Q.A flashlight has $8$ batteries out of which $3$ are dead. If two batteries are selected without replacement and tested, the probability that both are dead is
(A) $\dfrac{33}{56}$
(B) $\dfrac{9}{64}$
(C) $\dfrac{1}{14}$
(D) $\dfrac{3}{28}$

Showing the 12 most recent of 15 on this concept.

1. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.Urn A contains 6 white and 2 black balls; urn B contains 5 white and 3 black balls and urn C contains 4 white and 4 black balls. If an urn is chosen at random and a ball is drawn at random from it, then the probability that the ball drawn is white is (A) $\dfrac{3}{8}$ (B) $\dfrac{5}{8}$ (C) $\dfrac{1}{2}$ (D) $\dfrac{3}{4}$

   ›Reveal solutionSolution

   The problem is a law of total probability scenario: pick an urn uniformly at random, then draw a ball. The overall probability of white is the average of the three urn-specific white probabilities, which works out to $\frac{15}{24} = \frac{5}{8}$.

   We have three urns, each equally likely to be chosen. Within each urn, the chance of drawing a white ball is simply the fraction of white balls in that urn. Because the urn is chosen first (at random) and then the ball is drawn, the total probability of white is the weighted average of the individual white probabilities, with weights equal to the probability of picking each urn.

   Step-by-step reasoning

   1. Identify the probabilities of choosing each urn.

      Since an urn is chosen at random from three urns, each has probability $\frac{1}{3}$.

   2. Find the probability of drawing a white ball from each urn.

      • Urn A: 6 white, 2 black → total 8 balls. So $P(\text{white} \mid A) = \frac{6}{8} = \frac{3}{4}$.
      • Urn B: 5 white, 3 black → total 8 balls. So $P(\text{white} \mid B) = \frac{5}{8}$.
      • Urn C: 4 white, 4 black → total 8 balls. So $P(\text{white} \mid C) = \frac{4}{8} = \frac{1}{2}$.

   3. Apply the law of total probability.

      The overall probability of white is:

   $P(\text{white}) = P(A)\cdot P(\text{white}\mid A) + P(B)\cdot P(\text{white}\mid B) + P(C)\cdot P(\text{white}\mid C)$

   Substitute the values:

   $P(\text{white}) = \frac{1}{3}\cdot\frac{3}{4} + \frac{1}{3}\cdot\frac{5}{8} + \frac{1}{3}\cdot\frac{1}{2}$

   4. Simplify each term.

      • First term: $\frac{1}{3}\cdot\frac{3}{4} = \frac{3}{12} = \frac{1}{4}$
      • Second term: $\frac{1}{3}\cdot\frac{5}{8} = \frac{5}{24}$
      • Third term: $\frac{1}{3}\cdot\frac{1}{2} = \frac{1}{6}$

   5. Add them with a common denominator.

      The least common multiple of 4, 24, and 6 is 24.

      $$$$

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2. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
   Q.If two cards are drawn at random simultaneously from a well shuffled pack of 52 playing cards, then the probability of getting a card having a composite number and a card having a number which is a multiple of 3 is (A) $\dfrac{94}{663}$ (B) $\dfrac{62}{663}$ (C) $\dfrac{102}{663}$ (D) $\dfrac{64}{663}$

   ›Reveal solutionSolution

   Composite-number cards: $20$; multiple-of-$3$ cards: $12$; cards that are both: $8$. Counting unordered pairs with one card of each type gives $204$ favourable out of $\binom{52}{2}=1326$, i.e. $\dfrac{102}{663}$ — option (C).

   Classify the ranks (numbers $1$–$10$).

   • Composite: $4,6,8,9,10$ → $5$ ranks → $20$ cards.
   • Multiple of $3$: $3,6,9$ → $3$ ranks → $12$ cards.
   • Both ($6,9$): $2$ ranks → $8$ cards.
   • Composite-only ($4,8,10$): $12$ cards; multiple-of-3-only ($3$): $4$ cards.

