JKBOSE 10th Class Maths Solutions chapter – 15 Probability
JKBOSE 10th Class Maths Solutions chapter – 15 Probability
J&K 10th Class Maths Solutions chapter – 15 Probability
Jammu & Kashmir State Board JKBOSE 10th Class Maths Solutions
Experiment : Experiment mean an operation which can produce some welldefined outcome(s).
Random Experiment: In an experiment when repeated under identical condition do not produce the same outcome every time but the outcome in a trial is one of the several possible outcomes.
Sample Space: The total number of possible outcomes of random experiment is known as sample space.
Elementary Event: If a random experiment is performed then each outcomes is known as an elementary event.
Sure Event: An event associated with random experiment is called a certain event if it always occurs whenever the experiment is performed.
Impossible event. An event associated with random experiment is called an impossible event. If it never occurs when every experiment is performed.
Favourable Events: The cases which ensure the occurrence of an event, are called favourable cases to that event.
TEXT BOOK EXERCISE 15.1
Q. 1. Complete the following statements :
(i) Probability of an event E+ Probability of the event ‘not E’ = ………..
(ii) The probability of an event that cannot happen is ……. Such an event is called ……….
(iii) The probability of an event that is certain to happen is …… Such an event is called…………
(iv) The sum of the probabilities of all the elementary events of an experiment is ………
(v) The probability of an event is greater than or equal to ……… and less than or equal to …………
Solution.
(i) Probability of an event E + Probability of the event ‘not E’ = 1
(ii) The probability of an event that cannot happen is 0. Such an event is called impossible event.
(iii) The probability of an event that is certain to happen is 1. Such event is called sure event.
(iv) The sum of the probabilities of all the elementary events of an experiment is 1.
(v) The probability of an event is greater than or equal to 0 and less than or equal to 1.
Q. 2. Which of the following experiments have equally likely outcomes? Explain.
(i) A driver attempts to start a car. The car starts or does not start.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
(iii) A trial is made to answer a true-false question. The answer is right or wrong.
(iv) A baby is born. It is a boy or a girl.
Solution.
(i) When a driver attempts to start a car the car starts normally. Only when there is some defect, the car does not start. So the outcome is not equally likely.
(ii) When a player attempts to shoot a basketball the outcome in this situation is not equally likely because the outcome depends on many factors such as the training of the player, quality of the gun used etc.
(iii) Since for a question there are two possibilities either right or wrong the outcome in this trial of true-false question is either true or false i.e. one out of the two and both have equal chances to happen. Hence, the two outcomes are equally likely.
(iv) A new baby (i.e. who took birth at a moment) can be either a boy or a girl and both the outcome have equally likely chances.
Q. 3. Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game.
Solution. When a coin is tossed there are only two possibilities i.e. Head or tail both are equally likely to happen. Result of the toss of a fair coin is completely unpredictable.
Q.4. Which of the following cannot be the probability of an event ?
(A) 2/3
(B) -1.5
(C) 15%
(D) 0.7.
Solution. As we know probability of event cannot be less than 0 and greater than 1
i.e. 0 ≤ P ≤ 1
∴ (B) – 1.5 is not possible.
Q. 5. If P(E) = 0.05, what is the probability of not E.
Solution. As we know P (E) + P (Ē) = 1
P (Ē) = 1 – P (E)
= 1 – 0.05
= 0.95.
Q. 6. A bag contains lemon flavoured candies only. Sara takes out one candy without looking into the bag. What is the probability that she takes out
(i) an orange flavoured candy ?
(ii) a lemon flavoured candy ?
Solution. (i) Since bag contains only lemon flavoured candies
∴ There is no orange candies
∴ It is impossible event.
∴ Probability of getting orange flavoured = 0
(ii) Since there are only lemon flavoured candies, it is sure event
∴ Probability of getting lemon flavoured
candy = 1/1 = 1
Q. 7. It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday ?
Solution. Let A is event that two students have same birthday
∴ Ā is event that 2 students not having same birthday is 0.992
∴ P (Ā) = 0.992
∴ P (A) = 1 – P (Ā) (P (A) + P(Ā) = 1)
= 1-0.992
= 0.008
∴ Probability that two students have same birthday = 0.008. Ans.
Q.8. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red ? (ii) not red?
