Q.A box contains 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles are drawn from the box, what is the probability that

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

Total ways to choose $3$ of $25$: $\binom{25}{3}=2300$.

$P(E)=\dfrac{200}{2300}=\dfrac{2}{23}$.

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.

"No teacher" means all $4$ are drawn from the $5+3=8$ non-teachers:

Step 3 — probability.

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.

Count. The number of ways to pick 3 green balls out of 23 is

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: $\dfrac{\binom{4}{3}}{\binom{7}{3}}=\dfrac{4}{35}$.
3. Product $=\dfrac16\cdot\dfrac{4}{35}=\dfrac{4}{210}=\dfrac{2}{105}$.

Common Mistakes

• Forgetting the box is depleted (using $\binom{10}{3}$ again for the second draw).
• Not simplifying $4/210$ to $2/105$.

ANSWER: A

$X = \{a,b\}$, $Y = \{1,3,4,5\}$, 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).

Subsets containing exactly three elements:

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 probability is $\dfrac{6}{36}=\dfrac16$.

"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:

Simplify: $\dfrac{56}{19600}=\dfrac{1}{350}$.

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.

Common Mistakes

• Reading '< 7' as '<= 7' and using {1,...,7}.
• Using permutations instead of combinations (both cancel, but mixing them errs).

ANSWER: E

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.
3. Probability = 28/66 = 14/33.

Common Mistakes

• Using replacement, which would give (8/12)^2.
• Forgetting to reduce 28/66.

ANSWER: D

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

• Using permutations instead of combinations.
• Miscounting the number of elements.

ANSWER: A

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

• Swapping the pools (choosing men from women's count) or using permutations instead of combinations.

ANSWER: C

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 $\tfrac{1}{25}\cdot\tfrac{1}{25}=\tfrac{1}{625}$.
2. There are 25 matching pairs $(i,i)$, so $P=25\cdot\tfrac{1}{625}=\tfrac{1}{25}$.

Common Mistakes

• Reporting $1/625$ (that is one specific pair, not any matching pair).

ANSWER: A