CBSE Class 9 Mathematics Probability Set 07

Question. If a dice is thrown once what is the probability of getting an even prime number.
(a) \( \frac{1}{6} \)
(b) \( \frac{1}{2} \)
(c) \( \frac{2}{3} \)
(d) 1
Answer: (a) \( \frac{1}{6} \)

Question. A card is drawn from a pack of 52 cards what is the probability of getting an non ace card.
(a) \( \frac{1}{13} \)
(b) \( \frac{12}{13} \)
(c) \( \frac{1}{4} \)
(d) None of the options
Answer: (b) \( \frac{12}{13} \)

Question. The minimum value of probability is
(a) \( \frac{1}{2} \)
(b) 0
(c) None of the options
Answer: (b) 0

Question. Performing an experiment once is called
(a) Trial
(b) Event
(c) Probability
(d) None of the options
Answer: (a) Trial

Question. What is the probability of a number greater than 6 for a single throw of a die?
(a) 0
(b) 1
(c) \( \frac{1}{2} \)
(d) None of the options
Answer: (a) 0

Question. If \( P(E) = \frac{3}{4} \) what is value of \( P(\overline{E}) \).
(a) \( \frac{3}{4} \)
(b) \( \frac{1}{4} \)
(c) 1
(d) None of the options
Answer: (b) \( \frac{1}{4} \)

Question. A card is drawn from a pack of 52 playing cards. What is the probability of getting an king of black colour
(a) \( \frac{1}{26} \)
(b) \( \frac{4}{52} \)
(c) \( \frac{1}{4} \)
(d) None of the options
Answer: (a) \( \frac{1}{26} \)

Question. A coin is tossed 2 times what is the probability of getting at most 2 heads.
(a) \( \frac{3}{4} \)
(b) \( \frac{1}{2} \)
(c) \( \frac{1}{4} \)
(d) None of the options
Answer: (d) None of the options

Question. A bag contains 5 white, 4 red and 3 black balls. A ball is drawn from the bag, find the probability that it is not black
Answer: No. of white balls = 5
No. of red balls = 4
No. of black balls = 3
Total no. of balls = \( 5 + 4 + 3 = 12 \)
\( P(\text{Black balls}) = \frac{3}{12} = \frac{1}{4} \)
\( P(\text{the ball is not black}) = 1 - \frac{1}{4} = \frac{3}{4} \)

Question. A card is drawn from a 52 pack of cards. Find the probability that it is a queen
Answer: Total no. of cards = 52
No. of queens = 4
\( P(\text{getting a queen}) = \frac{4}{52} = \frac{1}{13} \)

Question. There are 500 tickets of a lottery out of which 10 are prize winning tickets. A person buys one ticket. Find the probability that he gets a prize winning ticket.
Answer: Total no. of lottery tickets = 500
No. of prize winning tickets = 10
\( P(\text{Prize winning tickets}) = \frac{10}{500} = \frac{1}{50} \)

Question. 1500 families with 2 children were selected randomly and the following data were recorded:

Compute the probability of a family, chosen at random, having:
(i) 2 girls (ii) 1 girl (iii) No girl
Also check whether the sum of these probabilities is 1.
Answer:
(i) No. of families having 2 girls = 475
\( \therefore P(\text{Family having 2 girls}) = \frac{475}{1500} = \frac{19}{60} \)
(ii) No of families having 1 girl = 814
\( \therefore P(\text{Family having 1 girl}) = \frac{814}{1500} = \frac{407}{750} \)
(iii) No. of families having no girl = 211
\( \therefore P(\text{Family having no girl}) = \frac{211}{1500} \)
Sum of all probabilities = \( \frac{19}{60} + \frac{407}{750} + \frac{211}{1500} = \frac{475 + 814 + 211}{1500} = \frac{1500}{1500} = 1 \)
Yes, the sum of probabilities is 1.

Question. An organization selected 2400 families at random and surveyed them to determine a relationship between income level and the number of vehicles in a family. The information gathered is listed in the table below:

Suppose a family is chosen. Find the probability that the family chosen is:
1. earning Rs. 10000 – 13000 per month and owning exactly 2 vehicles.
2. earning Rs. 16000 or more per month and owning exactly 1 vehicle.
3. earning less than Rs. 7000 per month and does not own any vehicle.
4. earning Rs. 13000 – 16000 per month and owning more than 2 vehicles.
5. not more than 1 vehicle.
Answer:
(i) \( P(\text{earning Rs. 10000 – 13000 per month and owning exactly 2 vehicles}) = \frac{29}{2400} \)
(ii) \( P(\text{earning Rs. 16000 or more per month and owning exactly 1 vehicles}) = \frac{579}{2400} \)
(iii) \( P(\text{earning Rs. 7000 per month and does not own any vehicles}) = \frac{10}{2400} = \frac{1}{240} \)
(iv) \( P(\text{earning Rs. 13000 – 16000 per month and owning more than 2 vehicles}) = \frac{25}{2400} = \frac{1}{96} \)
(v) \( P(\text{owning not more than 1 vehicle}) = \frac{2062}{2400} = \frac{1031}{1200} \)

