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

Showing the 12 most recent of 55 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.3 dice are thrown. Then the probability of getting 11 as the sum of all the numbers appeared on the faces of the three dice is (A) $\dfrac{13}{216}$ (B) $\dfrac{5}{108}$ (C) $\dfrac{1}{8}$ (D) $\dfrac{1}{16}$

   ›Reveal solutionSolution

   Count the ordered triples $(a,b,c)$ with $1\le a,b,c\le6$ summing to $11$ using inclusion–exclusion on the unrestricted stars-and-bars count, then divide by $216$. Answer: $\frac18$.

   Concept and Intuition

   With three dice there are $6^3=216$ equally likely outcomes. To count how many give a fixed sum, substitute $a'=a-1$ etc. so each variable ranges over $0$–$5$, turn it into a stars-and-bars problem, and use inclusion–exclusion to remove the outcomes where some die would need to show more than $6$.

   Step-by-Step Solution

   1. We need the number of solutions to $a+b+c=11$ with each of $a,b,c\in\{1,\dots,6\}$.
   2. Substitute $a'=a-1,b'=b-1,c'=c-1$, each in $\{0,\dots,5\}$: $a'+b'+c'=8$.
   3. Unrestricted non-negative solutions to $a'+b'+c'=8$: $\binom{8+2}{2}=\binom{10}{2}=45$.
   4. Subtract cases where some variable exceeds 5 (i.e. $\ge6$): set $a''=a'-6\ge0$, giving $a''+b'+c'=2$, with $\binom{2+2}{2}=\binom42=6$ solutions; same for $b'$ or $c'$ exceeding 5. Three such variables, so subtract $3\times6=18$.
   5. No two variables can simultaneously exceed 5 here (would need sum $\ge12>8$), so no double-count correction needed.
   6. Valid count: $45-18=27$.
   7. Probability $=\dfrac{27}{216}=\dfrac18$.

   Common Mistakes

   • Forgetting the upper bound (each die face is at most 6) and just using the raw stars-and-bars count of 45 instead of 27.
   • Sign or combinatorial slips in the inclusion–exclusion subtraction step.

   ✓Final answer

   The correct option is (C) — $\dfrac{1}{8}$.

   ANSWER: C

2. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.If five unit squares are selected at random from a chess board, then the probability that they all lie on a diagonal is (A) $\dfrac{112}{{}^{64}C_5}$ (B) $\dfrac{56}{{}^{64}C_5}$ (C) $\dfrac{448}{{}^{64}C_5}$ (D) $\dfrac{224}{{}^{64}C_5}$

   ›Reveal solutionSolution

   Count 5-square subsets lying wholly on one diagonal (either direction) of the $8\times8$ board, divide by $\binom{64}{5}$. Answer: $\dfrac{224}{\binom{64}{5}}$.

   Concept and Intuition

   An $8\times 8$ board has diagonals running in two perpendicular directions ($\nwarrow\!\searrow$ and $\nearrow\!\swarrow$). In each direction there are $15$ diagonals whose lengths run $1,2,\dots,7,8,7,\dots,2,1$. Five squares can only be chosen from a diagonal of length at least $5$.

   Step-by-Step Solution

   1. In one direction, diagonals of length $\ge 5$ have lengths: $5,6,7,8,7,6,5$ (two diagonals each of length $5,6,7$ and one of length $8$).
   2. Number of ways to pick 5 squares from a diagonal of length $L$ is $\binom{L}{5}$.
   3. Sum for one direction: $2\binom{5}{5}+2\binom{6}{5}+2\binom{7}{5}+\binom{8}{5} = 2(1)+2(6)+2(21)+56 = 2+12+42+56=112$.
   4. The board has this same structure in the other diagonal direction too, doubling the count: $112\times 2 = 224$.
   5. Total ways to choose any 5 squares from the 64 squares $= \binom{64}{5}$.
   6. Required probability $= \dfrac{224}{\binom{64}{5}}$.

   Common Mistakes

   • Counting only one diagonal direction and forgetting the board has diagonals both ways.
   • Including diagonals shorter than 5 squares (they contribute $\binom{L}{5}=0$, but a careless sum can miscount which lengths qualify).

   ✓Final answer

   The correct option is (D) — $\dfrac{224}{{}^{64}C_5}$.

