Q.A person has undertaken a construction job. The probabilities are 0.65 that there will be a strike, 0.80 that the construction job will be completed on time if there is no strike, and 0.32 that the construction job will be completed on time if there is a strike. Determine the probability that the construction job will be completed on time.

Showing the 12 most recent of 69 on this concept.

1. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
   Q.2 aero planes I and II bond a target in succession. The probabilities of I and II scoring a hit correctly is 0.3 and 0.2 respectively. The second plane will bomb only if first misses the target. The probability that the target is hit by the $2^{nd}$ plane is (A) 0.06 (B) 0.14 (C) 0.32 (D) 0.7

   ›Reveal solutionSolution

   The 2nd plane gets a chance only after the 1st fails, so $P = 0.7\times 0.2 = 0.14$. Answer: (B).

   Concept and Intuition

   This is a sequential (conditional) experiment: the second trial happens only when the first fails. The event 'the target is hit by the 2nd plane' is therefore a compound event —

   $\{\text{I misses}\}\cap\{\text{II hits}\}$

   and because the two planes' performances are independent, the probability of the intersection is the product of the probabilities.

   A useful picture is a probability tree:

            ┌── I hits (0.3)  ────────────────► target hit by plane I  (0.3)
   Start ───┤
            └── I misses (0.7) ─┬── II hits (0.2) ──► hit by plane II  (0.7 × 0.2 = 0.14)
                                └── II misses (0.8) ► target not hit   (0.7 × 0.8 = 0.56)
   

   The three leaves sum to $0.3+0.14+0.56=1$ ✓ — a good check that the model is complete.

   Step-by-Step Solution

   1. Let $H_1$ = plane I hits, with $P(H_1)=0.3$, so $P(\overline{H_1})=1-0.3=0.7$.
   2. Let $H_2$ = plane II hits (given it bombs), with $P(H_2)=0.2$.
   3. Plane II bombs only if plane I missed. Hence

   $P(\text{target hit by 2nd plane}) = P(\overline{H_1}\cap H_2) = P(\overline{H_1})\,P(H_2)$

   4. Substitute:

   $= 0.7\times 0.2 = 0.14$

   5. ⇒ option (B).

   Common Mistakes

   • Answering $0.2$ — that is the conditional probability $P(H_2\mid \overline{H_1})$, i.e. the chance the 2nd plane hits given it actually bombs, not the unconditional probability asked for.
   • Answering $0.3\times0.2=0.06$ (distractor A) — that would be 'both hit', but plane II never bombs if plane I has already hit.
   • Answering $0.3+0.2-0.06=0.44$ or the 'at least one hit' value — the question asks specifically for a hit by the second plane.

   ✓Final answer

   The correct option is (B) — 0.14.

   ANSWER: B

2. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.For a biased die, the probabilities for different faces to turn up are given below

   Face	1	2	3	4	5	6
   Probability	0.1	0.32	0.21	0.15	0.05	0.17

   The die is tossed and you are told that either face 1 or 2 has turned up. Then the probability that it is face 1 is (A) $\dfrac{10}{33}$ (B) $\dfrac{5}{21}$ (C) $\dfrac{8}{21}$ (D) $\dfrac{1}{42}$

   ›Reveal solutionSolution

   Conditioning on "face 1 or face 2" just means renormalizing the two individual probabilities so they add to 1.

   Concept and Intuition

   $P(\text{face 1}\mid\text{face 1 or 2})=\dfrac{P(\text{face 1})}{P(\text{face 1})+P(\text{face 2})}$, since these two events are mutually exclusive and their union is the conditioning event.

   Step-by-Step Solution

   1. $P(1)=0.1$, $P(2)=0.32$.
   2. $P(1\text{ or }2)=0.1+0.32=0.42$.
   3. $P(1\mid1\text{ or }2)=\dfrac{0.1}{0.42}=\dfrac{10}{42}=\dfrac{5}{21}$.

   Common Mistakes

   • Dividing by the total probability $1$ instead of the restricted event's probability $0.42$.
   • Arithmetic slip simplifying $10/42$ (should reduce to $5/21$).

   ✓Final answer

   The correct option is (B) — $\dfrac{5}{21}$.

   ANSWER: B

3. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.The probability that a person goes to college by car is $\dfrac{1}{5}$; by bus $\dfrac{2}{5}$ and by train is $\dfrac{3}{5}$ respectively. The probabilities that he reaches the college late if he takes car, bus, train are $\dfrac{2}{7}, \dfrac{4}{7}$ and $\dfrac{1}{7}$ respectively. If he reaches the college in time, the probability that he travelled by car is (A) $\dfrac{6}{29}$ (B) $\dfrac{24}{29}$ (C) $\dfrac{5}{29}$ (D) $\dfrac{23}{29}$

   ›Reveal solutionSolution

   Bayes' theorem: weight each mode's "on-time" probability by its usage probability, then take the car-share of the total.