   Favourable pairs — one composite card and one multiple-of-3 card (two distinct cards). …

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3. TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQ
   Q.If two cards are drawn one after the other without replacement from a well-shuffled ordinary deck of 52 cards, then the probability that both of them are aces is (A) $\dfrac{1}{169}$ (B) $\dfrac{1}{221}$ (C) $\dfrac{168}{169}$ (D) $\dfrac{220}{221}$

   ›Reveal solutionSolution

   The probability of drawing two aces without replacement is the product of the probability of the first ace and the conditional probability of the second ace given the first. The answer is $\frac{1}{221}$.

   The key idea here is conditional probability without replacement. When you draw cards without putting the first one back, the deck changes — so the probability of the second event depends on what happened first. This is not the same as drawing with replacement, where each draw is independent.

   Many students mistakenly treat the draws as independent and simply square $\frac{4}{52}$, getting $\frac{1}{169}$ — but that’s wrong because the deck shrinks. Let’s walk through it correctly.

   1. First draw: There are 4 aces in a deck of 52 cards. So the probability that the first card is an ace is

   $P(\text{first ace}) = \frac{4}{52} = \frac{1}{13}.$

   2. Second draw (without replacement): If the first card was an ace, there are now only 3 aces left, and only 51 cards remaining in the deck. So the probability that the second card is also an ace, given that the first was an ace, is

   $P(\text{second ace} \mid \text{first ace}) = \frac{3}{51} = \frac{1}{17}.$

   3. Multiply the probabilities: The probability that both events happen is the product of the probability of the first and the conditional probability of the second: $P(\text{both aces}) = \frac{4}{52} \times \frac{3}{51} = \frac{1}{13} \times \frac{1}{17} = \frac{1}{221}.$ …

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4. TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
   Q.A bag contains four balls. Two balls are drawn randomly and found them to be white. The probability that all the balls in the bag are white is (A) $\dfrac{1}{2}$ (B) $\dfrac{3}{5}$ (C) $\dfrac{1}{4}$ (D) $\dfrac{2}{3}$

   ›Reveal solutionSolution

   Take the number of white balls $W\in\{0,1,2,3,4\}$ as equally likely a priori, update on the evidence "two drawn balls are white" via Bayes, and the posterior $P(W=4)=\tfrac{3}{5}$.

   Setup. A bag of $4$ balls; each is white or non-white. Let $W$ be the (unknown) count of white balls, with prior $P(W=w)=\tfrac{1}{5}$ for $w=0,1,2,3,4$. Two balls are drawn without replacement and both are white — call this event $E$.

   Step 1 — likelihoods. Total ways to choose $2$ of $4$ is $\binom{4}{2}=6$. Drawing two whites is impossible for $W<2$.

   $P(E\mid W=4)=\frac{\binom{4}{2}}{\binom{4}{2}}=1,\quad P(E\mid W=3)=\frac{\binom{3}{2}}{6}=\frac{1}{2},\quad P(E\mid W=2)=\frac{\binom{2}{2}}{6}=\frac{1}{6}.$

   Step 2 — total probability of the evidence. …

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5. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
   Q.A, B, C, D cut a pack of 52 well shuffled playing cards successively in the same order. If the person who cuts a spade first, wins the game and the game continues until this happens, then the probability that A wins the game is (A) $\dfrac{74}{175}$ (B) $\dfrac{44}{175}$ (C) $\dfrac{54}{175}$ (D) $\dfrac{64}{175}$

   ›Reveal solutionSolution

   The probability that A wins is the sum of an infinite geometric series where A wins on the 1st, 5th, 9th, … cut. The result is $\frac{64}{175}$, which corresponds to option (D).

   Concept and Intuition

   This is a classic “first success” problem with a twist: the players cut in a fixed order (A, then B, then C, then D, then back to A, and so on). The game stops as soon as someone cuts a spade. Since the pack is well-shuffled, each cut is independent with probability $\frac{13}{52} = \frac14$ of being a spade.

   We want the probability that A is the first to cut a spade. That means A must succeed on her turn, and all players before her in that round must have failed. Because the game can go through many full cycles, we sum over all possible rounds where A wins.