   ANSWER: D

3. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.Out of the first 20 consecutive natural numbers, 3 numbers are chosen at random. If these 3 numbers are in arithmetic progression with a common difference $d \in N$, then the probability of getting those 3 numbers whose common difference is a prime number, is (A) $\dfrac{4}{9}$ (B) $\dfrac{1}{3}$ (C) $\dfrac{23}{45}$ (D) $\dfrac{29}{90}$

   ›Reveal solutionSolution

   Count all 3-term APs in $\{1,\ldots,20\}$ by common difference, then the fraction with a prime common difference. Answer: $\dfrac{23}{45}$.

   Concept and Intuition

   A 3-term AP from $1$ to $20$ is determined by its first term $a$ and common difference $d\ge1$, with the constraint $a+2d\le 20$. For a fixed $d$, $a$ can range from $1$ to $20-2d$, giving $20-2d$ possible APs (valid only while $20-2d\ge1$, i.e. $d\le 9$).

   Step-by-Step Solution

   1. Total number of 3-term APs: $\displaystyle\sum_{d=1}^{9}(20-2d) = (18+16+14+12+10+8+6+4+2) = 90$.
   2. Prime values of $d$ in the range $1$ to $9$: $d=2,3,5,7$.
   3. Count for each prime $d$: $d=2 \Rightarrow 16$; $d=3 \Rightarrow 14$; $d=5 \Rightarrow 10$; $d=7 \Rightarrow 6$.
   4. Sum of favourable cases: $16+14+10+6 = 46$.
   5. Since we are told the chosen triple is already an AP, this is a conditional probability over the 90 APs: $P = \dfrac{46}{90} = \dfrac{23}{45}$.

   Common Mistakes

   • Treating $d=1$ as prime (it is not) or omitting $d=2$ (the only even prime).
   • Using $\binom{20}{3}$ as the denominator instead of the actual count of AP triples — the problem conditions on the numbers already forming an AP.

   ✓Final answer

   The correct option is (C) — $\dfrac{23}{45}$.

   ANSWER: C

4. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.Let $P$ and $Q$ are two sets such that $n(P) = 27$, $n(Q) = 17$, $n(P \cap Q) = 5$. If $x$ is the number of ways of selecting 7 elements from $P$ such that all the elements of $P \cap Q$ are in each selection and $y$ is the number of ways of selecting 10 elements from $Q$ such that no element of $P \cap Q$ is present in any selection, then $x + y + 1 =$ (A) $231$ (B) $248$ (C) $297$ (D) $298$

   ›Reveal solutionSolution

   With the 5 common elements forced in (for x) or forced out (for y), both counts reduce to simple combinations from the remaining pool: x=C(22,2)=231, y=C(12,2)=66, so x+y+1=298.

   Concept and Intuition

   When a selection must contain a fixed subset, "use up" that subset first and count the remaining choices from what's left. When a selection must avoid a fixed subset entirely, simply remove that subset from the pool before counting.

   Step-by-Step Solution

   1. $P\setminus(P\cap Q)$ has $27-5=22$ elements; $Q\setminus(P\cap Q)$ has $17-5=12$ elements.
   2. For $x$: all 5 elements of $P\cap Q$ must be present in every 7-element selection from $P$, so we only choose the remaining $7-5=2$ elements, and they must come from the 22 elements of $P$ not in $Q$ (choosing from elements already in $P\cap Q$ again would double-count them). $x=\binom{22}{2}=\frac{22\cdot21}{2}=231$.
   3. For $y$: no element of $P\cap Q$ may appear, so all 10 elements must come from the 12 elements of $Q$ not in $P$. $y=\binom{12}{10}=\binom{12}{2}=\frac{12\cdot11}{2}=66$.
   4. $x+y+1 = 231+66+1=298$.

   Common Mistakes

   • For $x$, forgetting to subtract the forced 5 from both the total count needed (7) and the pool size (27), and instead computing $\binom{27}{7}$ or a similarly wrong combination.
   • For $y$, choosing from all of $Q$ (17) instead of restricting to $Q\setminus(P\cap Q)$ (12).

   ✓Final answer

   The correct option is (D) — 298.

   ANSWER: D

5. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.In each of its move, a pawn in a chess board can move one step either horizontally or vertically to its adjacent cell from its current position. If a pawn is initially located at the South-West corner cell of the chess board, then the number of ways it can reach the North-East corner cell of the chess board with minimum number of moves, is (A) $^{64}C_2$ (B) $2 \times {}^{8}C_2$ (C) $^{14}C_7$ (D) $^{16}C_8$

   ›Reveal solutionSolution

   A minimum-move path from one corner of the 8×8 board to the diagonally opposite corner needs 7 rightward + 7 upward unit moves (14 total), and the count of such paths is C(14,7).

