how to solve the following question? Tickets numbered 1 to 20 are mixed up together and then a ticket is drawn at random. What is the probability that the ticket has a number, which is the multiple of 3 or 7?

about 3 years ago 0 Answer 394 views

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A card is drawn from an ordinary pack and the gambler bets that it's a spade or an ace. What are the odds against winning the bet?

about 3 years ago 0 Answer 477 views

in this test how is the answer for the question: what is the probability of a coin getting tail and the answer came 1 i nstead of  the correct answer 1/2

 

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What is the probability of occurring four Wednesdays in 23 consecutive days?

about 5 months ago 1 Answer 92 views

What is the probability of occurring four Wednesdays in 23 consecutive days?

about 5 months ago 0 Answer 1 views

What is the probability of occurring four Wednesdays in 23 consecutive days?

about 5 months ago 0 Answer 0 views

There are 100 cards in a bag on which numbers from 1 to 100 are written. A card is taken out from the bag at random. Find the probability that the number on the selected card 

(i)  Is divisible by a and is a perfect square

(ii) Is a prime number greater than 80.

 

Sample space = {1,2,3.....99,100}

(i) Number divisible  by and perfect square are

A  = {9,36,81}

n (A) = 3

therefore Requared probability  P (A)  = frac{n(B)}{n (s)} = frac{3}{100}

(ii) Prime numbers greater than 80 and less than 100 are B  = {83,89, 97}

therefore n (B) = 3

therefore Required probability P (B) = frac{n(B)}{n(s)} = frac{3}{100}

A die is rolled 200 times and its outcomes are recorded below:

Outcome        1     2      3      4      5      6
Frequency    25   35    40    28    42    30


(i)  An even prime Find probability getting 

(ii) A multiple of 3.

(i) An even prime number i.e., '2'
therefore P (getting an even prime number) = frac{35}{200}

                                                        = frac{7}{40}

(ii) Multiple of 3  i.e., 3 and  6 

therefore (getting multiple of 3) = frac{40+30}{200}

                                      = frac{70}{200}

                                      = frac{7 }{2 0}

Teachers and students are selected randomly to make two teams of 20 members each on sports day to participate in the event of '' tug of war''. The  numbers of volunteers are as follows:

 

      Teachers                                                Students 

Male        Female                                   male       female
  12             18                                       20            10 

 

Find the probability that the person choosen at randomly 

(i) is male 

(ii) is a female student 

 Teachers                                                Students 

Male        Female                                   male       female
  12             18                                         20             10 

 

Total number of volunteers = 12+18+20+10

                                          = 60

Total number of males = 12+20 = 32

(i)  P  (Person is male)   =  frac{32}{60}=frac{8}{15}

(ii) P (Person is female student)   =  frac{10}{60} = frac{1}{6}

A die is thrown 1000 times the frequencies of outcomes 1,2,3,4,5 and 6  as given below:

Outcome         1             2             3          4                5             6

Frequency       179         150        157      149            175        190

A die is thrown once again . Find the probability of outcome '' greater than3'.

 

Total number of outcomes = 100

Number of outcomes  greater than 3

= (149+175+190)

= 514

thereforeRequired probability = frac{514}{1000}=0.514

On a particular day, the number of vehicles passing through a crossing is given below:

Vehicle           2 wheeler         3 wheeler      4 wheeler         

Frequency           57                 33                        30

A particular vehicle is choosen at random. what is the probability that it is not a four wheeler

 

 Total number of vehicles = 57+33+30=120

 vehicles which are not four wheelers

= 57+33 = 90

thereforeP (Choosen vehicle is not four wheeler )

= frac{90}{120}=frac{3}{4}

The table given below shows the marks obtained by 80 students of a class in a test with maximum marks 100: 

Marks                  0-20      20-40      40-60        60-80     Above 80

No.of students     8              16           40               10              6

A student is choosen at random. Find the probability that he gets:

(i) less than 40 marks 

(ii) 60% OR more marks.