   Step-by-step reasoning

   1. Probability of a spade on any single cut There are 13 spades in a 52-card deck, so

   $p = \frac{13}{52} = \frac14, \quad q = 1-p = \frac34.$

   2. When does A win? A wins if she cuts a spade on her first turn (round 1), or if everyone fails in the first round, then everyone fails again in the second round, …, and then A succeeds on her turn in some later round.
      • Round 1: A cuts first. She wins immediately with probability $p = \frac14$.
      • Round 2: For A to win in round 2, all four players must fail in round 1 (probability $q^4$), then A must succeed on her turn in round 2 (probability $p$). So probability = $q^4 \cdot p$.
      • Round 3: All four fail in round 1, all four fail in round 2, then A succeeds in round 3: probability = $(q^4)^2 \cdot p$.
      • In general, A wins on round $k$ ($k = 1,2,3,\dots$) with probability

   $(q^4)^{k-1} \cdot p.$

   3. Sum over all rounds The total probability that A wins is the infinite geometric series: $P(A) = p + q^4 p + (q^4)^2 p + \cdots = p \sum_{k=0}^\infty (q^4)^k.$ …

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6. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
   Q.Bag ‘P’ contains 3 white, 2 red, 5 blue balls and bag ‘Q’ contains 2 white, 3 red, 5 blue balls. A ball is chosen at random from ‘P’ and is placed in ‘Q’. If a ball is chosen from bag ‘Q’ at random, then the probability that it is a red ball is (A) $\dfrac{9}{50}$ (B) $\dfrac{13}{45}$ (C) $\dfrac{16}{55}$ (D) $\dfrac{12}{35}$

   ›Reveal solutionSolution

   Transferring a ball from P into Q changes Q's composition, so apply the Law of Total Probability over the three possible transfer colours. The probability of then drawing red from Q is $\dfrac{16}{55}$, option (C).

   Setup. Bag P has 3 white, 2 red, 5 blue (10 balls). Bag Q has 2 white, 3 red, 5 blue (10 balls). One ball is moved from P to Q, so Q then holds $11$ balls, and its red count depends on the transferred colour.

   Transfer probabilities from P.

   $P(\text{white})=\tfrac{3}{10},\quad P(\text{red})=\tfrac{2}{10},\quad P(\text{blue})=\tfrac{5}{10}.$

   Probability of red from Q after each transfer (Q now has 11 balls):

   • White moved $\Rightarrow$ Q red $=3 \Rightarrow \tfrac{3}{11}$. …

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7. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.There are two boxes each containing 10 balls. In each box, few of them are black balls and rest are white. A ball is drawn at random from one of the boxes and found that it is black. If the probability that the black ball drawn is from the second box is $\dfrac{1}{5}$, then number of black balls in the first box is (A) 5 or 10 (B) 2 or 7 (C) 4 or 8 (D) 3 or 6 or 9

   ›Reveal solutionSolution

   We use Bayes’ theorem to relate the conditional probability that the black ball came from the second box to the unknown numbers of black balls. Solving the resulting equation gives two possible values for the number of black balls in the first box, which match one of the given options.

   Concept and intuition

   This is a classic Bayes’ theorem problem: we have two boxes, each with a certain number of black and white balls. A black ball is drawn, and we are told the probability it came from the second box is $\frac{1}{5}$. That means the black ball is more likely to have come from the first box. The key is to set up the conditional probability using Bayes’ rule, letting the unknown number of black balls in the first box be $x$ and in the second box be $y$. Since each box has 10 balls total, the number of white balls is $10-x$ and $10-y$ respectively. The probability of choosing either box is $\frac{1}{2}$. Then we solve for possible integer values of $x$ and $y$ that satisfy the given condition.

   Step-by-step solution

   1. Define variables

      Let the first box have $x$ black balls (and $10-x$ white).

      Let the second box have $y$ black balls (and $10-y$ white).

      Both $x$ and $y$ are integers from 0 to 10.

   2. Write the probability of drawing a black ball from each box

      • $P(\text{black} \mid \text{Box 1}) = \frac{x}{10}$
      • $P(\text{black} \mid \text{Box 2}) = \frac{y}{10}$
      • The probability of picking either box is $P(\text{Box 1}) = P(\text{Box 2}) = \frac{1}{2}$.

   3. Apply Bayes’ theorem

      We want $P(\text{Box 2} \mid \text{black}) = \frac{1}{5}$.