   Concept and Intuition

   This is a classic lattice-path counting problem. Moving strictly right or up (never left/down, since that would not be a "minimum" path) from one corner to the diagonally opposite corner of an $n\times n$ grid of cells requires exactly $(n-1)$ rightward and $(n-1)$ upward moves. Every minimum path is just some interleaving/ordering of these moves, so the count is a binomial coefficient — choose the positions (among all moves) that are "right" moves (the rest are "up").

   Step-by-Step Solution

   1. A chessboard is an $8\times 8$ grid of cells. Moving from the SW corner cell to the NE corner cell requires crossing 7 cell-boundaries horizontally and 7 vertically (since there are 8 cells but only 7 "gaps" between them in each direction).
   2. Minimum number of moves = $7+7 = 14$ (any extra move, e.g. left then right, would only add moves, so the minimum path uses only right/up moves).
   3. Each minimal path is a sequence of 14 moves, 7 of which are "right" (R) and 7 of which are "up" (U), in some order.
   4. The number of distinct orderings of 7 R's and 7 U's among 14 positions is $\binom{14}{7}$.

   Common Mistakes

   • Miscounting the grid gaps as 8 instead of 7 (an 8×8 board has 8 cells per side but only 7 boundaries to cross to go from one edge cell to the opposite edge cell).
   • Confusing this with $^{64}C_2$ (which would be a different, unrelated count — e.g., ways to pick 2 cells out of 64 — not relevant here).

   ✓Final answer

   The correct option is (C) — $^{14}C_7$.

   ANSWER: C

6. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.A committee of 8 members is to be formed from 5 teaching staff, 4 office staff and 6 students so as to include atleast two from each category. Then the total number of ways of forming the committee is (A) 4100 (B) 3950 (C) 3200 (D) 3500

   ›Reveal solutionSolution

   Enumerate all ways to split 8 committee seats among the three groups with at least 2 from each (and not exceeding each group's size), then sum the products of combinations for each valid split. Total: 4100.

   Concept and Intuition

   This is a constrained partition-and-choose problem: we first find every integer triple $(t,o,s)$ satisfying the "at least 2 from each, totalling 8" constraint (respecting the group sizes as upper bounds too), then for each triple compute the number of ways to actually pick that many people from each group using combinations, and add up over all valid triples (since the triples are mutually exclusive committee compositions).

   Step-by-Step Solution

   1. We need $t+o+s=8$ with $2\le t\le5$, $2\le o\le4$, $2\le s\le6$.
   2. Fix $t=2$: then $o+s=6$ with $2\le o\le4$: $(o,s)=(2,4),(3,3),(4,2)$ — all satisfy $s\le6$.
   3. Fix $t=3$: then $o+s=5$: $(o,s)=(2,3),(3,2)$ [$o=4\Rightarrow s=1$ invalid].
   4. Fix $t=4$: then $o+s=4$: $(o,s)=(2,2)$ [$o=3\Rightarrow s=1$ invalid].
   5. Fix $t=5$: then $o+s=3$, but $o\ge2,s\ge2$ forces $o+s\ge4$ — no solutions.
   6. So the valid triples $(t,o,s)$ are: $(2,2,4),(2,3,3),(2,4,2),(3,2,3),(3,3,2),(4,2,2)$ — six of them.
   7. Compute $\binom{5}{t}\binom{4}{o}\binom{6}{s}$ for each:
      • $(2,2,4)$: $\binom52\binom42\binom64=10\cdot6\cdot15=900$
      • $(2,3,3)$: $10\cdot4\cdot20=800$
      • $(2,4,2)$: $10\cdot1\cdot15=150$
      • $(3,2,3)$: $10\cdot6\cdot20=1200$
      • $(3,3,2)$: $10\cdot4\cdot15=600$
      • $(4,2,2)$: $5\cdot6\cdot15=450$
   8. Sum: $900+800+150+1200+600+450=4100$.

   Common Mistakes

   • Missing an upper-bound constraint (e.g. allowing $o=5$ when only 4 office staff exist) and thereby including an invalid triple.
   • Forgetting a valid triple, especially the asymmetric ones like $(3,2,3)$ vs $(2,3,3)$ — these are genuinely different committees and both must be counted.

