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Virat
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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.

a has one apple and two orangeshalf yearly exam question paper - mathematics of maybe: introduction to probabilityimportant questions from chapter 7 introduction to probability (new book - 2026)ncert class 9 ganita manjari chapter 7 question answersample question from class 9 math chapter 7 introduction to probabilitytree diagram showing all possible pairs of fruitstwo fruit baskets a and b
Virat
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(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)}.

ganita manjari class 9 exercise 7.2 solutionsganita manjari class 9 exercise 7.3 solutionsganita manjari class 9 exercise 7.4 solutionsncert class 9 ganita manjari chapter 7 question answersample space of all possible snack and drink combinationsselecting samosa as a snackthere are 3 popular snacks availablevillage fair
Virat
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In an illustrative 20-toss trial getting 11 heads and 9 tails, experimental probability of heads is 11 / 20 = 0.55. For the next toss, because a coin has no memory, tails probability remains 1 / 2 = 0.5.

a coin 20 times and record the result each time (heads or tailscalculate the experimental probabilityganita manjari class 9 exercise 7.2 solutionsncert class 9 ganita manjari chapter 7 question answersample question from class 9 math chapter 7 introduction to probability