Probability: Solved Examples- 2 | CSAT Preparation - UPSC PDF Download

Definitions and Fundamental Formulae

• Sample space is the set of all possible outcomes of a random experiment.
• Event is any subset of the sample space.
• Conditional probability of an event A given event B (with P(B) > 0) is defined by the formula:
  P(A | B) = P(A ∩ B) / P(B).
• Multiplication rule (useful for sequential events):
  P(A ∩ B) = P(A) × P(B | A) = P(B) × P(A | B).
• Law of total probability: If {E1, E2, ..., En} is a partition of the sample space with P(Ei) > 0 for each i, then for any event A,
  P(A) = Σ P(A | Ei) × P(Ei), summed over i = 1 to n.
• Independent events: A and B are independent if P(A | B) = P(A) (equivalently P(A ∩ B) = P(A)P(B)).

Solved Examples

Example 1: An urn contains 10 white and 15 black balls. Two balls are drawn in succession without replacement. What is the probability that first is white and the second is black?

Solution:

Let A = event that the first ball is white, and B = event that the second ball is black.

We need P(A ∩ B) = P(A) × P(B | A).

P(A) = 10/25.

P(B | A) = 15/24.

P(A ∩ B) = (10/25) × (15/24).

P(A ∩ B) = (2/5) × (5/8).

P(A ∩ B) = 1/4.

Example 2: A standard die is rolled. Let C be the event that the outcome is an odd number and D be the event that the outcome is not more than 3. Find the probability C knowing that D has happened?

Solution:

If D has occurred, the possible outcomes are {1, 2, 3}.

The intersection C ∩ D is {1, 3} because these are odd and ≤ 3.

P(D) = 3/6 = 1/2.

P(C ∩ D) = 2/6 = 1/3.

P(C | D) = P(C ∩ D) / P(D).

P(C | D) = (1/3) / (1/2) = 2/3.

Example3: Hunar wrote two sections of the CAT paper: Verbal and QA in the same order. The probability of her passing both sections is 0.6. The probability of her passing the verbal section is 0.8. What is the probability of her passing the QA section given that she has passed the Verbal section?

Solution:

Let P(V) denote the probability of passing Verbal, and P(QA) denote the probability of passing QA.

Given P(V) = 0.8 and P(QA ∩ V) = 0.6.

P(QA | V) = P(QA ∩ V) / P(V).

P(QA | V) = 0.6 / 0.8.

P(QA | V) = 0.75.

Example 4: Amika has 2 pouches. One pouch contains 3 golden and 4 silver crystals. The second pouch contains 2 golden and 3 silver crystals. One pouch is selected at random and from the selected pouch, one crystal is drawn. Find the probability that the crystal drawn is golden.

Solution:

Let E1 = event that pouch 1 is chosen, and E2 = event that pouch 2 is chosen.

P(E1) = 1/2 and P(E2) = 1/2 because one pouch is selected at random.

P(golden | E1) = 3/7.

P(golden | E2) = 2/5.

By the law of total probability, P(golden) = P(golden | E1)P(E1) + P(golden | E2)P(E2).

P(golden) = (3/7) × (1/2) + (2/5) × (1/2).

P(golden) = 3/14 + 1/5.

P(golden) = (15/70) + (14/70) = 29/70.

Example 5: A year is selected at random. What is the probability that it contains 53 Mondays if every fourth year is a leap year?( It is a very famous question and appears frequently in many competitive exams.)

Solution:

Let L = event that a selected year is a leap year, and NL = event it is a non-leap year.

P(L) = 1/4 and P(NL) = 3/4 by the given assumption (every fourth year is leap).

In a non-leap year there are 365 = 52×7 + 1 days; so exactly one weekday occurs 53 times. Thus P(53 Mondays | NL) = 1/7.

In a leap year there are 366 = 52×7 + 2 days; so exactly two weekdays occur 53 times. Thus P(53 Mondays | L) = 2/7.

By the law of total probability, P(53 Mondays) = P(53 Mondays | L)P(L) + P(53 Mondays | NL)P(NL).

P(53 Mondays) = (2/7) × (1/4) + (1/7) × (3/4).

P(53 Mondays) = (2/28) + (3/28) = 5/28.

Example 6: There are three cartons, each containing a different number of soda bottles. The first carton has 10 bottles, of which four are flat, the second has six bottles, of which one is flat, and the third carton has eight bottles of which three are flat. What is the probability of a flat bottle being selected when a bottle is chosen at random from one of the three cartons?

Solution:

Let A be the event that the selected bottle is flat.

Let X be the event that carton 1 is chosen; Y that carton 2 is chosen; Z that carton 3 is chosen.

Assume one of the three cartons is chosen uniformly at random, so P(X) = P(Y) = P(Z) = 1/3.

P(A | X) = 4/10 = 2/5.

P(A | Y) = 1/6.

P(A | Z) = 3/8.

By the law of total probability, P(A) = P(A | X)P(X) + P(A | Y)P(Y) + P(A | Z)P(Z).

P(A) = (2/5) × (1/3) + (1/6) × (1/3) + (3/8) × (1/3).

P(A) = 4/30 + 1/18 + 3/24.

Converting to a common denominator 360: 4/30 = 48/360, 1/18 = 20/360, 3/24 = 45/360.

P(A) = (48 + 20 + 45) / 360 = 113/360.

Remarks and Typical Applications

• When to use conditional probability: whenever information about one event is known, and you need probabilities for another event under that information.
• The law of total probability is useful when an event can occur in several disjoint ways (partition), and you know the conditional probabilities for each way.
• Bayes' theorem (not illustrated in detail here) follows from the definitions and the law of total probability and allows reversing conditional probabilities: P(Ei | A) = P(A | Ei)P(Ei) / P(A).
• These ideas are widely used in reliability, signal detection, quality control, and decision-making problems in civil, electrical and computer engineering contexts.