Question. The marks obtained by 30 students is given in the following table:

Find the Probability that a student secures
(i) 60 marks (ii) 75 marks (iii) Less than 60 marks
Answer: Total no. of students = 30
No. of students securing 60 marks = 4
(i) \( \therefore P(\text{Students securing 60 marks}) = \frac{4}{30} = \frac{2}{15} \)
(ii) No. of students securing 75 marks = 2
\( \therefore P(\text{Students securing 75 marks}) = \frac{2}{30} = \frac{1}{15} \)
(iii) No. of students securing less than 60 marks = \( 5 + 7 = 12 \)
\( P(\text{Students securing less than 60 marks}) = \frac{12}{30} = \frac{2}{5} \)

Question. A tyre manufacturing company kept a record of the distance covered shows the results of 1000 tyres

If you buy a tyre of this company. What is the Probability that
(i) it will need to be replaced before it has covered 4000 km
(ii) it will last more than 9000 km
(iii) it will need to be replaced after it has covered somewhere between 4000 km and 14000 km
Answer:
(i) No. of tyres which covered distance less than 4000 km = 20
Total no. of tyres = 1000
Required probability \( P(E) = \frac{20}{1000} = \frac{1}{50} \)
(ii) No. of tyres needed to replaced more then 9000 km = \( 325 + 445 = 770 \)
Required Probability = \( \frac{770}{1000} = \frac{77}{100} = 0.77 \)
(iii) No. of tyres needed to replaced between 4000 km, to 14,000km = \( 210 + 325 + 445 = 980 \)
Required probability = \( \frac{980}{1000} = 0.98 \)

Question. The ages of 30 workers in a factory are as follows

Find the probability that the age of a works lies in the interval
(i) 27-29
(ii) 29-35
(iii) 21-27
Answer:
I Part
The no. of workers lies in the interval 27-29 are = 6
Total no. of workers = 30
Required probability = \( \frac{6}{30} = \frac{1}{5} \)
II Part
No. of workers having age between 29 - 35 = \( 5 + 4 + 3 = 12 \)
Total no. of workers = 30
Required Probability = \( \frac{12}{30} = \frac{2}{5} \)
III Part
No. of workers having age between 21 -27 = \( 3 + 4 + 5 = 12 \)
Total no. of workers = 30
Required Probability = \( \frac{12}{30} = \frac{2}{5} \)

Question. A die is thrown once. Find the probability of getting
(i) a prime number
(ii) a number less then 5
Answer: When a die is thrown, then outcomes are \( 1, 2, 3, 4, 5, 6 \)
(i) Prime numbers are \( = 2, 3, 5 \)
\( \therefore \) Frequency of happening prime number is \( 3 \)
\( \therefore \) The probability of getting prime number \( = \frac{3}{6} = \frac{1}{2} \)
(ii) Numbers less than \( 5 \) are \( 1, 2, 3, 4 \)
\( \therefore \) Frequency of happening of a no. less than \( 5 \) is \( 4 \)
\( \therefore \) Probability of getting a number less than \( 5 = \frac{4}{6} = \frac{2}{3} \)

Question. A die is thrown 450 times and outcomes are noted in the frequency distribution table given below.

Find the probability of the occurrence of the event.
(i) 4 (ii) a number < 3 (iii) 7
Answer: Total no. of times die thrown \( = 450 \)
(i) No. of times the no. 4 comes up \( = 70 \)
\( P(\text{getting no. 4}) = \frac{70}{450} = \frac{7}{45} \)
(ii) No. of times the die turns up 1 or 2 \( = 90 + 60 = 150 \)
\( P(1 \text{ or } 2) = \frac{150}{450} = \frac{1}{3} \)
(iii) No. of times the die turn up 7 \( = 0 \)
\( P(7) = \frac{0}{450} = 0 \)

Question. From a well-shuffled pack of 52 cards, a card is drawn at random, find the probability that it is:
(i) A spade
(ii) Black
(iii) Ace of diamond
Answer: Total no. of cards \( = 52 \)
(i) No. of spade \( = 13 \)
\( P(\text{Spade}) = \frac{13}{52} = \frac{1}{4} \)
(ii) No. of black cards \( = 26 \)
\( P(\text{black cards}) = \frac{26}{52} = \frac{1}{2} \)
(iii) No. of ace of diamond \( = 1 \)
\( P(\text{ace of diamond}) = \frac{1}{52} \)

