Day 3–10 days of Statistics (HackerRank)

Day 3 of this challenge series comprises three parts/challenges that focus on the concepts of Conditional Probability, Permutation and Combination.

This challenge series does not involve any code input; instead, it is presented in multiple-choice question (MCQ) format, making it easily approachable.

To ensure a solid foundation, let’s begin by clarifying the fundamental concept before delving into the challenges.

Conditional Probability -

Probability of an event occurring, assuming that one or more other events have already occurred.

Formula,

Formula for Conditional Probability

The steps involved in solving conditional probability can be modified as follows:

1. Identify the given condition or event and determine its probability within the sample space.
2. Use the given condition to create a new reduced sample space that includes only the outcomes satisfying the given condition.
3. Calculate the probability of the desired event within the new sample space.
4. Apply the formula for conditional probability, which is the probability of the desired event divided by the probability of the given condition.
5. Simplify or express the result as a fraction, decimal, or percentage as required.

Permutation (arrangement of elements in a specific order)

For example, if you have three items A, B, and C, the permutations would include ABC, ACB, BAC, BCA, CAB, and CBA. Each arrangement is considered a different permutation.

Formula for Permutation

Combination (the selection of elements without considering the order)

Using the same example of three items A, B, and C, the combinations would include ABC, but also ACB, BAC, BCA, CAB, and CBA. In combinations, these different arrangements are considered the same combination because they contain the same elements.

Formula for Combination

Let’s begin with the first challenge.

Challenge #1 — Conditional Probability

Question -

• Suppose a family has 2 children, one of which is a boy. What is the probability that both children are boys?

Answer — 1/3

Explanation -

• To calculate the probability of both children being boys, we need to find the ratio of the number of outcomes where both children are boys (event B) to the total number of outcomes in the sample space (event A).
• Given: Sample space A = {gb, bg, bb} with n(A) = 3 Event B = {bb} with n(B) = 1
• The probability of event B (both children being boys) can be calculated as follows:
• P(B) = n(B) / n(A)
• Substituting the values: P(B) = 1 / 3
• Therefore, the probability of both children being boys is 1/3.

Challenge #2 — Cards of same suit

Question -

• You draw 2 cards from a standard 52-card deck without replacing them. What is the probability that both cards are of the same suit?

Answer — 12/51

Explanation-

1. Determine the sample space, which represents all possible outcomes. In this case, the sample space is the number of ways to choose 2 cards from a 52-card deck, denoted as n(S) = 52C2.
2. Identify the desired event, which is the probability of drawing 2 cards of the same suit. In this case, the event A represents choosing 2 cards from the same suit. The number of ways to choose 2 cards from a single suit is given by 13C2, and since there are 4 suits (hearts, diamonds, clubs, and spades) in a deck, we multiply by 4 to account for all the suits. So, the number of favorable outcomes is n(A) = 13C2 * 4.
3. Calculate the probability by dividing the number of favorable outcomes (n(A)) by the sample space (n(S)). Thus, P(A) = n(A)/n(S) = (13C2 * 4) / (52C2) = 12/51

Challenge #3 — Drawing Marbles

Question -

• A bag contains 3 red marbles and 4 blue marbles. Then, 2 marbles are drawn from the bag, at random, without replacement. If the first marble drawn is red, what is the probability that the second marble is blue?

Answer — 2/3

Explanation -

• Sample space S = {1R, 2R, 3R, 1B, 2B, 3B, 4B} with n(S) = 7 We need to find the probability of drawing a blue marble as the second marble.
• To calculate this probability, we need to remove the first red marble from the sample space because it is already specified. Therefore, our modified sample space will have n(S) = 6.
• Let’s define event A as the event of drawing a blue marble: A = {1B, 2B, 3B, 4B} with n(A) = 4
• Now, we can calculate the probability of drawing a blue marble as the second marble (event A):
• P(A) = n(A) / n(S) = 4 / 6 = 2 / 3
• Therefore, the probability of drawing a blue marble as the second marble is 2/3.

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