Question

Given a family of two children (assume boys and girls equally likely, that is, probability 1/2 for each), what is the probability that both are boys? That at least one is a girl? Given that at least one is a girl, what is the probability that both are girls? Given that the first two are girls, what is the probability that an expected third child will be a boy?

Short answer

Both boys: 1/4. At least one girl: 3/4. Both girls if one is girl: 1/3. Third child is boy: 1/2.

Step-by-step solution

Identify Possible Outcomes

List all possible outcomes for a family with two children. Since each child can either be a boy (B) or a girl (G), the possible outcomes are: BB, BG, GB, GG.

Calculate Probability of Both Children Being Boys

There is only one outcome where both children are boys (BB). Out of the four equally likely outcomes, this gives the probability: \[ P(\text{Both boys}) = \frac{1}{4} \]

Calculate Probability of At Least One Girl

Three outcomes include at least one girl (BG, GB, GG). Therefore, the probability is: \[ P(\text{At least one girl}) = \frac{3}{4} \]

Calculate Probability both are Girls Given At Least One Girl

Given that at least one child is a girl, remaining outcomes are: BG, GB, GG. Out of these, only GG satisfies both children being girls. So, the conditional probability is: \[ P(\text{Both girls} | \text{At least one girl}) = \frac{1}{3} \]

Calculate Probability Expected Third Child Will Be Boy Given First Two Are Girls

The gender of the expected third child is independent. Hence, the probability of it being a boy is simply: \[ P(\text{Third child is boy}) = \frac{1}{2} \]

Key concepts

Conditional Probability

Conditional probability helps us find out the likelihood of an event occurring, given that another event has already occurred. This concept is crucial when we deal with more complex probability scenarios. Let's break it down further with our example: Given that at least one child is a girl, what is the probability that both are girls?
To tackle this, we can use the formula for conditional probability:
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]
Here, A is the event 'both children are girls,' and B is the event 'at least one child is a girl.'
From our solution steps, we know:

• Event A (\text{both are girls}) has only one favorable outcome: GG.
• Event B (\text{at least one girl}) has three favorable outcomes: BG, GB, GG.
• Therefore, the probability of A and B occurring together (which is GG) is: \[ P(A \cap B) = \frac{1}{4} \]
  And the probability of B occurring is: \[ P(B) = \frac{3}{4} \]
  Thus, putting it all together in the conditional probability formula: \[ P(A|B) = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3} \]
  This demonstrates how probabilistic outcomes change based on additional information.

Combinatorial Analysis

Combinatorial analysis allows us to count possible outcomes and their arrangements. This is especially useful in determining probabilities in various scenarios. For the example of two children, we need to list all possible combinations. There are 2 choices (boy or girl) for each of the 2 children, giving us:
\[ 2 \times 2 = 4 \]
Thus, resulting in the outcomes: BB, BG, GB, GG.
In another scenario, to calculate the probability of getting at least one girl in a two-child family:
We found that there are three favorable outcomes (BG, GB, GG) out of four possible, making the probability: \[ P(\text{At least one girl}) = \frac{3}{4} \]
This straightforward but powerful counting method helps simplify and solve many probability problems.

Independent Events

Independent events are those that do not affect each other's outcomes. Understanding this concept can simplify many problems. In our scenario, let's explore the question: Given that the first two children are girls, what is the probability that an expected third child will be a boy?
Here, the gender of the third child is independent of the gender of the first two children. No matter what the first two children are, the probability of the third child being a boy remains:
\[ P(\text{Third child is boy}) = \frac{1}{2} \]
This independence makes our calculation straightforward since the outcome of one child does not alter the probability of the next.