CBSE Class 9 Mathematics Probability Set 08

Question. Out of 35 students Participating in a debate 10 are girls. The Probability that winner is a boy is :
(a) \( \frac{1}{7} \)
(b) \( \frac{2}{7} \)
(c) \( \frac{3}{7} \)
(d) \( \frac{5}{7} \)
Answer: (d) \( \frac{5}{7} \)

Question. There are 5 balls, each of the colours white, blue, green, red and yellow in a bag. If 1 balls is drawn from the bag, then the Probability that the ball drawn is red is
(a) \( \frac{4}{5} \)
(b) \( \frac{1}{4} \)
(c) \( \frac{1}{5} \)
(d) \( \frac{1}{20} \)
Answer: (c) \( \frac{1}{5} \)

Question. If \( P(E) = 0.25 \) what is the value of \( P(\text{not } E) \)
(a) 0.5
(b) 1
(c) 0
(d) 0.75
Answer: (d) 0.75

Question. Sum of the probabilities of all events of a trial is
(a) less than 1
(b) greater than 1
(c) lies between 0 and 1
(d) 1
Answer: (d) 1

Question. A four digit number is to be formed by using the digits 2, 4, 7, 8. The probability that the number will start with 7 is
(a) \( \frac{3}{4} \)
(b) \( \frac{1}{4} \)
(c) \( \frac{1}{3} \)
(d) \( \frac{1}{7} \)
Answer: (b) \( \frac{1}{4} \)

Question. The probability of an event of a trial :
(a) is 1
(b) lies between 0 and 1 (both inclusive)
(c) is 0
(d) is greater than 1
Answer: (b) lies between 0 and 1 (both inclusive)

Question. A die is thrown once, the probability of getting a prime number on the die is :
(a) \( \frac{1}{6} \)
(b) \( \frac{1}{3} \)
(c) \( \frac{1}{2} \)
(d) \( \frac{2}{3} \)
Answer: (c) \( \frac{1}{2} \)

Question. If two coins are tossed, then the probability of getting no tail is :
(a) \( -\frac{1}{4} \)
(b) \( \frac{1}{4} \)
(c) \( \frac{1}{5} \)
(d) \( \frac{3}{4} \)
Answer: (b) \( \frac{1}{4} \)

2 Marks Questions

Question. A teacher analyses the performance of two sections of students in a mathematics test of 100 marks given in the following table:

1. Find the probability that a student obtained less than 20% in the mathematics test.
2. Find the probability that a student obtained 60 or above.
Answer: (i) No. of students obtaining marks less than 20 out of 100, i.e. 20% = 7
Total students in the class = 90
\( \therefore P(\text{A student obtained less than 20\%}) = \frac{7}{90} \)
\( \therefore P(\text{A student obtained marks 60 or above}) = \frac{23}{90} \)

Question. To know the opinion of the students about the subject statistics, a survey of 200 students was conducted. The data is recorded in the following table:

Find the probability that a student chosen at random:
(i) likes statistics (ii) dislikes it.
Answer: Total no. of students on which the survey about the subject of statistics was conducted = 200
i) No. of students who like statistics = 135
\( \therefore P(\text{a student likes statistics}) = \frac{135}{200} = \frac{27}{40} \)
ii) No. of students who do not like statistics = 65
\( \therefore P(\text{a student does not like statistics}) = \frac{65}{200} = \frac{13}{40} \)

Question. Refer Q.2, Exercise 14.2. What is the empirical probability than an engineer lives:
1. less than 7 km from her place of work?
2. more than or equal to 7 km from her place of work?
3. within \( \frac{1}{2} \) km from her place of work?
Answer: Total number of engineers = 40
i) No. of engineers living less than 7 km from her place of work = 9
\( \therefore P(\text{Engineer living less than 7 km from her place of work}) = \frac{9}{40} \)
ii) No. of engineers living more than or equal to 7 km from her place of work = 31
\( \therefore P(\text{Engineer living more than or equal to 7 km from her place of work}) = \frac{31}{40} \)
(iii) No. of engineers living within \( \frac{1}{2} \) km from her place of work = 0
\( \dots P(\text{Engineer living within } \frac{1}{2} \text{ km from her place of work}) = \frac{0}{40} = 0 \)

Question. Activity: Note the frequency of two wheelers, three wheelers and four wheelers going past during a time interval, in front of your school gate. Find the probability that any one vehicle out of the total vehicles you have observed is a two wheeler.
Answer: Let you noted the frequency of types of wheelers after school time (i.e. 3 pm to 3.30 pm) for half an hour.
Let the following table shows the frequency of wheelers.

