}\)

Question 11.
The probability of selecting a blue marble at random from a jar that contains only blue, black and green marbles is. The prolability of selecting a black marble at random from the same jar is. If the jar contains 11 green marbles, find the total number of marbles in the jar.
Solution :
Let the number of marbles of blue and black be x and y respectively.
Total number of marbles = x + y + 11
Number of all possible outcomes = x + y +11
According to question
P (black marbles) = \(\frac {1}{4}\) (given)

Putting the value of y in the equ. (2), we get
x = \(\frac{5+11}{4}=\frac{16}{4}\) = 4
Hence, total number of marbles in the jar
= 4 + 5 + 11
= 20

Question 12.
A card is drawn at random from a well shuffled deck of playing cards. Find the probability that the card drawn is :
(i) a card of space or an ace
(ii) a black king
(iii) neither a jack nor a king
(iv) either a king or a queen
Solution :
Total playing cards = 52
Number of all possible outcomes = 52
(i) Number of cards of spade or an ace = 13 + 3 = 16
Number of favourable outcomes = 16
∴ P (spade or an ace) = \(\frac {16}{52}\) = \(\frac {4}{13}\)

(ii) Number of black king cards = 2
Number of favourable outcomes = 2
∴ P (a black king) = \(\frac {2}{52}\) = \(\frac {1}{26}\)

(iii) Number of cards of Jack or King = 4 + 4 = 8
Number of cards other than Jack and King = 52 – 8 = 44
Number of favourable outcomes = 44
∴ P (neither jack or king) = \(\frac {44}{52}\) = \(\frac {11}{13}\)

(iv) Number of cards of either a king or a queen = 4 + 4 = 8
Number of cards other than Jack amd King = 8
Number of favourable outcomes = 8
∴ P (either a king or a queen) = \(\frac {8}{52}\) = \(\frac {2}{13}\).

Question 13.
The king, queen and jack of clubs are removed. The remaining cards are mixed together and then a card is drawn at random from it. Find the probability of getting (i) a face card (ii) a card of heart (iii) a card of clubs (iv) a queen of diamond.
Solution :
Since, king, queen and jack of clubs are removed from a deck of 52 cards.
∴ Number of all possible outcomes
= 52 – 3 = 49
(i) Number of face cards = 12 – 3 = 9
Number of favourable outcomes = 9
∴ P (a face card) = \(\frac {9}{49}\)

(ii) Number of cards of heart in deck = 13
Number of favourable outcomes = 13
∴ P (a card of heart) = \(\frac {13}{49}\)

(iii) Number of cards of clubs 13 – 3 = 10
Number of favourable outcomes = 10
∴ P (a card of clubs) = \(\frac {10}{49}\)

(iv) There is only one queen of diamond number of favourable outcomes = 1
∴ P (a queen of diamond) = \(\frac {1}{49}\)

Question 14.
All red face cards are removed from a pack of playing cards. The remaining cards are well shuffled and them a card is drawn at random from them. Find the probability that the drawn card is (i) a red card (ii) a face card and (iii) a card of clubs.
Solution :
Since, all red face cards are removed from a pack of playing cards.
Remaining cards = 52 – 6 = 46
Number of all possible outcomes = 46
(i) Remaining red cards = 26 – 6 = 20
Number of favourable outcomes = 20
P (a red card) = \(\frac {20}{46}\) = \(\frac {10}{23}\)

(ii) Number of face cards = 6
Number of favourable outcomes = 6
P (a face card) = \(\frac {6}{46}\) = \(\frac {3}{23}\)

(iii) Number of cards of clubs = 13
Number of favourable outcomes = 13
P (a card of club) = \(\frac {13}{46}\)

Fill in the Blanks

Question 1.
Experiments which have not fixed results experiment are called ______ experiment.
Solution :
Random

Question 2.
An event having only one _______ called an elementary event.
Solution :
outcome

Question 3.
The possible outcomes for a given event are called _______ outcomes.
Solution :
favourable

Question 4.
King, Queen and Jack are called ______ cards.
Solution :
face

Question 5.
The total number of non-face cards are ______ .
Solution :
40

Question 6.
Cards of spaced and clubs are ______ cards.
Solution :
black

Question 7.
Value of probability of an event cannot be _______ or greater than 1.
Solution :
negative.

Multiple Choice Questions

Choose the correct answer each of the following :

Question 1.
Cards bearing numbers 2, 3, 4, ……. 11 are kept in a bag. A card is drawn at random from the bag. The probability of getting a card with a prime number is :
(a) \(\frac {1}{2}\)
(b) \(\frac {2}{5}\)
(c) \(\frac {3}{10}\)
(d) \(\frac {5}{9}\)
Solution :
(a) \(\frac {1}{2}\)

Numbers are 2, 3, 4, 5, 6, 7, 8, 9, 10, 1
Prime numbers are, 2, 3, 5, 7, 11 = 5
p(E) = \(\frac{5}{10}=\frac{1}{2}\)
So, correct choice is (a)

Question 2.
If an event cannot occur, then its probability is :
(a) 1
(b) \(\frac {3}{4}\)
(c) \(\frac {1}{2}\)
(d) 0
Solution :
(d) 0

The event which can not occur is said to be impossible event and probability of impossible event is zero.
So correct option (d) is correct.

Question 3.
An event is very unlikely to happen. Its probability is closest to:
(a) 0.0001
(b) 0.001
(c) 0.01
(d) 0·1
Solution :
(a) 0.0001

In an event is vary unlikely to happen, its probability should be quite small.
Hence the option (a) is correct.

Question 4.
Which of the following cannot be probability of an event?
(a) 5%
(b) 0.9
(c) 1·1
(d) 0.1
Solution :
(c) 1·1

The probability lies between 0 and 1 so correct choice is (c).

Question 5.
The probability of certain event is :
(a) 1
(b) 0
(c) \(\frac {1}{2}\)
(d) None of these
Solution :
(a) 1

Question 6.
The probability of an impossible event is :
(a) \(\frac {1}{3}\)
(b) 0
(c) 1
(d) None of these
Solution :
(b) 0

Question 7.
The chance that a non leap year contains 53 sundays is:
(a) \(\frac {2}{7}\)
(b) \(\frac {1}{7}\)
(c) \(\frac {3}{7}\)
(d) \(\frac {1}{365}\)
Solution :
(b) \(\frac {1}{7}\)

Question 8.
The probability that in a family of 3 children there will be at least one boy is:
(a) \(\frac {1}{8}\)
(b) \(\frac {6}{8}\)
(c) \(\frac {4}{8}\)
(d) \(\frac {7}{8}\)
Solution :
(d) \(\frac {7}{8}\)