Questions: Probability

Q1. Define sample space and event with an example.

The sample space of a random experiment is the set of all possible outcomes. For a single throw of a die the sample space is {1, 2, 3, 4, 5, 6}. An event is a subset of the sample space, and it occurs when the outcome of the experiment lies in that subset. For the same die, the event of getting an even number is {2, 4, 6}, and the event of getting a number greater than four is {5, 6}. The event is a set, so unions, intersections and complements of events are defined exactly as for sets.

Hinglish: Sample space saare possible outcomes ka set hai, jaise die ke liye {1,2,3,4,5,6}. Event sample space ka subset hota hai, jaise even numbers {2,4,6}. Events par union, intersection aur complement sets ki tarah lagte hain.

Q2. State the three axioms of probability.

For any event A of a sample space S, the probability P(A) is at least 0 and at most 1. The probability of the sure event is 1, that is P(S) = 1. For mutually exclusive events A and B, the probability of their union is the sum of their probabilities. From these axioms it follows that the probability of the impossible event is 0 and that P(A') = 1 - P(A).

Hinglish: Teen axioms: probability 0 se 1 ke beech hoti hai, P(S) = 1 hota hai, aur mutually exclusive events ki probabilities add hoti hain. Inse P(phi) = 0 aur P(A') = 1 - P(A) nikalta hai.

Q3. A die is rolled. Find the probability of getting a prime number.

The sample space is {1, 2, 3, 4, 5, 6}, so there are 6 equally likely outcomes. The prime numbers in the sample space are 2, 3 and 5, so the favourable outcomes number 3. The probability is 3/6 = 1/2.

Hinglish: Sample space me 6 outcomes hain, prime numbers 2, 3 aur 5 hain, isliye probability 3/6 = 1/2 hai.

Q4. Two dice are rolled together. Find the probability that the sum of the numbers is 8.

Each die has 6 outcomes, so the ordered pairs number 36. The pairs with sum 8 are (2, 6), (3, 5), (4, 4), (5, 3) and (6, 2), five in all. The probability is therefore 5/36. Note that (4, 4) counts once while the other four pairs come in two orders, which is why the count is 5 and not 6.

Hinglish: Total outcomes 36 hain. Sum 8 wale pairs (2,6), (3,5), (4,4), (5,3), (6,2) hain, isliye probability 5/36 hai. (4,4) ek hi baar count hota hai.

Q5. What is the probability of drawing an ace or a king from a pack of 52 cards?

There are 4 aces and 4 kings, and the two sets are disjoint, so the favourable outcomes number 8. The probability is 8/52 = 2/13. Because the events are mutually exclusive, the probabilities simply add: 4/52 + 4/52 = 8/52.

Hinglish: 4 aces aur 4 kings hain, aur dono sets alag hain, isliye favourable outcomes 8 hue. Probability 8/52 = 2/13. Mutually exclusive hone ki wajah se probabilities seedha add hui.

Q6. Two events have probabilities 0.5 and 0.4, and the probability that both occur is 0.2. Find the probability that at least one occurs.

Use the general addition rule: P(A union B) = 0.5 + 0.4 - 0.2 = 0.7. The probability that neither occurs is 1 - 0.7 = 0.3. If the events had been mutually exclusive, the intersection probability would have been 0 instead of 0.2 and the union probability would have been 0.9.

Hinglish: Addition rule se P(A union B) = 0.5 + 0.4 - 0.2 = 0.7. At least one ka matlab union hota hai. Neither ki probability 1 - 0.7 = 0.3 hai.

Q7. A coin is tossed three times. Find the probability of getting exactly two heads.

The sample space has 2 cubed, that is 8, equally likely outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH and TTT. Exactly two heads appear in HHT, HTH and THH, three outcomes. The probability is 3/8. The same answer follows from the binomial count: three sequences with two heads and one tail, each with probability 1/8.

Hinglish: 3 coin tosses me 8 outcomes hote hain. Exactly two heads HHT, HTH, THH me aate hain, isliye probability 3/8 hai.

Q8. Prove that P(A') = 1 - P(A).

The event A and its complement A' are disjoint, and together they cover the whole sample space, because every outcome either belongs to A or does not. Therefore A union A' equals S, and by the addition rule for mutually exclusive events P(A) + P(A') = P(S). The probability of the sure event is 1, so P(A') = 1 - P(A). The result is useful whenever the complement is easier to count than the event itself.

Hinglish: A aur A' disjoint hain aur milke poora S banaate hain, isliye P(A) + P(A') = P(S) = 1. Isse P(A') = 1 - P(A) milta hai, jo counting ke time kaam aata hai.

Q9. From a class of 15 boys and 10 girls a student is chosen at random. Find the probability that the student is a girl.

The total number of students is 25, and every student is equally likely to be chosen. The favourable outcomes are the 10 girls. The probability is 10/25 = 2/5. The probability of choosing a boy is 15/25 = 3/5, and the two probabilities add to 1, as they should because boy and girl form a complementary pair in this class.

Hinglish: Total students 25 hain, girls 10 hain, isliye probability 10/25 = 2/5 hai. Boy choose karne ki probability 15/25 = 3/5 hai, aur dono ka sum 1 hota hai.

Q10. A bag has 4 red and 6 black balls. Two balls are drawn one after another without replacement. Find the probability that both are red.

For the first draw the probability of red is 4/10. If a red ball has been drawn, 3 red and 6 black remain, so the probability of red on the second draw is 3/9. The probability that both are red is the product, 4/10 times 3/9 = 12/90 = 2/15. The product appears because the second draw happens in the reduced bag that results from the first draw.

Hinglish: Pehle red ki probability 4/10 hai. Red nikalne ke baad bag me 3 red aur 6 black bachte hain, isliye doosri red 3/9. Dono ke liye product: 4/10 x 3/9 = 2/15.

Quiz below: five questions on sample spaces, axioms, the complement rule, the addition rule and equally likely outcomes. Try them before checking the formula sheet.