The answers are: (i) 0, (ii) sample space, (iii) 1 and (iv) 1/2 (or 0.5). These represent fundamental probability values: 0 denotes impossibility, 1 indicates certainty and an unbiased coin yields an even chance of 0.5.
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With replacement, total pens = 9. Outcomes for colors are RR, RB, RG, BR, BB, BG, GR, GB, GG. The probability of both picking the same color is P(RR) + P(BB) + P(GG) = 9/81 + 16/81 + 4/81 = ...
Labeling fruits as A1, O1, O2 for basket A and B1, M1 for basket B gives 6 equally likely outcomes. The sample space is {(A1, B1), (A1, M1), (O1, B1), (O1, M1), (O2, B1), (O2, M1)}. P(Apple, Banana) = 1/6.
(i) The sample space is S = {(Samosa, Chai), (Samosa, Lassi), (Pakora, Chai), (Pakora, Lassi), (Bhaji, Chai), (Bhaji, Lassi)}, giving 6 outcomes. (ii) The event selecting Samosa is E = {(Samosa, Chai), (Samosa, Lassi)}.
(i) S = {1H, 1T, 2H, 2T, 3H, 3T, 4H, 4T, 5H, 5T, 6H, 6T} with 12 outcomes. (ii) S = {-4, -3, -2, -1, 0, 1, 2, 3, 4}. (iii) By color, S = {Green, Red}; by distinct balls, S = ...