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Maths 3 Marks Question

Probability · Gujarat Board (GSEB) STD 9 (GSEB - English Medium). 7 questions.

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3 Marks Question

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7 questions · timed · auto-graded

Question 13 Marks
In a circket match, a batsman hits a boundary 6 times out of 30 balls he plays. Find the probability that he did not hit a boundary.
Answer
Total number of ball played = 30
Number of times boundary was hit = 6
⇒ Number of times boundary was not hit = 30 - 6 = 24
Therefore, 
Probability that the batsman did not hit the boundary
 $=\frac{\text{No. of times boundary was not hit}}{\text{Total number of balls played}}$
$=\frac{24}{30}$
$=\frac{4}{5}=0.8$
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Question 23 Marks
On one page of a telephone directory, there are 200 phone numbers. The frequency distribution of their unit's digits is given below:
Unit's digit
0
1
2
3
4
5
6
7
8
9
Frequency
19
22
23
19
21
24
23
18
16
15
One of the numbers is chosen at random from the page. What is the probability that the unit's digit of the chosen number is:
  1. 5?
  2. 8?
Answer
Total phone numbers on the directory page = 200

  1. Number of numbers with units digits 5 = 24

Let E1 be the event that the units digits of selected number is 5.

$\therefore$ Required probability = P(E1) $=\frac{24}{200}=0.12$

  1. Number of numbers with units digits 8 = 16

Let E2 be the event that the units digits of selected number is 8.

$\therefore$ Required probability = P(E2) $=\frac{16}{200}=0.08$

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Question 33 Marks
The following table shows the blood groups of 40 students of a class.
Blood group
A
B
O
AB
Number of students
11
9
14
6
One student of the class is chosen at random. What is the probability that the chosen students blood group is:
  1. O?
  2. AB?
Answer
Total number of students = 40

  1. Number of students with blood group O = 14

Let E1 be the event that the selected student's blood group is O. 

$\therefore$ Required probability = P(E1) $=\frac{14}{40}=0.35$

  1. Number of students with blood group AB = 6

Let E2 be the event that the selected student's blood group is AB.

$\therefore$ Required probability = P(E2) $=\frac{6}{40}=0.15$

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Question 43 Marks
The table given below shows the marks obtained by 30 students in a test.
Marks
(Class interval)
1 - 10
11 - 20
21 - 30
31 - 40
41 - 50
Number of students
(Frequency)
7
10
6
4
3
Out of these students, one is chosen at random. What is the probability that the marks of the chosen student:
  1. Are 30 or less?
  2. Are 31 or more?
  3. Lie in the interval 21 - 30?
Answer
Total number of students = 30
  1. Probability that the marks of the chosen student are 30 or less $=\frac{7+10+6}{30}=\frac{23}{30}$
  2. Probability that the marks of the chosen student are 31 or less $=\frac{4+3}{30}=\frac{7}{30}$
  3. Probability that the marks of the chosen student lie in the interval 21 - 30 $=\frac{6}{30}=\frac{1}{5}$
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Question 53 Marks
In a survey of 200 ladies, it was found that 142 like coffee, while 58 dislike it.
Find the probability that a lady chosen at random:
  1. Likes coffee
  2. Dislikes coffee.
Answer
Total number of tosses = 200

Number of ladies who likes coffee = 142

Number of ladies who dislikes coffee = 58

Let E1 and E2 be the events that the selected lady likes and dislikes coffee, respectively. Then,

  1. P(selected lady likes coffee) = P(E1) $=\frac{\text{Number of ladies who like coffee}}{\text{Total number of laies}}$

$=\frac{142}{200}=0.71$

  1. P(selected lady dislikes coffee) = P(E1) $=\frac{\text{Number of ladies who dislike coffee}}{\text{Total number of laies}}$

$=\frac{58}{200}=0.29$

Remark: In the given survey, the only possible outcomes are E1 and E2 and P(E1) + P(E2) = (0.71 + 0.29) = 1.

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Question 63 Marks
A coin is tossed 500 times and we get:
heads: 285 times and tails: 215 times.
When a coin is tossed at random, what is the probability of getting
  1. A head?
  2. A tail?
Answer
Total number of tosses = 500

Number of heads = 285

Number of tails = 215

  1. Let E be the event of getting a head.

P(getting a head) = $\text{P(E)}=\frac{\text{Number of heads coming up}}{\text{Total number of trials}}$

$=\frac{285}{500}=0.57$

  1. Let F be the event of getting a tail.

P(getting a tail) = $\text{P(E)}=\frac{\text{Number of heads coming up}}{\text{Total number of trials}}$

$=\frac{215}{500}=0.43$

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Question 73 Marks
It is known that a box of 800 electric bulbs contains 36 defective bulbs.One bulb is taken at random out of the box. What is the probability that the bulb chosen is non-defective?
Answer
Total number of electric bulbs = 800
Number of defective bulbs = 36
⇒ Number of non-defective bulbs = 800 - 36 = 764
Hence,
Probability that the bulb chosen is non-defective $=\frac{\text{Number of non-defective bulbs}}{\text{Total number of bulbs}}$
$=\frac{764}{800}$
$=\frac{191}{200}$
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3 Marks Question - Maths STD 9 Questions - Vidyadip