Solution. Number of Red balls = 3
Number of Black balls = 5
Total number of balls = 3 + 5 = 8
One ball is drawn at random
(i) Probability of getting Red ball
= Number of favourable cases / Total number of cases
P (Red ball) = 3/8 Ans.
(ii) Probability of getting no red ball
= 1 – P (Red ball)
= 1 – 3/8 = 5/8 And. [P(Ā) = 1 – P(E)
Q.9 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? (2015 (S)) (ii) white (iii) not green ?
Solution. Number of red marbles = 5
Number of white marbles = 8
Number of green marbles = 4
Total number of marbles = 5 + 8 + 4 = 17
Since, one marble is taken out
(i) There are 5 Red marbles
Probability of drawing Red marble
= Number of favourable cases / Total number of cases = 5/17 Ans.
(ii) Since there are 8 white marbles
Probability of drawing white marble
= Number of favourable cases/Total number of outcomes = 8/17 Ans.
(iii) There are 4 green marbles
Probability of drawing green marbles
= Number of favourable cases/Total number of outcomes = 4/17
∴ Probability of not drawing green marbles
= 1 – Probability of green marbles
Q. 10. A piggy bank contains hundred 50p coins, fifty ₹ 1 coins, twenty ₹ 2 coins and ten ₹ 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 50 p coin ? (ii) will not be a ₹ 5 coin ?
Solution. Number of 50 coins = 100
Number of ₹ 1 coins = 50
Number of ₹ 2 coins = 20
Number of ₹ 5 coins = 10
∴ Total number of coins
= 100 + 50 + 20 + 10 = 180
(i) Since there are 100; 50′ p coin
Probability of getting 50 p coins
(ii) Number of ₹ 5 coins = 10
∴ Probability of getting ₹ 5 coin
Q. 11. 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 = 5
Number of female fish = 8
Total number of fish in the tank= 5 + 8 = 13
Probability of getting a male fish
Q. 12. 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. (i) Total number of outcomes = {1, 2, 3, 4, 5, 6, 7, 8}
(iii) Numbers greater than 2 are {3, 4, 5, 6, 7, 8}
P (a number less than 9) = 1. Ans.
Q. 13. A die is thrown once. Find the probability of getting
(i) a prime number
(ii) a number lying between 2 and 6;
(iii) an odd number.
Solution. When dice is thrown number of possible outcomes
S = {1, 2, 3, 4, 5, 6}
(i) Prime numbers are {2, 3, 5}
∴ Probability of getting prime number
(ii) Numbers lying between 2 and 6={3,4,5}
Probability of getting number between 2 and
6
(iii) The odd numbers are = { 1, 3, 5}
Probability of getting an odd number
P (odd number) = 1/2 Ans.
Q. 14. One card is drawn from a wellshuffled deck of 52 cards. Find the probability of getting
(i) a king of red colour
(ii) a face card
(iii) a red face card
(iv) the jack of hearts
(v) a spade
(vi) the queen of diamonds.
Solution. There are 52 cards in a pack
(i). There are two red kings i.e. king of heart and king of diamond
Probability of getting red king = 2/52 = 1/26
P (Red king): = 1/26 Ans.
(ii) There are 12 face cards
i.e. 4 Jack, 4 Queens and 4 Kings
Probability of getting face card = 12/52
∴ P (A face card) = 3/13 Ans.
(iii) Since there are 6 Red face cards i.e. 2 Jacks; 2 Queens and 2 Kings
∴ Probability of getting 6 Red face cards = 6/12
P (Red face card) = 3/26 Ans.
(iv) There is only one Jack of Heart
∴ Probability of getting Jack of Heart 1/52
P (A Jack card) = 15/12 Ans.
(v) Since there are 13 spade cards
∴ Probability of getting a spade card = 13/12
P (A spade card) = 1/4 Ans.
(vi) Since there is only one queen of diamonds
∴ Probability of getting queen of spade card = 1/52
P (A queen of spade) = 1/52 Ans.
Q. 15. Five cards the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.
(i) What is the probability that the card is the queen ?
(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace ? (b) a queen ?
Solution. Five cards are ten, jack, queen, king and ace
(i) Probability of getting queen = 1/5
∴ P (A queen) = 1/5. Ans.
(ii) If the queen is drawn and put aside then there are 4 cards left-Ten, a Jack, a king and an ace.
(a) Probability of getting an ace = 1/4
P (An Ace) = 1/4. Ans.