 

Total number of students =80

(i) Number of students getting less than 40 marks 

=8+16= 24

P (less than 40 marks) = frac{24}{80} = frac{3}{10}

(ii) Number of students getting 60 or more than 60

P(60 or more than 60 marks) 

frac{16}{80}= frac{1}{5}

The following table shows the marks obtained by 30 students in a class test: 

 

Marks obtained       70      58     60      52     65   75   68

No. of students       3        5       4        7       6      2     3

 

Find the probability that students score:

(i) 60 marks

(ii) Less than 60 marks

 

 

Total number of students  = 30

(i) Students getting 60 marks = 4

Probability of getting 60 marks = frac{4}{30} = frac{2}{15}

(ii) Students getting less than 60 marks  = (5+7)= 12

Probability of getting less than 60 marks

= frac{12}{30}= frac{2}{5}

Following is the data about the months of birth of 40 students in class IX:

Feb, Jan, July, June, March, Feb, Feb, Feb, Nov, Jan, Jan, Dec, May, June, Jue, July, June Nov,

Dec, June, July, June, Aug, Dec, June, Mar, July, July, June, Dec, Sep, Marc, Jan, Dec, June, Dec, Sep, March, Jan, Nov.

One student is chosen at random. ind the probability that the student choosen:

(i)  Was born in June.

(ii) Was not born in the month of June.

 

Month Student
Jan 5
Feb 4
March 

4

May

1

June 9
July 5
Aug 1
Sep 2
Nov 3
Dec 6

 

 

(i) Let E1 be the event of the student born in June

                  therefore P(E_{1}) = frac{9}{40}

(ii) Let E2 be the event of the student not born in June 

                 

Favourable outcomes  = (40-9) = 31

therefore P(E_{2}) = frac{31}{40}

 

 

 

Given below is the frequency distribution of salary (in rupees) of 80 workers in a factory.

 

if a  worker is selected at random, find the probability thtat his salary is

(i) Less than  Rs. 3000

(ii) More  than or equal to Rs. 1000

(iii) More than or equal to Rs. 2000 but less than Rs.4000

 

(i) The probability of getting a salary of less than Rs. 3000

= frac{12+18}{80}

= frac{30}{80} = frac{3}{8}

(ii) The probability of getting a salary more than or equal to Rs., 1000

=frac{ 12+18+22+28}{80}

= frac{80}{80}=1

(iii) The probability of getting salary more or equal to 2000 but less than Rs. 4000

= frac{18+22}{80}

= frac{ 40}{80} = frac{1}{2}

Fifty seeds were selected at random from each of 5 bags of seeds and were kept under standardised conditions favourable to germination. After 20 days, the number of seeds which had germinated in each collection was counted and recorded as follows:

 

What is the probability of the germination of :

(i) More than 40 seeds in a bag

(ii) Less than 41 seeds in a bag

(iii) 49 seeds in a  bag 

(i) P more than 40 seeds in a bag =frac{2}{5}

(ii) P(Less than 41 seeds in a  bag ) =frac{3}{5}

(iii)  P (49 seeds in a bag ) = 0.

Following distribution gives the weight of 38  students of a class:

Weight in kg             No. of students

31-35                           9

36-40                           5

41-45                          14

46-50                           3

51-55                           1

56-60                           2

61-65                           2

66-70                           1

71-85                           1 

Find the probability that the weight of a student in the class is 

(i)  At most  60 kg 

(ii) at least 36 kg 

(iii)  Not more than 50 kg

Total number of students = 38

(i) Number of students whose weight is at most 60 kg

    = 9+5+14+3+1+2 = 34 kg 

     Probability that weight of a student is at most 60 kg 

    =frac{34}{38}=frac{17}{19} 

(ii) No. of students whose weight is at least  36 kg

    = 5+14+3+1+2+2+1+1 = 29 kg 

    Probability that weight is at least 36 kg

    = frac{29}{38}

(iii) No. of students whose weight is not more than 50 kg 

      = 9+5+14+3 = 31

      Probability that the weight of a student is not more than 50 kg  

      = frac{31}{38}

 

 

The heights of the students of a class is measured and recorded as given below:

 

A student is selected at random. Find the probability that height of the student 

(i) More than 135 cm 

(ii) at least 145 cm 

(iii)  Less than 130 cm 

(iv) 125 cm or more but less than 140 cm.

 

Total student = 7+7+11+3+5+9+8 = 50

(i)  P (Height of student is more that 135 cm) = frac{25}{50} =frac{1}{2}

(ii) P (Height of student is at least 145 cm  = frac{17}{50}

(iii) P (Height of student is less than 130 cm = frac{14}{50} =frac{7}{25}

(iv)  P (Height of student is 125 or more but less than 140 = frac{21}{50}

 

In class IX of 50 students, the second language opted by the student is as also follow:

Sanskrit - 14

Japanese - 08

French - 12

Urdu -6

Rest of then opted for German 

A student is selected at random. Find the probability that the student 

(a) opts for French

(b) Does not o[ts forJapanese.

