Assume that each born child is equally likely to be a boy or a girl. If two families have two childr

Q: Wait, why are we treating all 4 children as a single group? Shouldn't we look at each family separately?

That is a brilliant intuition! You absolutely could calculate the probabilities for each family separately and combine them, but treating them as a single set of 4 independent trials is mathematically equivalent and significantly faster. Since each child's gender is independent, the total sample space of $2^4$ remains valid regardless of how you group them.

Q: Why do we subtract the cases for 0 and 1 girl from the total? Is there a faster way?

Subtracting the complement is the 'pro' move in combinatorics! Calculating $n(G=2) + n(G=3) + n(G=4)$ involves summing $\binom{4}{2} + \binom{4}{3} + \binom{4}{4}$, which is $6 + 4 + 1 = 11$. Both methods yield the same result, but the complement method is less prone to arithmetic errors in larger problems.

Q: Does the order of birth matter here?

In probability, we always assume the sample space accounts for distinct outcomes. Even if you don't 'see' the order, the math treats 'Girl-Boy' as distinct from 'Boy-Girl' to ensure each outcome is equally likely. If we didn't account for order, the outcomes would not have equal probabilities, and our simple fraction would fail.

Q: What if the question said 'at least one girl'? Would the denominator change?

Exactly! If the condition were 'at least one girl', the complement would only be the case of 0 girls. The denominator would then be $16 - \binom{4}{0} = 15$. The logic remains the same; only the boundary of your 'restricted universe' shifts.

Q: Is it possible for the conditional probability to be greater than 1?

Never! A conditional probability is a ratio of a subset to a larger set. Since the event 'all girls' is a subset of 'at least two girls', the numerator must be less than or equal to the denominator. If you ever get a value $>1$, it is a sign to re-check your counting!

Question

Assume that each born child is equally likely to be a boy or a girl. If two families have two children each, then the conditional probability that all children are girls given that at least two are girls is :

Options:

(A) 1/11

(B) 1/17

(C) 1/10

(D) 1/12

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To begin our journey, we define the universe of possibilities. We have two families, each with two children, resulting in a total of $n = 4$ children.

Each child represents an independent event with two possible outcomes: a Boy ($B$) or a Girl ($G$). Since each child has two possibilities, the total number of ways these four children can be arranged is:

$$2^4 = 16$$

This represents our sample space, $S$, the ground truth from which all our probabilities originate.

### The Art of the Constraint

The problem introduces a specific condition: we are given that "at least two are girls." In the language of set theory, this defines our restricted sample space, $B$. We are no longer considering all 16 possibilities; we are focusing exclusively on scenarios where the number of girls $G \ge 2$.

Rather than listing every possibility, we utilize the elegance of the complement. The total number of outcomes is 16. The outcomes that fail to satisfy our condition are those with zero girls or exactly one girl.

For zero girls, there is $\binom{4}{0} = 1$ way (the $BBBB$ scenario). For exactly one girl, there are $\binom{4}{1} = 4$ ways (the girl could be the first, second, third, or fourth child).

Thus, the number of outcomes that fail our condition is $1 + 4 = 5$. By subtracting this from our total, we find the size of our restricted universe:

$$n(B) = 16 - 5 = 11$$

This value, 11, serves as the bedrock of our final calculation.

### The Target and the Triumph

We seek the probability that all children are girls, given our condition. Let this event be $A$. The only way to have all four children be girls is the single outcome $\{GGGG\}$.

Therefore, the number of favorable outcomes within our restricted space is:

$$n(A \cap B) = 1$$

We now stand at the threshold of the final calculation. The conditional probability $P(A \mid B)$ is defined as the ratio of the favorable outcomes within our restricted space to the total size of that space:

$$P(A \mid B) = \frac{n(A \cap B)}{n(B)} = \frac{1}{11}$$

### The Takeaway

The final result is **$\frac{1}{11}$**. It is clean, precise, and mathematically sound.

The trap in this problem is not the complexity of the math, but the temptation to overcomplicate the counting process. By using the complement, we transformed a tedious task into a swift, logical deduction.

In the JEE, your most powerful tool is the ability to simplify the problem before you begin calculating. Keep practicing, keep questioning, and maintain your curiosity.

Animated Solution for Mathematics - Probability: Assume that each born child is equally likely to be a boy or a girl. If two families have two children each, then the conditional probability that all children are girls given that at least two are girls is :

Select Answer:

Visualized Solution

Understanding the Scenario

Calculating Total Outcomes

Defining the Condition

Using the Complement

Case: Zero Girls

Case: Exactly One Girl

The Restricted Sample Space

The Target Event

Final Calculation

The Sigma Insight: Conditional Probability

Analyzing the Setup

The Art of the Constraint

The Target and the Triumph

The Takeaway

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