Let S be the sample when a die is thrown.

Then S = {1, 2, 3, 4, 5, 6}, n(S) = 6 

Let A be the event of getting a prime number. 

A = {2, 3, 5}, n(A) = 3 

Let B be the event of getting a number greater than or equal to 3. 

B = {3, 4, 5, 6}, n(B) = 4

(i) P(a prime number) = \(\frac{n(A)}{n(S)}=\frac{3}{6}=\frac{1}{2}\)

(ii) P(a number ≥ 3) = \(\frac{n(B)}{n(S)}=\frac{4}{6}=\frac{2}{3}\)

Sample space, S = 6 

(i) Number of event of getting a prime number, n(E) = 3

∴ p = \(\frac{n(E)}{n(S)} \) = \(\frac{3}{6} \) = \(\frac{1}{2}\)

(ii) p =   \(\frac{n(E)}{n(S)} \) = \(\frac{2}{6} \) = \(\frac{1}3\)

(iii) p =   \(\frac{n(E)}{n(S)} \) = \(\frac{4}{6} \) = \(\frac{2}3\)

(iv) p =   \(\frac{n(E)}{n(S)} \) = \(\frac{1}{6} \)

(v) p =   \(\frac{n(E)}{n(S)} \) = \(\frac{1}{6} \)

(vi) p =   \(\frac{n(E)}{n(S)} \) = \(\frac{3}{6} \) = \(\frac{1}{2}\)

Given: A die is thrown.

The total number of outcomes is six, n (S) = 6

By using the formula,

P (E) = favourable outcomes / total possible outcomes

(i) Let E be the event of getting a prime number

E = {2, 3, 5}

n (E) = 3

P (E) = n (E) / n (S)

= 3 / 6

= 1/2

(ii) Let E be the event of getting 2 or 4

E = {2, 4}

n (E) = 2

P (E) = n (E) / n (S)

= 2 / 6

= 1/3

(iii) Let E be the event of getting a multiple of 2 or 3

E = {2, 3, 4, 6}

n (E) = 4

P (E) = n (E) / n (S)

= 4 / 6

= 2/3

Given: A dice is thrown once

Required to find:

(i) Probability of getting a prime number

(ii) Probability of getting 2 or 4

(iii) Probability of getting a multiple of 2 or 3.

(iv) Probability of getting an even number

(v) Probability of getting a number greater than five.

(vi) Probability of lying between 2 and 6

Total number on a dice is 6 i.e., 1, 2, 3, 4, 5 and 6.

(i) Prime numbers on a dice are 2, 3, and 5. So, the total number of prime numbers is 3.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, probability of getting a prime number = \(\frac{3}{6}\) = \(\frac{1}{2}\)

(ii) For getting 2 and 4, clearly the number of favourable outcomes is 2.

We know that Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting 2 or 4 = \(\frac{2}{6}\) = \(\frac{1}{3}\)

(iii) Multiple of 2 are 3 are 2, 3, 4 and 6.

So, the number of favourable outcomes is 4

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting an multiple of 2 or 3 = \(\frac{4}{6}\) = \(\frac{2}{3}\)

(iv) An even prime number is 2 only.

So, the number of favourable outcomes is 1.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting an even prime number = \(\frac{1}{6}\)

(v) A number greater than 5 is 6 only.

So, the number of favourable outcomes is 1.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a number greater than 5 = \(\frac{1}{6}\)

(vi) Total number on a dice is 6.

Numbers lying between 2 and 6 are 3, 4 and 5

So, the total number of numbers lying between 2 and 6 is 3.

We know that, Probability = Number of favourable outcomes/ Total number of outcomes

Thus, the probability of getting a number lying between 2 and 6 = \(\frac{3}{6}\) = \(\frac{1}{2}\)

(i) Total no of possible outcomes = 6 {1, 2, 3, 4, 5, 6}

E ⟶ Event of getting a prime no.

No. of favorable outcomes = 3 {2, 3, 5}

P(E) = (No. of favorable outcomes)/(Total no. of possible outcomes)

P(E) = 3/6 = 1/2

(ii) E ⟶ Event of getting 2 or 4.

No. of favorable outcomes = 2 {2, 4}

Total no. of possible outcomes = 6

Then, P(E) = 2/6 = 1/3

(iii) E ⟶ Event of getting a multiple of 2 or 3

No. of favorable outcomes = 4 {2, 3, 4, 6}

Total no. of possible outcomes = 6

Then, P(E) = 4/6 = 2/3

(iv) E ⟶ Event of getting an even prime no.

No. of favorable outcomes = 1 {2}

Total no. of possible outcomes = 6 {1, 2, 3, 4, 5, 6}

P(E) = 1/6

(v) E ⟶ Event of getting a no. greater than 5.

No. of favorable outcomes = 1 {6}

Total no. of possible outcomes = 6

P(E) = 1/6

(vi) E ⟶ Event of getting a no. lying between 2 and 6.

No. of favorable outcomes = 3 {3, 4, 5}

Total no. of possible outcomes = 6

P(E) = 3/6 = 1/2