(C) Either opt for Sanskrit or German 

 

(a) Prob. that a student selected is opts 

     French language = =frac{12}{50} =frac{6}{25}

(b) Prob. that a student selected does not opt for Japanese = 1- selected student opts Japanese

     = 1 - frac{8}{50} = 1 - frac{8}{50} =frac{42}{50} = frac{21}{25}

(C) Prob. that selected student either opts for Sanskrit or  for German = prob. of student opts 

     Sanskrit + Prob.of student opts. German

therefore    No. of student who opted German 

      = 50- (14+08+12+6)

      = 50-40

      = 10

therefore  Prob. that selected student either opt for Sanskrit or for German 

     = frac{14}{50} + frac{10}{50}

     = frac{24}{50} + frac{12}{50}

There are 30 scouts and 20 guides in a school. In another school, there are 20 scouts and 15 guides. From each school, one student among them is to be selected for participation in a seminar.

a. What is the total number of possible selection?

b. What is the probability of both being scouts?

c. What is the probability of both being Guides?

d. What is the probability of one scout and one guide?

Name of school Scout Guide Total
A 30 20 50
B 20 15 35

Total number of possible selection  = m x n  (Fundamental  counting theorem)

a. m x n  = 50  x 35 = 1750.

 

b. Probability of both being scouts

   P(S) = frac{F}{N} ; , , , , , F = 30times 20 = 600;, , , , , N = 1750.

  P(S) = frac{F}{N} = frac{600}{1750} = frac{12}{35}

 

c. Probability of both being guide = m x n 

                                                 = 20 x 15 = 300 (F)

   P(G) = frac{F}{N} = frac{300}{1750} = frac{6}{35}

 

d.Probability of one scout and one guide 

     1 - P(S) + P(G) (Formula)

ie., 1 - left ( frac{12}{35} + frac{6}{35} right ) = 1 - frac{18}{35} = frac{17}{35}

A box contains 8 black beads and 12 white beads. Another box contains 9 black beads and 6 white beads. One bead from each box is taken.

a. What is the probability that both beads are black?

b. What is the probability of getting one black bead and one white bead?

 

BOX - 1

Number of black beads = 8

Number of white beads = 12

BOX - 2

Number of black beads = 9

Number white beads = 6

a. The probability of getting blackhead from the box - 1   = frac{8 }{20} = frac{2}{5}

    The probability of getting white beads from the box -2 = frac{9 }{15} = frac{3}{5}

    Hence the probability of getting both black = frac{2 }{5} times frac{3}{5}   = frac{6 }{25}

b. The probability of getting 1 black bead from the box - 1 = frac{2 }{5}

    The Probability of getting I white bead from the box -2 = frac{6 }{15}

                                                                          = frac{2 }{ 5}

    Hence the probability of getting the 1 black and 1 white bead

                                                                               = frac{2 }{ 5} times frac{2 }{ 5} = frac{4}{25}

What is the probability that there are 53 Wednesday in a leap year?

In a leap year number of days = 366 days

                                              = 52 weeks  + 2 days 

therefore Two days can be SM, MT, TW, WTH, THF, FS, SS, =7

therefore Out of these 7 calenders, two calenders will have 53 wednesdays

therefore P (53 Wednesdays in a leap year) = frac{2}{7}

From a pack of 52 playing cards, Jacks,, Queens and Kings of red colour are removed. From the remaining, a card is drawn at random. Find the probability that drawn card is :

i. A black king, 

ii. A card of red colour

iii. A card of black colour

Total cards  = 52

Cards removed  = 6

Card .... = 52 - 6 = 46

Total black king  = 2

Probability of drawing black king = frac{2}{46} = frac{1}{23}

Total red card  = 26 - 6 = 20

Probability of drawing red colour card = frac{20}{46} = frac{10}{23}

Total card of black colour = 26

Probability of drawing black colour card = frac{26}{46} = frac{13}{23}

A box consists of 100 shirts of which 88 are good, 8 have minor defects and 4 have major defects. Ramesh, a shopkeeper will buy only those shirts which are good but 'Kewal' another shopkeeper will not buy shirts with major defects A shirt is taken out of the box at random. What is the probability that:

i. Ramesh will buy the selected shirt?

ii. 'Kewal' will buy the selected shirt?

i. Number of good shirts = 88

   P (Ramesh buys the shirt) = frac{88}{100}, , or , , frac{22}{25}

ii. Number of shirts without Major defect  = 96

   P(Kewal buys a shirt) = frac{96}{100}, , or , , frac{24}{25}

A bag contains 18 balls out of which x balls are red.

i. If one ball is drawn at random from the bag, what is the probability that it is not red?

ii. If 2 more red balls are put in the bag, the of drawing a red ball will be frac{9}{8} times the probability of drawing a red ball in the first case. Find the value of x.