Probability that a two wheelers passes after this interval = \( \frac{125}{200} = \frac{5}{8} \)

Question. Activity: Ask all the students in your classroom to write a 3-digit number. Choose any student from the room at random. What is the probability that the number written by him is divisible by 3, if the sum of its digits is divisible by 3.
Answer: Let number of students in your class is 24.
Let 3-digit number written by each of them is as follows:
837, 172, 643, 371, 124, 512, 432, 948, 311, 252, 999, 557, 784, 928, 867, 798, 665, 245, 107, 463, 267, 523, 944, 314
Numbers divisible by 3 are = 837, 432, 948, 252, 999, 867, 798 and 267
Number of 3-digit numbers divisible by 3 = 8
\( \therefore P(\text{3-digit numbers divisible by 3}) = \frac{8}{24} = \frac{1}{3} \)

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: Number of bags containing more than 5 kg of wheat flour = 7
Total number of bags = 11
\( \therefore P(\text{a bag containing more than 5 kg of wheat flour}) = \frac{7}{11} \)

Question. Find the probability that any of these bags chosen at random contains more than 5 kg of flour.
Answer: Number of bags containing more than 5 kg of wheat flour = 7
Total number of wheat flour bags = 11
\( \therefore \text{P (a bag containing more than 5 kg of wheat flour)} = \frac{7}{11} \)

Question. In Q.5, Exercise 14.2, you were asked to prepare a frequency distribution table, regarding the concentration of sulphur dioxide in the air in parts per million of a certain city for 30 days. Using this table, find the probability of the concentration of sulphur dioxide in the interval 0.12 – 0.16 on any of these days.
Answer: From the frequency distribution table we observe that:
No. of days during which the concentration of sulphur dioxide lies in interval 0.12 – 0.16 = 2
Total no. of days during which concentration of sulphur dioxide recorded = 30
\( \therefore \text{P (day when concentration of sulphur dioxide (in ppm) lies in 0.12 – 0.16)} = \frac{2}{30} = \frac{1}{15} \)

Question. In Q.1, Exercise 14.1 you were asked to prepare a frequency distribution table regarding the blood groups of 30 students of a class. Use this table to determine the probability that a student of this class selected at random has blood group AB.
Answer: From the frequency distribution table we observe that:
Number of students having blood group AB = 3
Total number of students whose blood group were recorded = 30
\( \dots \text{P (a student having blood group AB)} = \frac{3}{30} = \frac{1}{10} \)

Question. A die is thrown 1000 times with the frequencies for the outcomes 1, 2, 3, 4, 5 and 6 as given in the following table:

Find the probability of getting each outcome.
Answer: (i) No. of outcome getting no. 1 = 179
\( P(1) = \frac{179}{1000} = 0.179 \)
(ii) Probability of outcome 2
\( P(2) = \frac{150}{1000} = 0.15 \)
(iii) Probability of outcome 3
\( P(3) = \frac{157}{1000} = 0.157 \)
(iv) Probability of outcome 4
\( P(4) = \frac{149}{1000} = 0.149 \)
(v) Probability of outcome 5
\( P(5) = \frac{175}{1000} = 0.175 \)
(vi) Probability of outcome 6
\( P(6) = \frac{190}{1000} = 0.19 \)

Question. Two coins are tossed 729 times and the out comes are:
No tail: 189, One tail: 297, Two tails: 243
Find the Probability of the occurrence of each of these events.
Answer: No. of total trials = 729
E1, E2 and E3 are events getting no tail, one tail and two tails, then
\( P(E_1) = \frac{189}{729} = \frac{7}{27} \)
\( P(E_2) = \frac{297}{729} = \frac{11}{27} \)
\( P(E_3) = \frac{243}{729} = \frac{1}{3} \)

