(i) A die is thrown once. Find the probability of getting : (a) a prime number (b) a number between 3 and 6 (c) a number greater than 4 (d) a number at most 4 (e) a factor of 6.

Explanation

(i) As 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))=(3)/(6)=(1)/(2)`
(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))=(2)/(6)=(1)/(3)`
(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))=(2)/(6)=(1)/(3)`
(d) Let A be the event of getting a number at most 4.
`therefore" "A= {1,2,3}" "rArr" "n(A)=4,n(S)=6`
`therefore" Required probability "=(n(A))/(n(S))=(4)/(2)=(2)/(3)`
(e) Let A be the event of getting a factor of 6.
`therefore" "A={1,2,36}" "rArr" "n(A)=4, n(A)=6`
`therefore" Required probability "=(4)/(6)=(2)/(3)`
(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))=(5)/(36)`
(b) Let A be an event ''a total of `10^(n)`. Favourable cases for total of 10 are (6,4), (4,6), (5,5).
`i.e.," "n(A)=5`
`P(A)=(n(A))/(n(S))=(3)/(36)=(1)/(12)`
(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))=(6)/(36)=(1)/(6)`
(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))=(4)/(36)=(1)/(9)`
(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 are : (1,1), (1,2), (2,1), (1,3),
`i.e.," "n(A)=15`
`"Hence, "P(A)=(n(A))/(n(S))=(15)/(36)=(5)/(12)`
(b) Let A be an event ''product less than 16'' i.e., 1,2,3,4,5,6,8,9,10,12,15.
Favourable cases for a product less than 16 are : (1,1), (1,2), (2,1), (1,3), (3,1), (1,4), (4,1), (2,2), (1,5), (5,1), (1,6), (6,1), (2,3), (3,2), (2,4), (4,2), (3,3), (2,5), (5,2), (2,6), (6,2), (3,4), (4,3), (3,5), (5,3).
`i.e., " "n(A)=25`
`"Hence, "P(A)=(n(A))/(n(S))=(25)/(36)`
(c) Let A be an event ''a doublet of odd numbers''.
Favourable cases for a doublet of odd numbers are (1,1), (3,3), (5,5)
`i.e.," "n(A)=3`
`"Hence "P(A)=(n(A))/(n(S))=(3)/(36)=(1)/(12)`