Question. The central Board of secondary education has a waiting list of examinations of 150 Persons. Out of these, 60 are women and 90 are men. One examiner is to selected to replace an examiner who has not reported at the centre find the probability that the examiner selected is a:
(i) woman
(ii) man
Answer:
(i) No. of trials \( = 150 \)
No. of women \( = 60 \)
\( \therefore P(\text{The examiner selected is a woman}) = \frac{60}{150} = \frac{2}{5} \)
(ii) Number of men \( = 90 \)
\( \dots P(\text{The examiner selected is a man}) = \frac{90}{150} = \frac{3}{5} \)
\( P(\text{woman}) + P(\text{man}) = \frac{2}{5} + \frac{3}{5} = 1 \)

Question. Two coins are tossed 250 times and the outcomes are:
(i) No head = 70 (ii) one head = 85 (iii) Two heads = 95,
Find the probability of the occurrence of each of these events.
Answer: Total no. of times coin tossed \( = 250 \)
(i) No. of times no head comes up \( = 70 \)
\( P(\text{No Head}) = \frac{70}{250} = \frac{7}{25} \)
(ii) No. of times one head comes up \( = 85 \)
\( P(\text{one head}) = \frac{85}{250} = \frac{17}{50} \)
(iii) No. of times two head comes up \( = 95 \)
\( P(\text{Two head}) = \frac{95}{250} = \frac{19}{50} \)

Question. Out of 100 balls in a bag 25 are green, 30 are yellow and 45 are white. Find the Probability that a ball drawn from the bag is (i) green (ii) yellow (iii) white
Answer: Total number of balls are \( 100 \)
(i) No. of green balls \( = 25 \)
\( P(G) = \frac{25}{100} = \frac{1}{4} \)
(ii) No. of yellow balls \( = 30 \)
\( P(Y) = \frac{30}{100} = \frac{3}{10} \)
(iii) No. of white balls \( = 45 \)
\( P(W) = \frac{45}{100} = \frac{9}{20} \)

Question. Eleven bags of wheat flour, each marked 5 kg actually contained the following weights of flour (in kg)
4.97, 5.05, 5.08, 5.03, 5.00, 5.06, 5.08, 4.98, 5.04, 5.07, 5.00
Find the probability that any of these bags chosen at random contains more than 5 kg of flour.
Answer: Total no. of bags \( = 11 \)
No. of bags contains more than \( 5\text{ kg} \) of flour \( = 7 \)
\( P(E) = \frac{7}{11} \)

Question. 1500 families with 2 children were selected randomly and the following data were recorded

Compute the probability of a family, chosen at random, having.
(i) 2 girls (ii) 1 girl (iii) No girl
Also check whether the sum of these probabilities is
Answer:
Total no. of Families \( = 1500 \)
(i) No. of family having 2 girls \( = 475 \)
\( P(E) = \frac{475}{1500} = \frac{95}{300} = \frac{19}{60} \)
(ii) No. of families having 1 girl \( = 814 \)
\( P(E) = \frac{814}{1500} = \frac{407}{750} \)
(iii) No. of families having no girl \( = 211 \)
\( P(E) = \frac{211}{1500} \)

Question. Fifty seeds were selected at random from each of 5 bags of seeds and were kept under standardized condition favorable to germination. After 20 days, the number of seeds which had germinated in each collection were counted and recorded as follows.

What is the probability of germination of
More than 40 seeds in a bag? (ii) 49 seeds in a bag (iii) More than 35 seeds in a bag
Answer:
i) No. of bags in which more than 40 seeds germinated out of 50 seeds is 3
\( \therefore \text{Required probability } P(E) = \frac{3}{5} = 0.6 \)
ii) No. of bags in which 49 seeds germinated \( = 0 \)
\( \text{Required probability } P(E) = \frac{0}{5} = 0 \)
iii) No. of bags in which more than 35 seeds germinated \( = 5 \)
\( \text{Required probability } P(E) = \frac{5}{5} = 1 \)

Question. It is known that a box of 550 bulbs contain 22 defective bulbs. One bulb is taken out at random from the box. Find the probability of getting
(i) Defective bulbs (ii) Good bulbs
Answer: Total number of bulbs \( = 550 \)
Number of defective bulbs \( = 22 \)
\( \therefore \text{No. of good bulbs} = 550 - 22 = 528 \)
(i) \( P(\text{defective bulbs}) = \frac{22}{550} = 0.04 \)
(ii) \( P(\text{good bulbs}) = \frac{528}{550} = 0.96 \)

