For the following experiments write down the sample space S.

(A) Rolling a die and tossing a coin together.

(B) Choosing a random integer between -5 and +5.

(C) A box containing 5 green and 7 red balls. One ball is drawn at random.                                      [Page No. 167]

(A) Each die outcome (1 to 6) pairs with each coin outcome (H, T), giving $6 \times 2 = 12$ outcomes.

$S = \{(1, H), (2, H), (3, H), (4, H), (5, H), (6, H), (1, T), (2, T), (3, T), (4, T), (5, T), (6, T)\}$ .

Hence, $n(S) = 12$ .

(B) Interpreting 'between -5 and +5' inclusively (including endpoints):

$S = \{-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5\}$ .

$n(S) = 11$ .

If the endpoints are excluded (strictly between):

$S = \{-4, -3, -2, -1, 0, 1, 2, 3, 4\}$ .

$n(S) = 9$ .

(C) If we distinguish only colours:

$S = \{\text{Green, Red}\}$ , $n(S) = 2$ .

If we label balls individually $S = \{G1, G2, G3, G4, G5, R1, R2, R3, R4, R5, R6, R7\}$ ,

$n(S) = 12$ .

For probability calculations, both are valid but if we use $\{\text{Green, Red}\}$ , the outcomes are not equally likely (Red is more likely because there are more red balls).