Question

Question 3 A die is thrown, find the probability of following events:

(i) A prime number will appear,

(ii) A number greater than or equal to 3 will appear,

(iii) A number less than or equal to one will appear,

(iv) A number more than 6 will appear,

(v) A number less than 6 will appear.

Class X1 - Maths -Probability Page 404

Answer 1

when a die is thrown , then 1 to 6 any number will appear .
So, total number of possible outcomes n(S) = 6

(i) Let E is the event getting a prime number .
Then, n( E) = 3 { because in one dice there are three prime number 1, 3 , 5 }
∴ P(E) = n(E)/n(S) = 3/6 = 1/2

(ii) Let E is the event getting a number greater than or equal to 3
Then, n(E) = 4 [ E ≥ 3 means E = { 3, 4 , 5 , 6 }]
∴ P(E) = n(E)/n(S) = 4/6 = 2/3

(iii) Let E is the event getting less than or equal to one .
Then n(E) = 1 [ E ≤ 1 means E = {1} ]
∴ P(E) = n(E)/n(S) = 1/6

(iv) Let E is the event getting a number more than 6.
Then n(E) = 0 [ E ≥ 6 means E = Ф ]
∴ P(E) = n(E)/n(S) = 0/6 = 0

(v) Let E is the event getting a number less than 6
Then n(E) = 5 [ ∵ E < 6 ⇒E = { 1, 2, 3, 4, 5 }]
∴ P(E) = n(E)/n(S) = 5/6

Answer 2

Answer:

(i) 1 / 2

(ii) 2 / 3

(iii) 1 / 6

(iv) 0

(v) 5 / 6

Step-by-step explanation:

An experiment consists of throwing a die.

∴ The sample space of the experiment is given by S = {1, 2, 3, 4, 5, 6}

(i) Let E be the event that a prime number will appear.

∴ P(E) = n(E) / n(S) = 3 / 6 = 1 / 2

(ii) Let F be the event that a number ≥ 3 will appear.

F ={3, 4, 5, 6}

∴ P(E) = n(F) / n(S) = 4 / 6 = 2 / 3

(iii) Let G be the event that a number ≤ 1 will appear.

G={l}.

∴ P(G) = n(G) / n(S) = 1 / 6

(iv) Let H be the event that a number more than 6 will appear.

H = ⌽

∴ P(H) = n(H) / n(S) = 0 / 6 = 0

(v) Let I be the event that a number less than 6 will appear.

I = (1, 2, 3, 4, 5}

∴ P(I) = n(I) / n(S) = 5 / 6