Question. A bag contains 15 cards bearing numbers 1, 2, 3, 4, ..........., 14, 15. A card is drawn from the bag. Find the Probability that it bears:
(i) a Prime number (ii) A number divisible by 2
Answer: Total number of cards = 15
No. of total trials = 15
(i) Among 1, 2, 3, 4,......., 14, 15, prime number are 2, 3, 5, 7, 11, 13
Number of favourable outcomes = 6
\( P (\text{Prime number}) = \frac{6}{15} = \frac{2}{5} \)
(ii) Among 1, 2, 3, 4, ............., 14, 15
No. divisible by 2 are 2, 4, 6, 8, 10, 12, 14
\( \therefore \text{Number of outcomes} = 7 \)
\( P (\text{no. divisible by 2}) = \frac{7}{15} \)

Question. A coin is tossed 400 times and outcomes are
Tail: 230 Head:170
Answer: (i) Total outcomes = 400
Head = 170
\( P(H) = \frac{170}{400} = \frac{17}{40} \)
(ii) Tail = 230
\( P(T) = \frac{230}{400} = \frac{23}{40} \)

Question. A survey of 200 students was conducted to check the opinion of students about the topic geometry. It was found that 175 students do not like geometry. Find the probability of the students who like geometry.
Answer: Total no. of Students = 200
The no. of students do not like geometry = 175
The no. of students who like geometry = 200-175=25
\( \therefore P (\text{no. of students who like geometry}) = \frac{25}{200} = \frac{5}{40} = \frac{1}{8} \)

Question. Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes.

Compute the probability of 2 heads coming up.
Answer: Total number of tosses = 200
Number of outcomes of 2 heads = 72
\( P (\text{2 heads}) = \frac{72}{200} = \frac{9}{25} \)

Question. The heights of 70 students are given in the following table.

Find the probability that a student has height.
(i) 169 cm (ii) Less than 150 cm
Answer: (i) Total no. of students = 70
No. of total trial =70
The no. of students has height 169 cm = 0
\( P (\text{a student has a height 169 cm}) = \frac{0}{70} = 0 \)
(ii) No. of students has height less than 150 cm = 0
\( P (\text{a student has a height less than 150 cm}) = \frac{0}{70} = 0 \)

Question. A bag contains 20 cards numbered from 1 to 20 one card is drawn from the bag. Find the probability that it bears a prime number.
Answer: Total no. of cards = 20
No. marks on the cards are 1, 2, 3, 4, 5, 6, .............., 20
Prime numbers are {2, 3, 5, 7, 11, 13, 17, 19}
Total prime numbers = 8
\( \therefore P (\text{a prime no.}) = \frac{8}{20} = \frac{2}{5} \)

Question. Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes

Compute the probability of 2 heads coming up
Answer: Total no. of tosses = 200
No. of outcomes of 2 heads = 72
\( P (\text{2 heads}) = \frac{72}{200} = \frac{9}{25} \)

Question. A die is thrown once. Find the probability of getting (i) an odd number (ii) a number greater than one.
Answer: If a die is thrown once, total possible outcomes are \( S = \{1, 2, 3, 4, 5, 6\} \)
i) \( E = \{1, 3, 5\} \)
\( n(E) = 3 \), \( n(S) = 6 \)
\( E = \text{an odd no} \)
\( \therefore P(E) = \frac{n(E)}{n(S)} = \frac{3}{6} = \frac{1}{2} \)
(ii) a no. greater than one

Question. Two coins are tossed 340 times and the outcomes are
(i) Two tail =115 (ii) one tail =100 (iii) no tail =125
Find the probability of occurrence of (i) one tail (ii) three tail
Answer: Total possible outcomes = 340
(i) Occurrence of one tail = 100
\( \therefore P(E) = \frac{100}{340} = \frac{5}{17} \)
(ii) Occurrence of three tails = 0
\( \therefore P(E) = \frac{0}{340} = 0 \)

Question. To know the option of the students about the subject mathematics a survey of 200 students was conducted. The obtained data is given below.

Find the probability that a student chosen at random
(i) like mathematics
(ii) does not like it
Answer: Total number of students are 200
(i) No. of students like mathematics = 135
\( P(E) = \frac{135}{200} = \frac{27}{40} \)
(ii) No. of students dislike mathematics = 65
\( P(\bar{E}) = \frac{65}{200} = \frac{13}{40} \)

Question. Out of 17 boys and 13 girls of a class, 1 student is to be selected. Find the probability of selecting a girl
Answer: Total no. of students = \( 17 + 13 = 30 \)
No. of girls = 13
\( P(\text{a student selecting a girl}) = \frac{13}{30} \)