      Bayes’ theorem says:

   $P(\text{Box 2} \mid \text{black}) = \frac{P(\text{black} \mid \text{Box 2}) \cdot P(\text{Box 2})}{P(\text{black})}$

   The total probability of drawing a black ball is:

   $P(\text{black}) = P(\text{Box 1}) \cdot \frac{x}{10} + P(\text{Box 2}) \cdot \frac{y}{10} = \frac{1}{2} \cdot \frac{x}{10} + \frac{1}{2} \cdot \frac{y}{10} = \frac{x+y}{20}$

   4. Plug into Bayes’ formula

   $\frac{1}{5} = \frac{\frac{y}{10} \cdot \frac{1}{2}}{\frac{x+y}{20}} = \frac{\frac{y}{20}}{\frac{x+y}{20}} = \frac{y}{x+y}$

   So we have:

   $\frac{y}{x+y} = \frac{1}{5}$

   5. Solve the equation Cross-multiply:

   $5y = x + y \quad \Rightarrow \quad 4y = x \quad \Rightarrow \quad x = 4y$

   Since $x$ and $y$ are integers between 0 and 10, and $x = 4y$, the possible pairs are:

   • $y = 0$ → $x = 0$
   • $y = 1$ → $x = 4$
   • $y = 2$ → $x = 8$
   • $y = 3$ → $x = 12$ (invalid, since $x \le 10$) …

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8. TG EAPCET 2021Set eng-2021-08-05-AN1 markMCQ
   Q.From $3n$ consecutive integers three integers are selected at random. The probability that their sum is divisible by $3$ is (A) $\dfrac{3 \cdot nC_3 + n^2}{3nC_3}$ (B) $\dfrac{2 \cdot nC_3 + n^3}{3nC_3}$ (C) $\dfrac{3n^2 - 3n + 2}{(3n-1)(3n-2)}$ (D) $\dfrac{3n^2 - 3n + 2}{(3n-1)(3n+2)}$

   ›Reveal solutionSolution

   The key idea is to classify the $3n$ consecutive integers by their remainder modulo 3, then count favourable triples whose sum is divisible by 3. The probability simplifies to $\dfrac{3n^2 - 3n + 2}{(3n-1)(3n-2)}$, which matches option (C).

   Any set of $3n$ consecutive integers contains exactly $n$ numbers from each residue class modulo 3: $n$ numbers congruent to $0$, $n$ to $1$, and $n$ to $2$. This is because the residues cycle $0,1,2$ repeatedly, and with $3n$ terms the cycle completes exactly $n$ times.

   When we pick three integers, their sum is divisible by 3 precisely when the sum of their residues modulo 3 is $0$ modulo 3. So we need to count all triples (order doesn’t matter) whose residues add to a multiple of 3.

   1. List the residue patterns that work.

      The possible residue triples $(r_1, r_2, r_3)$ with $r_i \in \{0,1,2\}$ that sum to $0 \pmod{3}$ are:

      • $(0,0,0)$ — all three from the 0‑class.
      • $(1,1,1)$ — all three from the 1‑class.
      • $(2,2,2)$ — all three from the 2‑class.
      • $(0,1,2)$ — one from each class (in any order).

      No other combination works: e.g. $(0,0,1)$ sums to $1$, $(1,2,2)$ sums to $2$, etc.

   2. Count the number of favourable unordered triples.

      • For $(0,0,0)$: choose any 3 from the $n$ numbers in the 0‑class → $\binom{n}{3}$ ways.
      • For $(1,1,1)$: similarly $\binom{n}{3}$ ways.
      • For $(2,2,2)$: again $\binom{n}{3}$ ways.
      • For $(0,1,2)$: pick one from each class → $n \cdot n \cdot n = n^3$ ways.

      Total favourable outcomes = $3\binom{n}{3} + n^3$.

   3. Total number of ways to choose any 3 integers from $3n$.

      That is $\binom{3n}{3}$.

   4. Write the probability.

   $P = \frac{3\binom{n}{3} + n^3}{\binom{3n}{3}}$

   This matches option (A) in form, but we should simplify to see if it matches one of the other options too.

