Chapter 15. PROBABILITY 

Chapter 15. Probability

Exercise 15.1 Complete solutions

Important Notes : 1. The theoretical (classical) probability of an event E , written as P(E) , is defined as   2. The probability of a sure event or certain event is 1. 3. The probability of an impossible event is 0. 4. The probability of an event  is a number  such that 5. An event having only one outcome of the experiment is called an elementary event . 6. The sum of the probabilities of all the elementary events of an experiment is 1. 7. For any event , , where  stands for ‘not  ’.  and  are called complementary events. 8. For one coin :  The sample space  H , T 9. For two coins :    The sample space  HH , HT , TH , TT   10. For three coins : The sample space HHH , HHT , HTT , HTH , THT , THH , TTH , TTT 11. For one dice : The sample space  1, 2 , 3 , 4 , 5 , 6   12. For two dice : The sample space (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1), (3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)   13. The playing card consists of 52 cards which are divided into 4 suits of 13 cards each : (i) Spades (black colour) : Ace , king , queen , jack , 10 ,9 , 8 , 7 , 6 , 5 , 4 , 3 and 2 (ii­) Clubs  (black colour) : Ace , king , queen , jack , 10 ,9 , 8 , 7 , 6 , 5 , 4 , 3 and 2 (iii) Hearts (Red colour) : Ace , king , queen , jack , 10 ,9 , 8 , 7 , 6 , 5 , 4 , 3 and 2 (iv) Diamonds (Red colour) : Ace , king , queen , jack , 10 ,9 , 8 , 7 , 6 , 5 , 4 , 3 and 2 (v)  Kings, queens and jacks are called face cards. 14. Prime number 1 to 100 are : 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 , 41 , 47 , 53 , 59 , 61 , 67 , 71 , 73 , 79 , 83 , 89 , 97 .

Class 10 Maths Chapter 15. PROBABILITY Exercise 15.1 Solutions

1. Complete the following statements :

(i) Probability of an event E + Probability of the event ‘ not E’  .

(ii) The probability of an event that is certain to 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) 1     

[ We have ,   ]

  (ii) 0 , impossible event  

(iii)  1 , sure event or certain event .

(iv) 1   

[ The sum of the probabilities of all the elementary events of an experiment is 1 .]

(v)  0 , 1 

[The probability of an event is greater than or equal to 0 and less than or equal to 1 .]

2. Which of the 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) The experiments have no equally likely outcomes . Because, we can not justify to assume that each outcome is a likely to occur as of the other .  

(ii) The experiments have no equally likely outcomes . Because, we can not justify to assume that each outcome is a likely to occur as of the other . 

(iii) The experiments have equally likely outcomes . Because, we can justify to assume that each outcome is a likely to occur as of the other . 

(iv) The experiments have equally likely outcomes .Because, we can justify to assume that each outcome is a likely to occur as of the other . 

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 we toss a coin, the outcomes head and tail are equally likely . So, the result of an individual coin toss is completely unpredictable .

4. Which of the following cannot be the probability of an event ?

 (A)         (B)        (C)       (D)   

Solution :  (B)  – 1.5

[The probability of an event E is a number such that  .]

5. If  , what is the probability of ‘ not E’ ?

Solution : We have ,  

  

  

6. A bag contains lemon flavoured candies only . Malini 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 : Total number of possible outcome   

(i) The number of an orange flavoured candy in the bag   

P(getting an orange flavoured candy)  

(ii) The number of a lemon flavoured candy in the bag  

P(getting a lemon flavoured candy)       

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 :  Here,  

We have ,  

        

        

                     

Therefore, the probability that the 2 students have the same birthday is 0.008 .

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 : Total number of balls  = 3 + 5 = 8

(i) The number of red balls  = 3    .  

P( getting red balls)   

(ii) The number of balls are not red = 8 – 3 = 5 . 

P(getting not red balls)  

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 probability that the marble taken out will be 

(i) red ?    (ii) white ?  (iii) not green ?

Solution : Total number of balls  = 5 + 8 + 4 = 17

(i) The number of red balls  = 5    .  

P( getting the red balls)    

(ii)  The number of white balls  = 8    .  

 P( getting the white balls)    

(iii) The number of balls are not green = 17 – 4 = 13 . 

  P(getting not red balls)  

10. A piggy 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 : Total number of coins

(i) The number of a 50 p coins  

P(getting a 50 p coin)

(ii) The number of coin will not be a Rs 5 coin

P( getting not be a Rs 5 coin)  

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 (see Fig. 15.4) . What is the probability that the fish taken out is a male fish ?