   ✓Final answer

   The correct option is (A) — 4100.

   ANSWER: A

7. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.If three cards are drawn randomly from a well shuffled pack of 52 cards, then the probability that all the three bearing a prime number is (A) $\dfrac{35}{1105}$ (B) $\dfrac{28}{1105}$ (C) $\dfrac{21}{1105}$ (D) $\dfrac{18}{1105}$

   ›Reveal solutionSolution

   Identify how many of the 52 cards carry a prime-numbered rank, then use the hypergeometric (combinations) count. Answer: $\dfrac{28}{1105}$.

   Concept and Intuition

   A standard deck's ranks are A, 2–10, J, Q, K. "Bearing a prime number" restricts us to the numeral ranks that are themselves prime: $2, 3, 5, 7$ (Ace = 1 is not prime, and face cards carry no number). Each of these 4 ranks appears in all 4 suits, giving 16 qualifying cards. The probability that all 3 drawn cards come from this set of 16 is a straightforward ratio of combinations.

   Step-by-Step Solution

   1. Prime ranks among 1–13 (card numerals, ignoring face cards/ace as non-numeric primes): $2, 3, 5, 7$ — that's 4 ranks.
   2. Each rank has 4 suits, so there are $4\times4=16$ prime-numbered cards in the deck.
   3. Number of ways to choose 3 cards all from these 16: $\binom{16}{3}=\dfrac{16\cdot15\cdot14}{6}=560$.
   4. Total ways to choose any 3 cards from 52: $\binom{52}{3}=\dfrac{52\cdot51\cdot50}{6}=22100$.
   5. Probability $=\dfrac{560}{22100}$. Simplify: divide by 20 to get $\dfrac{28}{1105}$.

   Common Mistakes

   • Including the Ace as a "prime number" card (1 is not prime) or including face cards.
   • Forgetting to simplify the final fraction, or miscounting $\binom{52}{3}$.

   ✓Final answer

   The correct option is (B) — $\dfrac{28}{1105}$.

   ANSWER: B

8. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.S is the set of all 5 digit numbers greater than 60,000 formed from the digits $1,2,3,4,5,6,7$. If a number is selected at random from S, then the probability that the sum of the first and last digits is more than 10 in the selected number is (A) $\dfrac{1}{7}$ (B) $\dfrac{1}{2}$ (C) $\dfrac{2}{7}$ (D) $\dfrac{5}{7}$

   ›Reveal solutionSolution

   This tests identifying which first digits make a 5-digit number (from digits 1-7) exceed 60,000, then computing a conditional count on the last digit while the middle three digits cancel out entirely.

   Concept and Intuition

   Since the available digits are $1$ through $7$ (no zero), any 5-digit number starting with $6$ or $7$ is automatically greater than $60{,}000$ (the smallest such number, $61111$, already exceeds it), while any number starting with $1$–$5$ is automatically at most $57777<60000$. So the "greater than 60000" condition depends only on the first digit being in $\{6,7\}$ — this fully defines set $S$. The three middle digits are free and identical in every case, so they contribute the same factor to both numerator and denominator and cancel.

   Step-by-Step Solution

   1. First digit must be $6$ or $7$ for the number to exceed $60{,}000$ (digits only go up to $7$, no zero).
   2. So $S$ = all numbers $d_1d_2d_3d_4d_5$ with $d_1\in\{6,7\}$ and $d_2,d_3,d_4,d_5\in\{1,\dots,7\}$ freely (repetition allowed); $|S|=2\times7^4$.
   3. Want: $P(d_1+d_5>10)$. The middle three digits don't affect this, so we can work with just $d_1,d_5$: each combination of $d_2,d_3,d_4$ contributes equally to numerator and denominator.
   4. Case $d_1=6$: need $d_5>4$, i.e. $d_5\in\{5,6,7\}$ — 3 out of 7 choices.
   5. Case $d_1=7$: need $d_5>3$, i.e. $d_5\in\{4,5,6,7\}$ — 4 out of 7 choices.
   6. Favourable count $=(3+4)\times7^3=7\times7^3=7^4$. Total count $=2\times7^4$.
   7. $P=\dfrac{7^4}{2\times7^4}=\dfrac12$.