   5. Simplify the expression.

   $\binom{n}{3} = \frac{n(n-1)(n-2)}{6}, \quad \binom{3n}{3} = \frac{3n(3n-1)(3n-2)}{6}$

   So

   $P = \frac{3 \cdot \frac{n(n-1)(n-2)}{6} + n^3}{\frac{3n(3n-1)(3n-2)}{6}} = \frac{\frac{n(n-1)(n-2)}{2} + n^3}{\frac{3n(3n-1)(3n-2)}{6}}$

   Multiply numerator and denominator by 6:

   $P = \frac{3n(n-1)(n-2) + 6n^3}{3n(3n-1)(3n-2)}$

   Factor $n$ out of the numerator:

   $$$$

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9. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.There are two boxes each containing 10 balls. In each box, few of them are black balls and rest are white. A ball is drawn at random from one of the boxes and found that it is black. If the probability that the black ball drawn is from the second box is $\dfrac{1}{5}$, then number of black balls in the first box is (A) 3 or 6 or 9 (B) 5 or 10 (C) 4 or 8 (D) 2 or 7

   ›Reveal solutionSolution

   Use Bayes’ theorem to relate the conditional probability that the black ball came from the second box to the unknown numbers of black balls in each box. The condition leads to a Diophantine equation whose integer solutions give the possible black-ball counts in the first box.

   The problem gives two boxes, each with 10 balls. Let the first box contain $x$ black balls (and $10-x$ white), and the second box contain $y$ black balls (and $10-y$ white). A box is chosen at random (each with probability $\frac12$), then a ball is drawn from that box. The drawn ball is black. We are told that the probability that this black ball came from the second box is $\frac15$.

   This is a classic Bayes’ theorem situation: we have a prior probability of choosing each box, a likelihood of drawing a black ball from each box, and we want the posterior probability that the ball came from the second box given that it is black.

   1. Set up the probabilities. The probability of drawing a black ball from the first box is $\frac{x}{10}$, and from the second box is $\frac{y}{10}$. The prior probability of choosing either box is $\frac12$. By the law of total probability, the overall probability of drawing a black ball is

   $P(\text{black}) = \frac12 \cdot \frac{x}{10} + \frac12 \cdot \frac{y}{10} = \frac{x+y}{20}.$

   2. Apply Bayes’ theorem. The probability that the black ball came from the second box is

   $P(\text{box 2} \mid \text{black}) = \frac{P(\text{box 2}) \cdot P(\text{black} \mid \text{box 2})}{P(\text{black})} = \frac{\frac12 \cdot \frac{y}{10}}{\frac{x+y}{20}} = \frac{y/20}{(x+y)/20} = \frac{y}{x+y}.$

   The problem states this equals $\frac15$. So

   $\frac{y}{x+y} = \frac15.$

   3. Solve the equation. Cross-multiplying:

   $5y = x + y \quad\Rightarrow\quad 4y = x \quad\Rightarrow\quad x = 4y.$

   Since $x$ and $y$ are integers between 0 and 10 (inclusive), and each box has exactly 10 balls, $y$ can be 0, 1, 2, or 3 (because $x = 4y \le 10$). …

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10. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
    Q.If 2 cards drawn at random from a well shuffled pack of 52 playing cards are from the same suit, then the probability of getting a face card and a card having a prime number is (A) $\dfrac{8}{13}$ (B) $\dfrac{2}{13}$ (C) $\dfrac{8}{221}$ (D) $\dfrac{32}{221}$

    ›Reveal solutionSolution

    Conditioned on both cards being from the same suit, within each $13$-card suit there are $3$ face cards and $4$ prime-number cards. The favourable-to-total ratio is $\dfrac{4\times 12}{4\times\binom{13}{2}}=\dfrac{2}{13}$, option (B).

    Set up the conditional experiment

    Two cards are drawn and we are told they are from the same suit. A single suit has $13$ cards, so the conditioned sample space is

    $4\ \text{suits}\times\binom{13}{2} = 4\times 78 = 312\ \text{equally-likely same-suit pairs.}$

    Within one suit:

    • Face cards: J, Q, K $\Rightarrow 3$ cards.
    • Prime-number cards: $2,3,5,7$ are prime $\Rightarrow 4$ cards (Ace $=1$ is not prime).