Solution : Total number of fish  =  

The number of male fish  

P(getting a male fish)   

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 (see Fig. 15.5) 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 possible outcome  = 8 

 (i)   The favourable number of elementary events  = 1  [ i.e., 8 ]

 P(getting 8)   

 (ii)   The favourable number of elementary events  = 4  [1 , 3 , 5 , 7 ]

 P(getting an odd number)  

(iii) The favourable number of elementary events  = 6  [ 3 , 4 , 5 , 6 , 7 , 8 ]

P(getting an odd number)   

(iv) The favourable number of elementary events  = 8  [1 , 2, 3 , 4 , 5 , 6 , 7 , 8 ]

P(getting a number less than 9)   

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 :  Total number of possible outcome  = 6 

(i)   The favourable number of elementary events  = 3  [ 2 , 3 , 5 ]

 P(getting a prime number)   

(ii)   The favourable number of elementary events  = 3  [3 , 4 , 5 ]

P(getting a number lying between 2 and 6)  

(iii) The favourable number of elementary events  = 3  [ 1 , 3 , 5 ]

P(getting an odd number)   

14. One card is drawn from a well-shuffled 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 : Total number possible outcome   .

(i) The number of a king of red colour  

P(getting a king of red colour)   

(ii)  The number of a face card   

 P(getting a king of red colour)  

(iii)  The number of a red face card  

 P(getting a red face card)  

(iv)  The number of the jack of hearts  

 P(getting the jack of hearts)  

(v)  The number of a spade  

P(getting a spade)  

(vi)  The number of the queen of diamonds  

P(getting the queen of diamonds)  

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: Total number of possible outcome  

(i) The number of favourable outcome  

 P(getting the queen)  

(ii) Total number of possible outcome   [Queen out]

(a)  P(getting an ace)

(b) P(getting a queen)

16. 12 defective pens are accidentally mixed with 132 good ones . It is not possible to just look at a pen and tell whethe 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 : Total number of pen

The number of good pen  .

P(getting the pen taken out is a good one)

17. (i) A lot of 20 bulbs contain 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)  Total number of bulbs  .

  The number of defective bulbs   .

   P(getting the bulbs is not defective)  

(ii) Total number of bulbs  [not replaced]

    The number of the bulbs is not defective  

      P(getting the bulb is not defective)  

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 : Total number of discs in the box  

(i) The number of a two-digit number in box  .

      P(getting a two-digit number)  

(ii) The number of a perfect square number in box  

      P(getting a perfect square number)  

(iii)  The number of  discs divisible by     

      P(getting a number divisible by 5)  

19. A child has a die whose six faces show the letters as given below :

   

The die is thrown once .

What is the probability of getting  (i) A  ?    (ii) D

Solution : Total number of possible outcome  .

(i) The number of favourable outcome  

P(getting A)  

(ii) The number of favourable outcome  

P(getting D)  

20. Suppose you drop a die random on the rectangular region shown in Fig. 15.6 . What is the probability that it will land inside the circle with diameter 1 m ?

              

Solution : The total area of the rectangular region  

Area of the circle

     [Radius =12 m ]  

  P(getting the circle)  

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 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 ball pens = 144 .  

The number of defective ball pens = 20 .

The number of good ball pens = 144 – 20 =124 

(i) P( getting a good ball pens)   .

(ii) P(getting a defective ball pens) 

22. Refer to Example 13. (i) complete the following table :

Event :  ‘Sum on 2 dice’	Probability
2
3
4
5
6
7
8
9
10
11
12

(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  . Do you agree with this argument ? justify your answer .

Solution : The sample space  {(1,1) , (1,2) , (1,3) , (1,4) ,

(1,5) , (1,6) , (2,1) , (2,2) , (2,3) , (2,4) , (2,5) , (2,6) , (3,1) ,

(3,2) , (3,3) , (3,4) , (3,5) , (3,6) , (4,1) , (4,2) , (4,3) , (4,4)

, (4,5) , (4,6) , (5,1) , (5,2) , (5,3) , (5,4) , (5,5) , (5,6) , (6,1)

, (6,2) , (6,3) , (6,4) , (6,5) , (6,6)}

 (i) We have ,

Event :  ‘Sum on 2 dice’	Probability
2
3
4
5
6
7
8
9
10
11
12

(ii)  No  . because, the eleven sums are not equally likely .

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: The sample space  { HHH , HHT , HTT , TTH , THH , THT , HTH , TTT }

Total number of possible outcome  .

The number of favourable outcome  

P(Hanif  will lose the game )

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 ?

[Hint : Throwing a die twice and throwing two dice simultaneously are treated as the same experiment ]

Solution: The sample space

   {(1,1) , (1,2) , (1,3) , (1,4) , (1,5) , (1,6) , (2,1) , (2,2) , (2,3) , (2,4) , (2,5) , (2,6) , (3,1) , (3,2) , (3,3) , (3,4) , (3,5) , (3,6) , (4,1) , (4,2) , (4,3) , (4,4) , (4,5) , (4,6) , (5,1) , (5,2) , (5,3) , (5,4) , (5,5) , (5,6) , (6,1) , (6,2) , (6,3) , (6,4) , (6,5) , (6,6)}

Total number of possible outcome

(i) The number of  favourable outcome  

   P(not 5)  

(ii) The number of  favourable outcome  

    P(at least 5 once)  

25. Which of the following arguments are correct and which are not 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   .

(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  .

 Solution :  (i) Total number of possible outcome  

 The favourable number of outcome  

 P(getting one outcome)  

 Therefore, the arguments is not correct .

 (ii) Total number of possible outcome  

  Total number of odd number  

 P(getting an odd number)  

Therefore, the arguments is correct .