A general problem-solving technique video — not a specific exam question.
Problem
A committee of 4 persons has to be chosen from 8 boys and 6 girls, consisting of at least one girl. Find the probability that the committee consists of more girls than boys.
Key insight. A 4-person committee can only have “more girls than boys” if it has 3 or 4 girls (since 2-2 is a tie, not “more”). That set of outcomes is automatically contained within “at least one girl,” so the numerator and denominator of the conditional probability can both be built from the same breakdown of boy-girl compositions — no separate counting needed for each.
Watch this explained step by step → · More on the Shiwam’s Classes channel
Approach
List every possible boy-girl composition of a 4-person committee: $(4,0), (3,1), (2,2), (1,3), (0,4)$. Compute the number of ways to form each. “At least one girl” excludes only $(4,0)$; “more girls than boys” includes only $(1,3)$ and $(0,4)$. Since the second set is contained in the first, the conditional probability is simply their ratio.
Solution
Step 1 — Count the ways to form each composition
$$(0,4): \binom{8}{0}\binom{6}{4} = 1\times15 = 15$$
$$(1,3): \binom{8}{1}\binom{6}{3} = 8\times20 = 160$$
$$(2,2): \binom{8}{2}\binom{6}{2} = 28\times15 = 420$$
$$(3,1): \binom{8}{3}\binom{6}{1} = 56\times6 = 336$$
$$(4,0): \binom{8}{4}\binom{6}{0} = 70\times1 = 70$$
Step 2 — Verify the total
$$15+160+420+336+70 = 1001 = \binom{14}{4} \checkmark$$
Step 3 — Count “at least one girl”
This excludes only the $(4,0)$ (all-boys) case:
$$1001 – 70 = 931$$
Step 4 — Count “more girls than boys”
This requires strictly more girls, i.e. 3 or 4 girls out of 4 — the $(1,3)$ and $(0,4)$ cases:
$$160+15 = 175$$
Step 5 — Apply the conditional probability formula
Since “more girls than boys” $\subseteq$ “at least one girl”:
$$P(\text{more girls}\mid\text{at least one girl}) = \frac{175}{931}$$
Step 6 — Simplify
$$\gcd(175,931) = 7 \implies \frac{175\div7}{931\div7} = \frac{25}{133}$$
Answer
$$\frac{25}{133}$$
Common mistakes
- Including the $(2,2)$ composition in “more girls than boys.” An equal split is not “more” — only strictly more girls than boys counts, which restricts to the $(1,3)$ and $(0,4)$ cases.
- Computing the denominator as the full $\binom{14}{4}=1001$ instead of $931$. The condition “at least one girl” is given information that changes the sample space for a conditional probability — it isn’t optional context to ignore.
Practise next
- A committee of 5 is chosen from 6 boys and 5 girls, given at least one girl. Find the probability the committee has more girls than boys, using the same composition-counting method.
Show answer
$\dfrac{181}{456}$. Committees of $5$ from $11$ people number $\binom{11}{5}=462$; those with no girl number $\binom65=6$, so the conditioning event has $462-6=456$ outcomes.
More girls than boys means at least $3$ girls:
$$\binom53\binom62+\binom54\binom61+\binom55\binom60=10(15)+5(6)+1=181.$$
So the probability is $\dfrac{181}{456}$. As always with conditioning, the denominator shrinks to the restricted count — using $462$ here is the standard slip.
Doubts are answered by Shiwam, usually within a day. Ask about this question specifically — a general question about the chapter is better asked in class.