   Common Mistakes

   • Assuming digits cannot repeat (the problem doesn't say "distinct digits"); if repetition were disallowed the counting would differ substantially.
   • Forgetting that first digit $=6$ and $=7$ give different thresholds for $d_5$ (4 vs 3 need to be exceeded), and averaging incorrectly.

   ✓Final answer

   The correct option is (B) — $\dfrac{1}{2}$.

   ANSWER: B

9. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.Let A be the set of 5 distinct prime numbers. S be the set of all possible products of two or more distinct elements of A. If one element $\alpha$ and another element $\beta$ are selected at random from A and S respectively, then the probability that $\beta$ is divisible by $\alpha$ is (A) $\dfrac{15}{26}$ (B) $\dfrac{25}{36}$ (C) $\dfrac{5}{8}$ (D) $\dfrac{12}{17}$

   ›Reveal solutionSolution

   This tests counting subsets of size $\ge2$ that contain a fixed element, out of all subsets of size $\ge2$ from a 5-element set, using the symmetry that this count is the same for every choice of the fixed element.

   Concept and Intuition

   $S$ consists of products of two-or-more of the 5 distinct primes — equivalently, $S$ is in bijection with subsets of $A$ of size $\ge2$ (each such subset gives one product, and all products are distinct since the primes are distinct). "$\beta$ divisible by $\alpha$" means the subset defining $\beta$ must include the prime $\alpha$. Because every prime plays a symmetric role, the count of qualifying subsets doesn't depend on which specific $\alpha$ was drawn — this symmetry is what makes the probability well-defined and easy to compute.

   Step-by-Step Solution

   1. Total subsets of $A$ (5 primes) of size $\ge2$: $\binom52+\binom53+\binom54+\binom55=10+10+5+1=26=|S|$.
   2. Fix any $\alpha\in A$. Subsets of size $\ge2$ containing $\alpha$ correspond to choosing at least $1$ more element from the remaining $4$ primes (then adjoining $\alpha$): $\binom41+\binom42+\binom43+\binom44=4+6+4+1=15$.
   3. This count, $15$, is the same regardless of which $\alpha$ was selected (full symmetry among the 5 primes), so the conditional count doesn't depend on the random choice of $\alpha$.
   4. Hence $P(\alpha\mid\beta)=\dfrac{15}{26}$ regardless of which prime is drawn as $\alpha$, so the overall probability is $\dfrac{15}{26}$.

   Common Mistakes

   • Forgetting that $S$ only includes products of two or more primes (not the primes themselves), so $|S|=26$, not $2^5-1=31$ or $2^5=32$.
   • Trying to average over specific primes with different probabilities instead of recognizing the symmetry that makes every prime give the same count.

   ✓Final answer

   The correct option is (A) — $\dfrac{15}{26}$.

   ANSWER: A

10. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
    Q.From the set of numbers $\{1,2,3,4,5,6,7,8,9,10,11,12\}$, two numbers are selected at random. The probability that the two numbers selected differ by a prime number is (A) $\dfrac{16}{33}$ (B) $\dfrac{1}{11}$ (C) $\dfrac{3}{11}$ (D) $\dfrac{11}{24}$

    ›Reveal solutionSolution

    Count pairs from $\{1,\dots,12\}$ whose difference is prime by summing, over each prime gap $d$, the $12-d$ pairs that achieve it; the probability comes out to $\dfrac{16}{33}$.

    Concept and Intuition

    For two numbers chosen from $1$ to $n$, the count of unordered pairs with a fixed difference $d$ (where $1\le d\le n-1$) is exactly $n-d$ — because the smaller number can be any of $1,2,\dots,n-d$ while the larger is fixed as (smaller)$+d$. Summing this count over all prime values of $d$ in range gives the total favourable pairs, without needing to enumerate every pair individually.

    Step-by-Step Solution

    1. Total ways to select 2 numbers out of 12: $\binom{12}{2}=\dfrac{12\times11}{2}=66$.
    2. Possible differences range from $1$ to $11$. The primes in this range are $2,3,5,7,11$.
    3. For difference $d$, the number of pairs is $12-d$:
       • $d=2$: $12-2=10$ pairs
       • $d=3$: $12-3=9$ pairs
       • $d=5$: $12-5=7$ pairs
       • $d=7$: $12-7=5$ pairs
       • $d=11$: $12-11=1$ pair
    4. Total favourable pairs $=10+9+7+5+1=32$.
    5. Probability $=\dfrac{32}{66}=\dfrac{16}{33}$.

