Day 3 - Conditionnal probability

Conditionnal probability

Problem

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

Mathematical explanation

Let's look at the possible outcomes :

We know that at least one of the children is a boy, so only "GG" is not possible. The event where the family has a new boy is then "BB". Hence the probability is :

Draw 2 cards from a deck

Problem

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

Mathematical explanation

There are 13 cards of each suit. Draw one card. It can be anything with probability of 1. Now there are 51 cards left and 12 of them are the same suit as the first card you drew. So the chance the second card matches the 1st is \(\frac{12}{51}\).

Drawing marbles

Problem

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?

Mathematical explanation

On the first draw, the probabilities are the following : we call B the event "a blue ball is drawn" and R the event "a red ball is drawn" * \(P(B)=\frac{4}{7}\) * \(P(R)=\frac{3}{7}\)

On the second draw, if a red ball has been drawn at first, the probabilities are : * \(P(B|R)=\frac{4}{6}\) * \(P(R|R)=\frac{2}{6}\)

Hence, the probability of drawing a blue ball if the first ball drawn was red is \(\frac{1}{3}\)