TOPICS
Unit-15(Examples)

Example-1 :-  Find the probability of getting a head when a coin is tossed once. Also find the probability of getting a tail.

Solution :-
  In the experiment of tossing a coin once, the number of possible outcomes is two — Head (H) and Tail (T). 
  Let E be the event ‘getting a head’. 
  The number of outcomes favourable to E, (i.e., of getting a head) is 1. 
  
  Therefore, 
  P(E) = P(head) = 1/2

  Similarly, if F is the event ‘getting a tail’, then
  P(F) = P(tail) = 1/2
    

Example-2 :-  A bag contains a red ball, a blue ball and a yellow ball, all the balls being of the same size. Kritika takes out a ball from the bag without looking into it. What is the probability that she takes out the
(i) yellow ball?
(ii) red ball?
(iii) blue ball?

Solution :-
  Let Y be the event ‘the ball taken out is yellow’, 
  B be the event ‘the ball taken out is blue’, 
  and R be the event ‘the ball taken out is red’.
  Now, the number of possible outcomes = 3.
         
(i) The number of outcomes favourable to the event Y = 1.
  P(Y) = 1/3

(ii) Similarly, P(B) = 1/3

(iii) Similarly, P(R) = 1/3
    

Example-3 :-  Suppose we throw a die once.
(i) What is the probability of getting a number greater than 4 ?
(ii) What is the probability of getting a number less than or equal to 4 ?

Solution :-
(i) Here, let E be the event ‘getting a number greater than 4’. 
  The number of possible outcomes is six : 1, 2, 3, 4, 5 and 6, and the outcomes favourable to E are 5 and 6. 
  Therefore, the number of outcomes favourable to E is 2. So,
  P(E) = P(number greater than 4) = 2/6 = 1/3
    
(ii) Let F be the event ‘getting a number less than or equal to 4’. 
  Number of possible outcomes = 6 Outcomes favourable to the event F are 1, 2, 3, 4. 
  So, the number of outcomes favourable to F is 4.
  P(F) = 4/6 = 2/3
    

Example-4 :-  One card is drawn from a well-shuffled deck of 52 cards. Calculate the probability that the card will
(i) be an ace,
(ii) not be an ace.

Solution :-
(i) There are 4 aces in a deck. Let E be the event ‘the card is an ace’. 
  The number of outcomes favourable to E = 4 
  The number of possible outcomes = 52 
  Therefore, P(E) = 4/52 = 1/13
    
(ii) Let F be the event ‘card drawn is not an ace’. 
  The number of outcomes favourable to the event F = 52 – 4 = 48.
  The number of possible outcomes = 52
  Therefore, P(F) = 48/52 = 12/13
    

Example-5 :-  Two players, Sangeeta and Reshma, play a tennis match. It is known that the probability of Sangeeta winning the match is 0.62. What is the probability of Reshma winning the match?

Solution :-
  Let S and R denote the events that Sangeeta wins the match and Reshma wins the match, respectively. 
  The probability of Sangeeta’s winning = P(S) = 0.62 (given) 
  The probability of Reshma’s winning = P(R) = 1 – P(S) 
  [As the events R and S are complementary] = 1 – 0.62 = 0.38
    

Example-6 :-  Savita and Hamida are friends. What is the probability that both will have
(i) different birthdays?
(ii) the same birthday? (ignoring a leap year).

Solution :-
  Out of the two friends, one girl, say, Savita’s birthday can be any day of the year. 
  Now, Hamida’s birthday can also be any day of 365 days in the year. 
  We assume that these 365 outcomes are equally likely. 

(i) If Hamida’s birthday is different from Savita’s, the number of favourable outcomes for her birthday is 365 – 1 = 364
  So, P (Hamida’s birthday is different from Savita’s birthday) = 364/365

(ii) P(Savita and Hamida have the same birthday) = 1 – P (both have different birthdays)
= 1 - 364/365
= 1/365

    

Example-7 :-  There are 40 students in Class X of a school of whom 25 are girls and 15 are boys. The class teacher has to select one student as a class representative. She writes the name of each student on a separate card, the cards being identical. Then she puts cards in a bag and stirs them thoroughly. She then draws one card from the bag. What is the probability that the name written on the card is the name of
(i) a girl?
(ii) a boy?

Solution :-
  There are 40 students, and only one name card has to be chosen. 