Question. Frequency distribution of marks obtained by 70 Students is given below:
Marks obtained 0-10 10-20 20-40 40-45 45-60 60-70 70-80
No. of students 4 8 20 10 12 6 10
Find the probability that the marks obtained by a student lies in the internal
(i) 0-40 (ii)0-80 (iii) 80-90
Answer: Total number of students = 70
(i) no. of students getting marks 0-40
\( = 4 + 8 + 20 = 32 \)
\( P(\text{students getting marks 0-40}) = \frac{32}{70} = 0.457 \)
(ii) no. of students getting marks 0-80
\( = 4 + 8 + 20 + 10 + 12 + 6 + 10 = 70 \)
\( P(\text{students getting marks 0-80}) = \frac{70}{70} = 1 \)
(iii) no. of students getting marks 80-90 = 0
\( P(\text{students getting marks 80-90}) = \frac{0}{70} = 0 \)

Question. A box contains 150 balls of red, blue and white colours out of these 50 balls are red, 40 balls are blue and 60 balls are white. One ball is drawn from the bag. Find the probability that the ball drawn is
(i) Red (ii) blue (iii) white
Answer: Total number of balls = 150
No. of red balls = 50
No. of blue balls = 40
No. of white balls = 60
(i) \( P(\text{red ball}) = \frac{50}{150} = \frac{1}{3} \)
(ii) \( P(\text{blue ball}) = \frac{40}{150} = \frac{4}{15} \)
(iii) \( P(\text{white ball}) = \frac{60}{150} = \frac{2}{5} \)

Question. A die is thrown 500 times. The frequency of the outcomes of the event 1,2,3,4,5 and 6 are recorded in the following frequency distribution table
Outcome 1 2 3 4 5 6
Frequency 85 75 80 9010070
Find the probability of the occurrence of an (i) even number (ii) odd number.
Answer: Total no. of out comes = 500
(i) Frequency of dice getting even number \( 75 + 90 + 70+ = 235 \)
\( P(\text{even number}) = \frac{235}{500} = \frac{47}{100} \)
(ii) \( P(\text{odd number}) = 1 - \frac{47}{100} = \frac{53}{100} \)

Question. The weekly pocket expenses of students are given below:
POCKET EXPENSES (in Rs.) 45 40 59 71 58 47 65
NO. OF STUDENTS 7 4 10 6 3 8 1
Find the probability that the weekly pocket expenses of a student are
(a) (i) Rs 59 (ii) more than Rs 59 (iii) less than Rs 59
(b) Find the sum of probabilities computed in (i), (ii), and (iii)s
Answer: (a) No. of students = 39
\( \therefore \) No. of trials = 39

(i) Number of students with weekly pocket expenses of Rs 59 = 10
\( \dots P(\text{the weekly pocket expenses of a student are Rs 59}) = \frac{10}{39} \)

(ii) No. of students with weekly pocket expenses of more than Rs 59 = 6+1=7
\( \therefore P(\text{the weekly pocket expenses of a student are more than Rs 59}) = \frac{7}{39} \)

(iii) Number of students with weekly pocket expenses of less than Rs 59 = 7+4+3+8=22
\( \therefore P(\text{the weekly pocket expenses of a student are less than Rs 59}) = \frac{22}{39} \)

(b) Sum of probabilities in (i),(ii), and (iii)
\( = \frac{10}{39} + \frac{7}{39} + \frac{22}{39} = \frac{39}{39} = 1 \)

Question. Cards marked 2 to 101 are placed in a box and mixed thoroughly. One card is drawn from the box. Find the probability that number on the card is (i) an even number (ii) a number less than 14 (iii) a number which is a perfect square (iv) a prime number less than 20. (v) an odd number.
Answer: Total number of cards = 100
even numbers are = 50
\( P(\text{even number}) = \frac{50}{100} = \frac{1}{2} \)

(ii) no. less than 14 are
\( = \{2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13\} \)
\( = 12 \)
\( P(\text{no. less than 14}) = \frac{12}{100} = 0.12 \)

(iii) Number which is a Perfect square
\( = \{1, 4, 9, 16, 25, 36, 49, 64, 81, 100\} \)
\( = 10 \)
\( P(\text{number which is a perfect square}) = \frac{10}{100} = \frac{1}{10} \)

(iv) Prime no. less than 20 are
\( = \{2, 3, 5, 7, 11, 13, 17, 19\} \)
\( = 8 \)
\( P(\text{Prime no. less than 20}) = \frac{8}{100} = \frac{2}{25} \)

(v) P (odd number) \( = 1 - P(\text{an even no.}) \)