Question -
Answer -
Let us assume that 1, 2, 3, 4, 5 and 6 are the possible outcomes when the die is thrown.
Here S = {1, 2, 3, 4, 5, 6}
∴n(S) = 6
(i) A prime number will appear,
Let us assume ‘A’ be the event of getting a prime number,
A = {2, 3, 5}
Then, n(A) = 3
P(Event) = Number of outcomes favorable to event/ Totalnumber of possible outcomes
∴P(A) = n(A)/n(S)
= 3/6
= ½
(ii) A number greater than or equal to 3 will appear,
Let us assume ‘B’ be the event of getting a number greaterthan or equal to 3,
B = {3, 4, 5, 6}
Then, n(B) = 4
P(Event) = Number of outcomes favorable to event/ Totalnumber of possible outcomes
∴P(B) = n(B)/n(S)
= 4/6
= 2/3
(iii) A number less than or equal to one will appear,
Let us assume ‘C’ be the event of getting a number less thanor equal to 1,
C = {1}
Then, n (C) = 1
P(Event) = Number of outcomes favorable to event/ Totalnumber of possible outcomes
∴P(C) = n(C)/n(S)
= 1/6
(iv) A number more than 6 will appear,
Let us assume ‘D’ be the event of getting a number more than6, then
D = {0)}
Then, n (D) = 0
P(Event) = Number of outcomes favorable to event/ Totalnumber of possible outcomes
∴P(D) = n(D)/n(S)
= 0/6
= 0
(v) A number less than 6 will appear.
Let us assume ‘E’ be the event of getting a number less than6, then
E= (1, 2, 3, 4, 5)
Then, n (E) = 5
P(Event) = Number of outcomes favorable to event/ Totalnumber of possible outcomes
∴P(E) = n(E)/n(S)
= 5/6