    Count the favourable pairs …

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11. TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQ
    Q.Out of 40 consecutive integers two integers are drawn at random. The probability that their sum is odd is (A) $\dfrac{15}{29}$ (B) $\dfrac{20}{39}$ (C) $\dfrac{25}{29}$ (D) $\dfrac{40}{39}$

    ›Reveal solutionSolution

    The sum of two integers is odd only when one is even and the other is odd. In 40 consecutive integers, there are exactly 20 evens and 20 odds, so the probability is $\frac{20 \times 20}{\binom{40}{2}} = \frac{20}{39}$.

    The key idea here is a simple parity rule: odd + even = odd, while odd + odd and even + even both give even sums. So the problem reduces to counting how many ways we can pick one even and one odd number from the set.

    In any block of 40 consecutive integers, the count of evens and odds is perfectly balanced — 20 each. This is true whether the block starts with an even or an odd number, because 40 is even. For example, 1 to 40 has 20 odds (1,3,...,39) and 20 evens (2,4,...,40); 2 to 41 also has 20 evens (2,4,...,40) and 20 odds (3,5,...,41). So the composition is fixed.

    Now we draw two distinct integers at random. The total number of ways to choose any two from 40 is $\binom{40}{2} = \frac{40 \times 39}{2} = 780$.

    1. Count favourable outcomes: We need one even and one odd.

       Number of ways to pick one even from 20 evens: $20$ ways.

       Number of ways to pick one odd from 20 odds: $20$ ways.

       Since the two draws are independent (order doesn't matter in combination), the total favourable pairs = $20 \times 20 = 400$.

    2. Compute probability:

       $P(\text{sum odd}) = \frac{400}{780} = \frac{40}{78} = \frac{20}{39}.$ …

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12. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
    Q.Among the 5 married couples, if the names of 5 men are matched with the names of their wives randomly, then the probability that no man is matched with name of his wife is (A) $\dfrac{9}{20}$ (B) $\dfrac{1}{5}$ (C) $\dfrac{11}{30}$ (D) $\dfrac{17}{60}$

    ›Reveal solutionSolution

    This is a classic derangement problem for 5 items. The probability that no man is matched with his own wife is the number of derangements of 5 items divided by 5! = 120, which equals 44/120 = 11/30. The correct option is (C).

    Concept & Intuition

    We have 5 men and their 5 wives. A random matching is a permutation of the wives' names assigned to the men. We want the probability that no man gets his own wife — i.e., the permutation has no fixed points. Such permutations are called derangements. The key idea: count all permutations (5! = 120), then count those with zero fixed points using inclusion–exclusion. The formula for derangements of $n$ items is

    $!n = n! \sum_{k=0}^{n} \frac{(-1)^k}{k!}.$

    For $n=5$, this gives the exact count.

    Step-by-step solution

    1. Total number of matchings There are 5 men and 5 wives. A matching is a bijection from men to wives — a permutation of the 5 wives. Total permutations:

    $5! = 120.$

    2. Define the event we want

       Let $A_i$ be the event that the $i$-th man is matched with his own wife. We want the probability that none of $A_1, A_2, \dots, A_5$ occur. That is, the number of permutations with no fixed points (derangements) divided by 120.

    3. Use inclusion–exclusion to count derangements

       Count permutations where at least one man gets his own wife, then subtract from total.

       • Number of permutations where a specific set of $k$ men get their own wives: the remaining $5-k$ wives can be arranged arbitrarily among the remaining $5-k$ men, so $(5-k)!$ ways.
       • There are $\binom{5}{k}$ ways to choose which $k$ men are fixed.

       By inclusion–exclusion, the number of permutations with at least one fixed point is:

    $\sum_{k=1}^{5} (-1)^{k+1} \binom{5}{k} (5-k)!.$

    So the number with no fixed points (derangements) is:

    $!5 = 5! - \sum_{k=1}^{5} (-1)^{k+1} \binom{5}{k} (5-k)!.$

    Equivalently, the standard derangement formula:

    $!5 = 5! \sum_{k=0}^{5} \frac{(-1)^k}{k!}.$

    4. Compute the sum

    $\sum_{k=0}^{5} \frac{(-1)^k}{k!} = 1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120}.$

    Compute step by step: …

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