Answer:
Step-by-step explanation:
s a die is rolled once, therefore there are six possible outcomes, i.e., 1,2,3,4,5,6.
(a) Let A be an event ''getting a prime number''.
Favourable cases for a prime number are 2,3,5,
i.e., n(A)=3
Hence P(A)=n(A)n(S)=36=12
(b) Let A be an event ''getting a number between 3 and 6''.
Favourable cases for events A are 4 or 5.
i.e., n(A)=2
P(A)=n(A)n(S)=26=13
(c) Let A be an event ''a number greater than 4''.
Favourable cases of events A are 5, 6.
i.e., n(A)=2
P(A)=n(A)n(S)=26=13
(d) Let A be the event of getting a number at most 4.
∴ A={1,2,3} ⇒ n(A)=4,n(S)=6
∴ Required probability =n(A)n(S)=42=23
(e) Let A be the event of getting a factor of 6.
∴ A={1,2,36} ⇒ n(A)=4,n(A)=6
∴ Required probability =46=23
(ii) Since, a pair of dice is thrown once, so there are 36 possible outcomes. i.e.,
(a) Let A be an event ''a total 6''. Favourable cases for a total of 6 are (2,4), (4,2), (3,3), (5,1), (1,5).
i.e., n(A)=5
Hence P(A)=n(A)n(S)=536
(b) Let A be an event ''a total of 10n. Favourable cases for total of 10 are (6,4), (4,6), (5,5).
i.e., n(A)=5
P(A)=n(A)n(S)=336=112
(c) Let A be an event ''the same number of the both the dice''. Favourable cases for same number on both dice are (1,1), (2,2), (3,3), (4,4), (5,5), (6,6).
i.e., n(A)=6
P(A)=n(A)n(S)=636=16
(d) Let A be an event ''of getting a total of 9''. Favourable cases for a total of 9 are (3,6), (6,3), (4,5), (5,4).
i.e., n(A)=4
P(A)=n(A)n(S)=436=19
(iii) We have, n(S) = 36
(a) Let A be an event ''a sum less than 7'' i.e., 2,3,4,5,6.
Favourable cases for a sum less than 7 ar