There’s no queen left
(b) Probability of getting a queen = 0/4 = 0
P (a queen) = 0. Ans.
Q. 16. 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 pens = 12 + 132 = 144
Probability of getting good pen = 132/144 = 11/12
P (a good pen) = 11/12. Ans.
Q. 17. (i) A lot of 20 bulbs contains 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective ?
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?
Solution. (i) Number of defective bulbs = 4
Number of good bulbs (Not defective) = 16
Total number of bulbs = 4 + 16 = 20
Probability of getting defective bulb
= 4/20 = 1/5. Ans.
(ii) When a defective bulb drawn is not being replaced, we are left with 19 bulbs
Now probability of getting not defective bulb
= 15/19
∴ P (Not defective bulb) = 15/19. Ans.
Q. 18. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn 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. From 1 to 90 there are 90 numbers in all and 81 two-digit numbers from 10 to 90
(i) Probability of getting two digit number
= 81/90
∴ P (two digit number) = 81/90 = 9/10. Ans.
(ii) Perfect square numbers are {1, 4, 9, 16, 25, 36, 49, 64, 81} there are 9 perfect square numbers between 1 to 90
Probability of getting perfect square
= 9/90 = 1/10
P (Perfect square) = 1/10. Ans.
(iii) Numbers divisible by 5 are {5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85,90}
There are 18 numbers divisible by 5
∴ Probability of number getting divisible by 5
= 18/90 = 1/5
∴ Required probability = 1/5. Ans.
Q. 19. A child has a die whose six faces show the letters as given below :
The die is thrown. What is the probability of getting (i) A ? (ii) D?
Solution. Number of faces of a die = 6
S = {A, B, C, D, E, A}
n (S) = 6
(i) Since there are two A’s
Q. 20. Suppose you drop a die at random on the rectangular region shown in Fig. What is the probability that it will land inside the circle with diameter 1 m ?
Solution. Length of rectangle (l) = 3 m
Width of rectangle (b) = 2 m
∴ Area of rectangle = 3 m × 2 m = 6 m²
Diameter of circle = 1 m
Probability of die to land on a circle
Q. 21. A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy it if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that
(i) She will buy it?
(ii) She will not buy it?
Solution. Total number of Pens in lot = 144
Number of defective Pens = 20
∴ Number of good Pens = 144 – 20 = 124
(i) Let ‘A’ is event showing she buy the pen
(ii) Ā is event showing that she will not buy the pen
Q. 22. Two dice, one blue and one grey are thrown at the same time. Write down all the possible outcomes.
(i) Complete the following table :
(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. When two dices are thrown total number of possible outcomes
n(S) = 36
Let A is event of getting sum as 3
∴ A = {(1,2) (2, 1)}
n (A) = 2
Let D is event of getting sum as 6
D = {(1, 5) (5, 1) (2, 4) (4, 2) (3, 3)}, n (D) = 5
∴ P (6) = 5/36
Let E is event of getting sum as 7
E = {(1, 6) (6, 1) (2, 5) (5, 2) (4, 3) (3,4)}
∴ P (E) = P (Sum as 7) = 3/36 = 1/6
Let F is event of getting sum as 8
F = {(2,6) (6, 2) (3,5) (4, 4) (5,3)}
∴ n (F) = 5
P (F) = P (sum as 8) = 5/36
Let G is event of getting sum as 9 when two dices are thrown
G = {(4, 5) (5, 4) (3, 6) (6,3)}
n(G) = 4
∴ P (G) = P (Sum as 8) = 4/36 = 1/9
Let H is event of getting sum as 10
H = {(6,4) (4, 6) (5,5)}
n (H) = 3
∴ P (H) =P (sum as 10): = 3/36 = 1/12
Let I is event of getting sum as 11
I = {(5,6) (6,5)}
n (1)=2 ∴ P (I) = 2/36 = 1/18
Let J is event of getting sum as 12
J= {(6, 6); n (J) = 1
∴ P (J) = 1/36
(ii) No, here all 11 possible outcomes are not equally likely
∴ Three probabilites are different.
Q. 23. A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e. three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.
