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Probability | Probability Exercise 19

Question 4

A die is thrown 300 times and the outcomes are noted as given below:

Outcome Frequency
1 60
2 72
3 54
4 42
5 39
6 33

When a die is thrown at random, what is the probability of getting a

(i) 3?

(ii) 6?

(iii) 5?

(iv) 1?

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It is given that

Number of trials = 300

Consider E1, E2, E3 and E4 as the events

(i) So we get

P (E1) = number of times 3 head/ number of trials

By substituting the values

P (E1) = 54/ 300

We get

P (E1) = 0.18

(ii) So we get

P (E2) = number of times 6 head/ number of trials

By substituting the values

P (E2) = 33/ 300

We get

P (E2) = 0.11

(iii) So we get

P (E3) = number of times 5 head/ number of trials

By substituting the values

P (E3) = 39/ 300

We get

P (E3) = 0.13

(iv) So we get

P (E4) = number of times 2 head/ number of trials

By substituting the values

P (E4) = 60/ 300

We get

P (E4) = 0.2

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Our top 5% students will be awarded a special scholarship to Lido.

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Question 4

A die is thrown 300 times and the outcomes are noted as given below:

Outcome Frequency
1 60
2 72
3 54
4 42
5 39
6 33

When a die is thrown at random, what is the probability of getting a

(i) 3?

(ii) 6?

(iii) 5?

(iv) 1?

Looking to do well in your science exam ? Learn from an expert tutor. Book a free class!

It is given that

Number of trials = 300

Consider E1, E2, E3 and E4 as the events

(i) So we get

P (E1) = number of times 3 head/ number of trials

By substituting the values

P (E1) = 54/ 300

We get

P (E1) = 0.18

(ii) So we get

P (E2) = number of times 6 head/ number of trials

By substituting the values

P (E2) = 33/ 300

We get

P (E2) = 0.11

(iii) So we get

P (E3) = number of times 5 head/ number of trials

By substituting the values

P (E3) = 39/ 300

We get

P (E3) = 0.13

(iv) So we get

P (E4) = number of times 2 head/ number of trials

By substituting the values

P (E4) = 60/ 300

We get

P (E4) = 0.2

Our top 5% students will be awarded a special scholarship to Lido.

subject-cta
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