P(red ball) = frac{x}{8}

i) P(no red ball) = 1 - frac{x}{8} = frac{18-x}{18}

ii) Total number of balls  = 18 + 2 = 20

    Red balls are = x  + 2 

    P(red balls) = frac{x+2}{20}

    Now, According to the question, 

    frac{x+2}{20} = frac{9}{8}times frac{x}{18}

    180x = 144x + 288

    36 x = 288

    x = frac{288}{6} = 8

 Cards marked with numbers 3,4,5...........,50 are placed in a bag and mixed throughly. One card is drawn at random from the bag. Find the probability that number on the card drawn is :

i. Divisible by 7

ii. A perfect square

iii. A multiple of 6.

Total number of cards  = 48

Probability of an event 

=frac{total, , , number, , , of , , favourable , , , outcomes}{Total, , number , , of , , outcomes}

Number of cards divisible by 7 = 7

P (Cards divisible by 7) = frac{7}{48}

Number of cards having a perfect square = 6

P(Cards having a perfect square) = frac{6}{48} = frac{1}{8}

Number of multiples of 6 from 3 to 50 = 8

P (Multiple of 6 from 3 to 50) = frac{8}{48} = frac{1}{6}

A game of chance consists of spinning an arrow on a circular board, divided into 8 equal parts, Which comes to rest poiniting at one of the numbers 1,2, 3......8, which are equally likely outcomes. What is the probability that the arrow will point at (i) and odd number (ii) a number greater than 3 (iii) a number less than 9.

i.  Favourable outcomes are  1,3,5,7, ie., Outcomes 

    therefore P (an odd number) = frac{4}{8} , , or, , frac{1}{2}

ii. Favourable outcomes are 4,5,6,7,8 ie., 5 outcomes

    P (a number greater than 3)  = frac{4}{8} , , or, , frac{1}{2}

iii. Favourable outcomes are 1,2,3......8

    P (a number less than 9) = frac{8}{8} = 1

All the red face card are removed from a pack of 52 playing cards. A card is drawn  at random from the remaining cards, afetr reshuffling them. Find the probability that the drawn card is 

i. Of red colour 

ii. A queen

iii. An ace

iv. A face card

i)

No. of crads remaining = 52 - 3 x 2

                                    = 52 - 6 = 46

No.of red cards = 26 - 6 = 20 

P (a red colour) = frac{20}{46} = frac{10}{23}

 

ii) 

No. of queen  = 4 - 2 = 2

P (a queen) =  frac{2}{46} = frac{1}{23}

 

iii)

No.of ace = 4

P (as ace) = frac{4}{46} = frac{2}{23}

 

iv)

No. of face cards = 12 - 6 = 6

P (a face card) = frac{6}{46} = frac{3}{23}

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.

i. A one digit number 

ii. A one number divisible by 5

iii. An odd number less than 30.

iv. A composite number between 50 and 70

Total number of cards = 65

i) P (one digit number) = frac{4}{65}

ii) No. divisible by 5 = 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70 = 13

   P (a number divisible by 5) = frac{13}{65} = frac{1}{5}

iii) Odd no.less than 30 = 7,9,11,13,15,17,19,21,23,25,27,29 = 12

     P (an odd number less than 30 ) = frac{12}{65}

iv) P (a composite number between 50 and 70) = frac{15}{65} = frac{3}{13}

A card is draw at a random from a well shuffled deck of playing cards. Find the probability that teh card drawn is:

i. A card of spade or an ace

ii. A black king

iii. Neither a jack nor a king 

iv. Either a king or a queen.

i)   Cards of spade or an ace  = 13 +  3 = 16

     Total no.of cards = 52

     P (Spade or an ace) = frac{16}{52} = frac{4}{13}

 

ii)  Black kings = 2

     P(a black king) = frac{2}{52} = frac{1}{26}

 

iii) Jack or king  =  4 + 4 = 8

     P (Neither jack nor king) = frac{52-8}{52} = frac{44}{52} = frac{11}{13}

 

iv) King or queen = 4 + 4 = 8

P(Either a king or a queen) = frac{8}{52}= frac{2}{13}