(i) The number of all possible outcomes is 40 The number of outcomes favourable for a card with the name of a girl = 25 .
  Therefore, P (card with name of a girl) = P(Girl) = 25/40 = 5/8

(ii) The number of outcomes favourable for a card with the name of a boy = 15 .
  Therefore, P(card with name of a boy) = P(Boy) = 15/40 = 3/8
    

Example-8 :-  A box contains 3 blue, 2 white, and 4 red marbles. If a marble is drawn at random from the box, what is the probability that it will be
(i) white?
(ii) blue?
(iii) red?

Solution :-
  The number of possible outcomes = 3 + 2 + 4 = 9 
  Let W denote the event ‘the marble is white’, 
  B denote the event ‘the marble is blue’ and 
  R denote the event ‘marble is red’. 
        
(i) The number of outcomes favourable to the event W = 2
  So, P(W) = 2/9

(ii) The number of outcomes favourable to the event B = 3
  So, P(B) = 3/9 = 1/3

(iii) The number of outcomes favourable to the event R = 4
  So, P(R) = 4/9
    

Example-9 :-  Harpreet tosses two different coins simultaneously (say, one is of Re 1 and other of Rs 2). What is the probability that she gets at least one head?

Solution :-
  The outcomes favourable to the event E, ‘at least one head’ are (H, H), (H, T) and (T, H). 
  So, the number of outcomes favourable to E is 3.
  Therefore, P(E) = 3/4
  i.e., the probability that Harpreet gets at least one head is 3/4.
    

Example-10 :-  In a musical chair game, the person playing the music has been advised to stop playing the music at any time within 2 minutes after she starts playing. What is the probability that the music will stop within the first half-minute after starting?

Solution :-
  Let E be the event that ‘the music is stopped within the first half-minute’.
  The outcomes favourable to E are points on the number line from 0 to 1/2.
  The distance from 0 to 2 is 2, while the distance from 0 to 1/2 is 1/2.
  Since all the outcomes are equally likely, we can argue that, of the total distance
  of 2, the distance favourable to the event E is 1/2.
  So, P(E) = (1/2)/2 = 1/4
    

Example-11 :-  A missing helicopter is reported to have crashed somewhere in the rectangular region shown in Figure. What is the probability that it crashed inside the lake shown in the figure? probability

Solution :-
  The helicopter is equally likely to crash anywhere in the region. 
  Area of the entire region where the helicopter can crash = (4.5 × 9) km² = 40.5 km²
  Area of the lake = (2.5 × 3) km² = 7.5 km²
  Therefore, P (helicopter crashed in the lake) = 7.5/40.5 = 75/405 = 5/27
    

Example-12 :-  A carton consists of 100 shirts of which 88 are good, 8 have minor defects and 4 have major defects. Jimmy, a trader, will only accept the shirts which are good, but Sujatha, another trader, will only reject the shirts which have major defects. One shirt is drawn at random from the carton. What is the probability that
(i) it is acceptable to Jimmy?
(ii) it is acceptable to Sujatha?

Solution :-
  One shirt is drawn at random from the carton of 100 shirts. 
  Therefore, there are 100 equally likely outcomes.

(i) The number of outcomes favourable (i.e., acceptable) to Jimmy = 88
  Therefore, P (shirt is acceptable to Jimmy) = 88/100 = 0.88

(ii) The number of outcomes favourable to Sujatha = 88 + 8 = 96
  So, P (shirt is acceptable to Sujatha) = 96/100 = 0.96
    

Example-13 :-  Two dice, one blue and one grey, are thrown at the same time. Write down all the possible outcomes. What is the probability that the sum of the two numbers appearing on the top of the dice is
(i) 8?
(ii) 13?
(iii) less than or equal to 12?

Solution :-
  The number of possible outcomes = 6 × 6 = 36. 
        probability
(i) The outcomes favourable to the event ‘the sum of the two numbers is 8’ denoted by E, are: 
  (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) (see Figure) i.e., the number of outcomes favourable to E = 5.
  Hence, P(E) = 5/36

(ii) As you can see from Figure, there is no outcome favourable to the event F, ‘the sum of two numbers is 13’.
  So, P(F) = 0/36 = 0

(iii) As you can see from Figure, all the outcomes are favourable to the event G, ‘sum of two numbers ≤ 12’.
  So, P(G) = 36/36 = 1
    
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