PROBABILITY ( LECTURE 5)

LESSON  5     PROBABILITY

  27TH APRIL 2020

Dear students 

GOOD MORNING  

LETS SOLVE FEW QUESTION S WHICH ARE IMPORTANT, FROM THE EXAMINATION POINT OF VIEW 

BEFORE WE START, QUICK REVISION  OF THE FOLLOWING

 PRIME NUMBER:-    A Number that is divisible only, by itself  and 1    (e.g 2,3,5,7 ,11........)

COMPOSITE  NUMBER:- Is a postive integer  which can be expressed as product of primes   (e.g  4,6,8,9,10,12..........)  

 UNIQUE NUMBER:- 1  is called a unique number because it is neither prime nor composite number.

 DO IT  IN THE  REGISTER 👇

Question 1

A box contains cards bearing numbers from 6 to 70. If one card is drawn at random from the box, find the probability that it bears

1. a one digit number.
2. a number divisible by 5.
3. an odd number less than 30.
4. a composite number between
   50 and 70.

Solution:
Number of cards in the box = 65

1. Cards bearing one digit number are 6, 7,8, 9 (i.e. 4 cards)
   Probability of card bears a one digit number=4/65
2. B : Number on the cards is divisible by 5.
   Cards favourable to B are 10,15,20,25,30,35,40,45,50,55,60,65,70 (i.e. 13 cards).
   P(B)=13/65=1/5
3. C: Cards with an odd number less than 30.
   Cards favourable to C are 7, 9,11,13,15,17,19, 21,23, 25, 27, 29 (i.e. 12 cards).
   P(C)=12/65
4. D : Card with composite number between 50 and 70.
   Cards favourable to D are 51,52,54,55,56,57,58,60,62,63,64,65, 66,68, 69 (i.e. 15 cards).
   P(D) =15/65=3/13

Question 2.

A bag. contains cards numbered from 1 to 49. A and C is drawn from the beg at random, after mixing the cards thoroughly. Find the probability that the number on the drawn card is

1. an odd number.
2. a multiple of 5.
3. a perfect square.
4. an even prime number

Solution:
Total number of cards in a bag = 49.

1. Number of odd number cards = 25
   .’.Required probability = 25/49
2. Number of multiple of 5 = 9 (i.e. 5,10,15, 20, 25, 30, 35,40, 45)
   .’.Required probability = 9/49
3. Number of perfect square = 7 (i.e. 1,4,9,16, 25, 36,49)
   .’. Required probability = 7/49=1/7
4. Number of prime number = 1 (i.e. 2)
   .’. Required probability = 1/49

Question 3.
A box contains cards numbered 3,5,7,9,…, 35,37. A card is drawn at random from the box. Find the probability that the number on the drawn card is a prime number.
Solution:
Total no. of cards in the box = 18
Number of ways to draw one card = 18
Number of cards with a prime number are 11 (i.e. 3,5, 7,11,13,17,19, 23, 29, 31, 37)
Number of ways to draw a card bearing a prime number = 11
Required probability =11/18

Question 4.

A die is thrown once. Find the probability of getting the following:

1. a prime number
2. a number lying between 2 and 5.
3. a composite number

Solution:
The possible outcomes are = 6 (i.e. 1, 2, 3, 4, 5, 6). .

1. Prime numbers on a die = {2, 3, 5}
   P(getting a prime number) =3/6=1/2
2. Number lying between 2 and 5 are 2 (i.e. 3, 4).
   P(getting a number lying between 2 and 5) =2/6=1/3
3. Composite numbers = (4,5)  NOTE  (1 is neither prime nor composite number )     P (getting  a composite number) =  2 / 5

Question 5.

A box contains 70 cards numbered from 1 to 70. If one card is drawn at random from the box, find the probability that it bears

1. a perfect square number.
2. a number divisible by 2 and 3

Solution:

Question 6.

Two dice are rolled once. Find the probability of getting such numbers on the two dice,

(i) whose product is 12.
(ii) is a doublet


Solution:
(i)When two dice are rolled, total number of cases = 36
Number of favourable cases whose product is 12 are given as {(2,6), (3,4), (4,3), (6,2)}, i.e. 4

(ii)  doublets  are =  (1,1,(2,2),(3,3)  ( 4,4,( 5,5) and (6,6)
P(a doublet) = 6 /36  = 1 / 6

 Question 7.
A card is drawn at random from a well-shuffled deck of playing cards. Find the probability that the card drawn is

1. a a card of spade or an ace.
2. a black king.
3.  neither a jack nor a king
4. either a king or a queen

Solution:

Total number of outcomes = 52

1.  A = Card is spade   = 13 and an ace = 4 - 1 = 3
   Cards favourable to A = 13 + 3 = 16
   P(A)=16/52=4/13
2. B = Card is black king
   Number of black kings = 2
   P(B)=2/52=1/26
3. C = Card is neither a jack nor a king
   Number of favourable cards to C = 52-4-4 = 44
   P(C) =44/52=11/13
4. D = Card is either a king or a queen
   Number of cards favourable to D = 4 + 4 = 8
   P(D) = 8/52=2/13

Question 8
A letter of English alphabet is chosen at random. Determine the probability that the chosen letter is a consonant.
Solution:
Total english alphabets = 26
Number of consonants in English alphabets = 21
.’. P(Choosing a consonant) =21/26

 NOW LETS SOLVE      EXERCISE 15.2

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 on another day. What is the probabilihy that t tlt will visit the shop on (i) the same day? (ii) consecutive days? (iii) different days?

Sol. Here, the number of all the possible outcomes
                                                                                     Total possible 

                                                                                       outcomes

 ( Tue,tue),(Tue,Wed),(Tue,Thu),(Tue,Fri),(tue,Sat)     = 5
( wed,tue),(wed,Wed),(wed,Thu),(wed,Fri),(wed,Sat  )=5
 (Thu,tue),(Thu,Wed),(Thu,Thu),(Thu,Fri),(Thu,Sat)   =5
(Fri,tue),(Fri,Wed),(Fri,Thu),(Fri,Fri),(Fri,Sat)              =5

(Sat,tue),(Sat,Wed),(Sat,Thu),(Sat,Fri),(Sat ,Sat)         =5

                             

total possible outcomes  = 5 × 5 = 25

        (i) For both customers visiting same day:

                Number of favourable outcomes = 5

                                                [∵ (Tue., Tue.), (Wed., Wed.), (Thu., Thu.), (Fri., Fri.), (Sat., Sat.)]

        

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 fiew 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?

Sol. The completed table is as under:

        ∴

 Number of all possible outcomes = 36

        (i) For total score being even:

                Favourable outcomes = 18

                                                [∵ The even outcomes are: 2, 4, 4, 4, 4, 8, 4, 4, 8, 4, 6, 6, 4, 6, 6, 8, 8]

                

        (ii) For the score being 6:

                In list of score, we have four 6' s.

                ∴ Favourable outcomes = 4

                

        (iii) For the score being at least 6:

                The favourable scores are:

                7, 8, 8, 6, 6, 9, 6, 6, 9, 7, 8, 8, 9, 9 and 12

                ∴ Number of favourable outcomes = 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.

Sol. Let the number of blue balls in the bag be x.

        ∴ Total number of balls = x + 5

        Number of possible outcomes = (x + 5).

        For a blue ball favourable outcomes = x

        ∴ Probability of drawing a blue ball

        

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 lx f ire. Find x.

Sol. ∵The total number of balls in the box = 12

        ∴ Number of possible outcomes = 12

        Case-I: For drawing a black ball

        Number of favourable outcomes = x

        ∴ Probability of getting a black ball 

        Case-II: When 6 more black balls are added

        Now, the total number of balls

                = 12 + 6

                = 18

        ⇒ Number of possible outcomes = 18

        Since, the number of black balls now

                = (x + 6).

        

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  Find the number of blue balls in the jar.

Sol. ∵There are 24 marbles in the jar.

        ∴ Number of possible outcomes = 24.

        Let there are x blue marbles in the jar.

        ∴ Number of green marbles = 24 – x

        ⇒ Favourable outcomes = (24 – x)

        ∴ Required probability for drawing a green marble

                

           

SOLVE IT 

Question 1.

A box contains 80 discs which are numbered from 1 to 80. If one disc is drawn at random from the box, find the probability that it bears a perfect square number                    . ANS 1/10

Question 2.
A ticket is drawn at random from a bag containing tickets numbered from 1 to 40. Find the probability that the selected ticket has a number which is a multiple of 5.              ANS 1/5

Question 3
Cards marked with numbers 5,6,7, , 74 are placed in a bag and mixed thoroughly. One card is drawn at random from the bag. Find the probability that the number on the card is a perfect square.            ANS 3/35

Question 4
A bag. contains cards numbered from 1 to 49. A and C is drawn from the beg at random, after mixing the cards thoroughly. Find the probability that the number on the drawn card is

1. an odd number.
2. a multiple of 5.
3. a perfect square.
4. an even prime number

 ANS  25/49.,  9/49,1/7,1/49

REVISE THE CHAPTER WELL  

TOMORROW THERE WILL BE A  SMALL TEST  

THANK YOU  AND  TAKE CARE OF YOURSELF