Solution. When a coin tossed three times, then possible out comes are
S = {HHH, HHT HTH, THH, HTT, THT, TTH, TTT}
n (S) = 8
Let A is event of getting all the three same results i.e., {HHH, TTT}
Q. 24. A die 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. When a die is thrown twice all possible outcomes are
n (S) = 36
Ler A is event that 5 will come up either time
n (A) = 11
∴ Ā is event that 5 will not come up either time.
n(Ā) = 36- 11 = 25
(i) ∴ Probability of not getting 5 up either time = 25/36
P (Ā) = 25/36
Probability that 5 will not come up at least once = 11/36
∴ P (A) = 11/36. Ans.
Q. 25. Which of the following arguments are correct? Give reasons for your answer :
(i) If two coins are tossed simultaneously there are three possible outcomes – two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is 1/3.
(ii) If a die is thrown, there are two possible outcomes. an odd number or an even number. Therefore, the probability of getting an odd number is 1/2.
Solution. (i) When two coins are tossed the possible outcomes are S = {HH, HT, TH, TT}
Probability of getting 2 Heads = 1/4
P (HH) = 1/4
Probability of getting two tails = 1/4
P (TT) = 1/4
Probability of getting one head and one tail = 2/4 = 1/2
∴ (i) argument is incorrect.
(ii) When a die is thrown possible outcomes are
S = (1, 2, 3, 4, 5, 6}
n (S) = 6
Odd numbers are 1, 3, 5
∴ Probability of getting odd number = 3/6 = 1/2
Even numbers are 2, 4, 6
∴ Probability of getting even number = 3/6 = 1/2
(ii) argument is correct.
TEXT BOOK EXERCISE 15.2
(OPTIONAL)
Q. 1. Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as an another day. What is the probability that both will visit the shop on (i) the same day ? (ii) consecutive days? (iii) different days ?
Solution. When Shyam and Ekta visiting a particular shop in the same week. Possible outcomes are:
Here T stands For Tuesday
W stands For Wednesday
Th stands For Thursday
F stands For Friday
S stands For Saturday
n (S) = 25
(i) Let A is event that Shyam and Ekta visit the shop on the same day
A = {(T, T) (W, W) (Th, Th) (F, F) (S, S)}
n (A) = 5
Probability that both will visit the shop on same day
(ii) Let B is event that both will visit consecutive days particular shop
B = {(T, W) (W, T) (W, Th) (Th, W) (Th, F) (F, Th) (F, S) (F, S)}.
n (B) = 8
∴ Probability that both will visit particular shop on consecutive days = 8/25. Ans.
(iii) Probability that both will visit the shop on different days
= 1 – Probability that both will visit the shop on consecutive days
Q. 2. A die is numbered in such a way that its faces show the numbers 1, 2, 2, 3, 3, 6. It is thrown two times and the total score in two throws is noted. Complete the following table which gives a few values of the total score on the two throws :
What is the probability that the total score is
(i) even? (ii) 6 ? (iii) at least 6 ?
Solution. The complete table is
Number of all possible out comes = 6 × 6 = 36
(i) Let A is event of getting total as even
A = {2, 4, 4, 4, 4, 4, 4, 4, 4, 6, 6, 6, 6, 8, 8, 8, 8, 12}
n (A) = 18
∴ Probability of getting an even number
(ii) Let B is event of getting sum as 6
B = {6, 6, 6, 6)
n (B) = 4
(iii) Let C is event of getting sum at least 6
C = {6, 6, 6, 6, 7, 7, 8, 8, 8, 8, 9, 9, 9, 9, 12}
n (C) = 15
Q. 3. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag.
Solution. Number of red balls = 5
Let number of blue balls = x
∴ Total number of balls = 5 + x
According to question,
Probability of drawing blue ball
= 2 Probability of Red ball
x = 10
∴ Number of blue balls = 10 Ans.
Q. 4. A box contains 12 balls out of which x are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball ?
If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find x.
Solution. Total number of balls in bag = 12
Number of black balls = x
If 6 more balls put in the box then total umber of balls in the box = 12 + 6 = 18
Number of black balls = x + 6
According to Question,
Probability of drawing black ball = 2
Probability of drawing black ball in first case
x + 6= 3x
6 = 3x – x
6 = 2x
x = 3
∴ Number of black balls = 3 Ans.
Q. 5. A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is 2/3 Find the number of blue marbles in the jar.
Solution. Total number of marbles in jar = 24
Let number of green marbles = x
∴ Number of blue marbles = 24 – x
When a marble is drawn
Probability of drawing green marble = 2/3
∴ Number of green marbles = 16
∴ Number of blue marbles = 24 – x = 24 – 16 = 8.
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