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Maths 4 Marks Questions

Probability · Gujarat Board (GSEB) STD 9 (GSEB - English Medium). 3 questions.

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Question 14 Marks
Two coins are tossed 400 times and we get:
Two heads: 112 times; one head: 160 times; 0 head: 128 times.
When two coins are tossed at random, what is the probability of getting
  1. 2 heads?
  2. 1 heads?
  3. 0 heads?
Answer
Total number of tosses = 400

Number of times 2 heads appear = 112

Number of times 1 head appears = 160

Number of times 0 head appears = 128

In a random toss of two coins, let E1, E2, E3 be the events of getting 2 heads, 1 head and 0 head, respectively. Then,

  1. P(getting 2 heads) = P(E1) $=\frac{\text{Number of times 2 heads appear}}{\text{Total number of trials}}$

$=\frac{112}{400}=0.28$

  1. P(getting 1 head) = P(E2) $=\frac{\text{Number of times 1 head appear}}{\text{Total number of trials}}$

$=\frac{160}{400}=0.4$

  1. P(getting 0 head) = P(E3) $=\frac{\text{Number of times 0 head appear}}{\text{Total number of trials}}$

$=\frac{128}{400}=0.32$

Remark: Clearly, when two coins are tossed, the only possible outcomes are E1, E2 and E3 and P(E1) + P(E2) + P(E3) = (0.28 + 0.4 + 0.32) = 1

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Question 24 Marks
A die is thrown 300 times and the outcomes are noted as given below:
Outcome
1
2
3
4
5
6
Frequency
60
72
54
42
39
33
When a dice is thrown at random, what is the probability of getting a
  1. 3?
  2. 6?
  3. 5?
  4. 1?
Answer
Total number of tosses = 300

In a random throw of a dice, let E1, E2, E3, E4 be the events of getting 3, 6, 5, and 1, respectively. Then,

  1. P(getting 3) = P(E1) $=\frac{\text{Number of times 3 appear}}{\text{Total number of trials}}$

$=\frac{54}{300}=0.18$

  1. P(getting 6) = P(E2) $=\frac{\text{Number of times 6 appear}}{\text{Total number of trials}}$

$=\frac{33}{300}=0.11$

  1. P(getting 5) = P(E3) $=\frac{\text{Number of times 5 appear}}{\text{Total number of trials}}$

$=\frac{39}{300}=0.13$

  1. P(getting 1) = P(E4) $=\frac{\text{Number of times 1 appear}}{\text{Total number of trials}}$

$=\frac{60}{300}=0.20$

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Question 34 Marks
Three coins are tossed 200 times and we get:
Three heads: 39 times; two heads: 58 times;
One head: 67 times; 0 head: 36 times.
When three coins are tossed at random, what is the probability of getting
  1. 3 heads?
  2. 1 heads?
  3. 0 heads?
  4. 2 heads?
Answer
Total number of tosses = 200

Number of times 3 heads appear = 39

Number of times 2 head appears = 58

Number of times 1 head appears = 67

Number of times 0 head appears = 36

In a random toss of three coins, let E1, E2, E3, E4 be the events of getting 3 heads, 2 heads, 1 head and 0 head, respectively. Then,

  1. P(getting 3 heads) = P(E1) $=\frac{\text{Number of times 3 heads appear}}{\text{Total number of trials}}$

$=\frac{39}{200}=0.195$

  1. P(getting 1 heads) = P(E2) $=\frac{\text{Number of times 1 head appear}}{\text{Total number of trials}}$

$=\frac{67}{200}=0.335$

  1. P(getting 0 head) = P(E3) $=\frac{\text{Number of times 0 head appear}}{\text{Total number of trials}}$

$=\frac{36}{200}=0.18$

  1. P(getting 2 heads) = P(E4) $=\frac{\text{Number of times 2 heads appear}}{\text{Total number of trials}}$

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

Remark: Clearly, when two coins are tossed, the only possible outcomes are E1, E2, E3 and E4 and P(E1) + P(E2) + P(E3) + P(E4) = (0.195 + 0.335 + 0.18 + 0.29) = 1

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