    Common Mistakes

    • Forgetting that $1$ is not prime, or mistakenly including $d=1$ or $d=4,6,8,9,10$ (composite/non-prime differences) in the count.
    • Miscounting the number of pairs for a given difference $d$ (it is $12-d$, not $12$ or $d$ itself).

    ✓Final answer

    The correct option is (A) $\dfrac{16}{33}$.

    ANSWER: A

11. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
    Q.5 letters are randomly selected from English alphabets and they are arranged in alphabetical order. The probability that the 5 letters selected and arranged has M in the middle place is (A) $\dfrac{9}{115}$ (B) $\dfrac{^{12}C_2 + \,^{13}C_2}{^{26}C_5}$ (C) $\dfrac{^{25}C_4}{^{26}C_5}$ (D) $\dfrac{9}{125}$

    ›Reveal solutionSolution

    Combinatorics/probability: M must be the 3rd smallest of 5 chosen letters; answer simplifies to $\frac{9}{115}$.

    Concept and Intuition

    Because the 5 selected letters are always written in alphabetical order, each subset of 5 letters corresponds to exactly one arrangement — so the sample space is simply the number of 5-letter subsets, $^{26}C_5$. For M (the 13th letter of the alphabet) to occupy the middle (3rd) position after sorting, exactly 2 of the other 4 letters must come alphabetically before M and exactly 2 must come after M.

    Step-by-Step Solution

    1. Total ways to choose 5 letters from 26: $^{26}C_5 = 65780$.
    2. Letters before M (A–L): 12 letters. Letters after M (N–Z): 13 letters.
    3. Favorable selections: choose 2 from the 12 before M and 2 from the 13 after M (M itself is fixed as chosen): $^{12}C_2\cdot{}^{13}C_2 = 66\times 78 = 5148$.
    4. Probability $=\dfrac{5148}{65780}$. Simplify: $65780 = 2^2\cdot5\cdot11\cdot13\cdot23$ and $5148 = 2^2\cdot3^2\cdot11\cdot13$, so the ratio reduces to $\dfrac{9}{5\cdot23} = \dfrac{9}{115}$.

    Common Mistakes

    • Treating the sample space as $26^5$ or $P(26,5)$ (permutations) instead of $^{26}C_5$, forgetting the fixed alphabetical order collapses each subset to one arrangement.
    • Mis-splitting letters before/after M as 13/12 instead of 12/13 (M is the 13th letter, so 12 letters precede it, 13 follow).

    ✓Final answer

    The correct option is (A) — $\dfrac{9}{115}$.

    ANSWER: A

12. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
    Q.From a group of 10 men and 5 women, a four member committee which includes atleast one woman is to be formed. Then the probability for the committee thus formed to have more women than men, is (A) $\dfrac{3}{11}$ (B) $\dfrac{2}{23}$ (C) $\dfrac{1}{11}$ (D) $\dfrac{21}{220}$

    ›Reveal solutionSolution

    Counting favorable (women outnumber men, i.e. 3W-1M or 4W-0M) committees against all committees with at least one woman gives probability $\dfrac{1}{11}$.

    Concept and Intuition

    This is conditional counting: restrict the sample space to committees satisfying "at least one woman," then find what fraction of those also satisfy the stronger condition "more women than men."

    Step-by-Step Solution

    1. Total 4-member committees from 15 people: $\binom{15}{4}=1365$.
    2. Committees with zero women (all men): $\binom{10}{4}=210$.
    3. Committees with at least one woman: $1365-210=1155$ — this is the sample space (denominator).
    4. "More women than men" in a 4-member committee means women count $>2$, i.e. 3 women & 1 man, or 4 women & 0 men.
    5. 3W,1M: $\binom{5}{3}\binom{10}{1}=10\times10=100$.
    6. 4W,0M: $\binom{5}{4}\binom{10}{0}=5\times1=5$.
    7. Favorable total $=100+5=105$.
    8. Probability $=\dfrac{105}{1155}=\dfrac{1}{11}$.

    Common Mistakes

    • Using $\binom{15}{4}$ (all committees) as the denominator instead of the conditioned "at least one woman" space.
    • Missing the 4W-0M case when counting favorable outcomes.

    ✓Final answer

    The correct option is (C) — $\dfrac{1}{11}$.

    ANSWER: C