Question

[{( 100 POINTS )}]

A die is thrown at random. Find the probability of getting :

(i) 2.

(ii) a number less than 3.

(iii) a composite number.

(iv) a number not less than 4.

I WILL PROMISE THAT I WILL BRAINLIEST YOUR ANSWER

Answer 1

Ello There!

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First, lets note down the number of outcomes we can derive from a die

1, 2, 3, 4, 5, 6

Total no: of outcomes is 6.

i) Probability of getting 2

2 is present in the die only once.

Therefore there is 1 favourable outcome.

There are a total of 6 outcomes.

P(getting 2) = $\frac{\mathsf{No: \ of \ favourable \ outcomes}}{\mathsf{Total \ no: \ of \ outcomes}}$

P(getting 2) = $\frac{1}{6}$

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ii) Probablity of getting a number less than 3.

The numbers less that 3 in a die are:

1 and 2

So there are 2 favourable outcomes.

P(getting a number less than 3) = $\frac{\mathsf{No: \ of \ favourable \ outcomes}}{\mathsf{Total \ no: \ of \ outcomes}}$

P(getting a number less than 3) = $\frac{2}{6}$

P(getting a number less than 3) = $\frac{1}{3}$

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iii) A composite number.

Composite numbers that can be obtained on rolling a die are:

4 and 6 , Therefore there are 2 favourable outcomes.

[Remember 1 is neither composite nor prime]

P(getting a composite number) = $\frac{\mathsf{No: \ of \ favourable \ outcomes}}{\mathsf{Total \ no: \ of \ outcomes}}$

P(getting a composite number) = $\frac{2}{6}$

P(getting a composite number) = $\frac{1}{3}$

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iv) A number not less than 4.

When a die is rolled, the number that are obtained which are not less than 4 are:

4, 5 and 6.

Therefore there are 3 favourable outcomes

P(getting a number not less than 4) = $\frac{\mathsf{No: \ of \ favourable \ outcomes}}{\mathsf{Total \ no: \ of \ outcomes}}$

P(getting a number not less than 4) = $\frac{3}{6}$

P(getting a number not less than 4) = $\frac{1}{2}$

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$\mathbb{THANK \ YOU}$

Answer 2

Step-by-step explanation:

answer in the above attachment