Q.Out of 100 students, two sections of 40 and 60 are formed. If you and your friend are among the 100 students, what is the probability that

Selecting one boy and two girls: $\binom{20}{1}\binom{5}{2}=20\cdot10=200$. …

Concept. For an equally-likely selection, probability $=\dfrac{\text{number of favourable committees}}{\text{total number of committees}}$, and each count is a combination since order does not matter.

Step 1 — total members and total committees.

The council has $5+4+3=12$ members. A committee of $4$ can be formed in

Step 2 — committees with no teacher. …

Setup. Total balls $=24$: one black, $24-1=23$ green. Three balls are drawn simultaneously (order irrelevant), and we want all three to be green. …

Concept and Intuition

Multiply the probability of drawing 3 white first (from 6W,4R) by the conditional probability of drawing 3 red next (from the depleted box 3W,4R).

Step-by-Step Solution

1. First draw all white: $\dfrac{\binom{6}{3}}{\binom{10}{3}}=\dfrac{20}{120}=\dfrac16$.
2. Remaining: 4 red, 3 white (7 total). Second draw all red: $\dfrac{\binom{4}{3}}{\binom{7}{3}}=\dfrac{4}{35}$. …

$X = \{a,b\}$, $Y = \{1,3,4,5\}$, so $X\times Y$ has $2\times 4 = 8$ elements.

The total number of subsets of an 8-element set is $2^8 = 256$ (the sample space). …

A difference of zero occurs exactly when both dice show the same number: $(1,1),(2,2),\dots,(6,6)$ — 6 outcomes out of 36. The pr …

"Divisible by both $2$ and $3$" $\Leftrightarrow$ divisible by $6$. In $1$–$50$: $6,12,18,24,30,36,42,48$ — that is $8$ numbers.

Choosing $3$ distinct numbers all from these: …

Concept and Intuition

The maximum of the three drawn numbers is below 7 exactly when every drawn number is at most 6. So the favourable draws are 3-subsets of the six labels 1-6.

Step-by-Step Solution

1. max < 7 is equivalent to all three numbers being in {1,2,3,4,5,6}.
2. Favourable ways = 6C3; total ways = 10C3.
3. Probability = 6C3 / 10C3. …

Concept and Intuition

Without replacement, the chance both are black is the number of ways to pick 2 black marbles over the number of ways to pick any 2 marbles.

Step-by-Step Solution

1. Total marbles = 12; ways to choose 2 = 12C2 = 66.
2. Ways to choose 2 black = 8C2 = 28. …

Concept and Intuition

Choosing a subset of a fixed size is a combination, order irrelevant.

Step-by-Step Solution

1. The set $\{2,4,\dots,18\}$ has 9 elements.
2. Number of 4-element subsets $=\binom{9}{4}$.
3. $\binom{9}{4}=\frac{9\cdot8\cdot7\cdot6}{24}=126$.

Common Mistakes …

Concept and Intuition

A committee is an unordered selection, so we use combinations. Independent choices of women and men multiply.

Step-by-Step Solution

1. Choose 3 women from 5: $\binom{5}{3}=10$.
2. Choose 5 men from 8: $\binom{8}{5}=56$.
3. Multiply: $10\times 56=560$.

Common Mistakes …

Concept and Intuition

With replacement, the two draws are independent. Whatever number appears first, the second draw must equal it, which has probability $1/25$ regardless of the first.

Step-by-Step Solution

1. Each ordered pair $(i,j)$ has probability $\tfrac{1}{25}\cdot\tfrac{1}{25}=\tfrac{1}{625}$. …