   Concept and Intuition

   This is a direct application of Bayes' theorem / total probability: to find $P(\text{cause}\mid\text{effect})$, build the denominator as the weighted sum over every possible cause of reaching the observed effect (here, arriving on time), then take the numerator's share of it.

   Step-by-Step Solution

   1. On-time probabilities per mode: car $1-\tfrac27=\tfrac57$; bus $1-\tfrac47=\tfrac37$; train $1-\tfrac17=\tfrac67$.
   2. Total on-time probability: $P(\text{ontime})=\tfrac15\cdot\tfrac57+\tfrac25\cdot\tfrac37+\tfrac35\cdot\tfrac67=\tfrac{5}{35}+\tfrac{6}{35}+\tfrac{18}{35}=\tfrac{29}{35}$.
   3. Joint probability of taking the car AND being on time: $\tfrac15\cdot\tfrac57=\tfrac{5}{35}$.
   4. Conditional probability: $P(\text{car}\mid\text{ontime})=\dfrac{5/35}{29/35}=\dfrac{5}{29}$.

   Common Mistakes

   • Trying to compute $P(\text{late})$ then subtracting from 1 using an assumed normalized partition — here the weighted-sum route for $P(\text{ontime})$ directly is the one that matches the exam's intended numbers.
   • Dividing by the wrong denominator (using $P(\text{late})$ instead of $P(\text{ontime})$).

   ✓Final answer

   The correct option is (C) — $\dfrac{5}{29}$.

   ANSWER: C

4. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
   Q.Two persons P and Q are considering to apply for a job. The probability that P applies for the job is 1/4, the probability that P applies for the job given that Q applies for the job is 1/2, and the probability that Q applies for the job given that P applies for the job is 1/3. Then the probability that P does not apply for the job given that Q does not apply for the job is (A) $4/5$ (B) $5/6$ (C) $7/8$ (D) $11/12$

   ›Reveal solutionSolution

   Chain the given conditional probabilities to find $P(Q)$ and $P(P\cap Q)$, then use the complement rule — the answer is (A) $4/5$.

   Concept and Intuition

   Conditional probability definitions let us cross-multiply to recover the joint probability $P(P\cap Q)$ from either conditional. Once $P(P), P(Q), P(P\cap Q)$ are all known, De Morgan's law converts "neither event" into the complement of the union.

   Step-by-Step Solution

   1. Given: $P(P)=\tfrac14$, $P(P|Q)=\tfrac12$, $P(Q|P)=\tfrac13$.
   2. $P(P\cap Q) = P(Q|P)\cdot P(P) = \tfrac13\times\tfrac14 = \tfrac1{12}$.
   3. Also $P(P\cap Q) = P(P|Q)\cdot P(Q) \Rightarrow \tfrac1{12} = \tfrac12\cdot P(Q) \Rightarrow P(Q)=\tfrac16$.
   4. $P(P\cup Q) = P(P)+P(Q)-P(P\cap Q) = \tfrac14+\tfrac16-\tfrac1{12} = \tfrac{3}{12}+\tfrac{2}{12}-\tfrac{1}{12}=\tfrac{4}{12}=\tfrac13$.
   5. By De Morgan's law, "neither P nor Q applies" is the complement of $P\cup Q$: $P(\overline P\cap\overline Q) = 1-\tfrac13=\tfrac23$.
   6. $P(\overline Q) = 1-P(Q) = 1-\tfrac16=\tfrac56$.
   7. $P(\overline P|\overline Q) = \dfrac{P(\overline P\cap\overline Q)}{P(\overline Q)} = \dfrac{2/3}{5/6} = \dfrac23\times\dfrac65 = \dfrac{12}{15}=\dfrac45$.

   Common Mistakes

   • Mixing up which conditional probability to use for computing the joint probability first (must use $P(Q|P)P(P)$, not $P(P|Q)P(P)$, to directly match given data).
   • Forgetting to convert to complements at the final step (computing $P(P|Q)$-type instead of $P(\overline P|\overline Q)$).

   ✓Final answer

   The correct option is (A) — $4/5$.

   ANSWER: A

5. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.An item is tested on a device for its defectiveness. The probability that such an item is defective is 0.3. The device gives accurate result in 8 out of 10 such tests. If the device reports that an item tested is not defective, then the probability that it is actually defective is (A) $\dfrac{2}{15}$ (B) $\dfrac{3}{29}$ (C) $\dfrac{3}{31}$ (D) $\dfrac{4}{51}$

   ›Reveal solutionSolution

   A direct Bayes'-theorem inversion problem; the item is actually defective with probability $\dfrac{3}{31}$ given a "not defective" report — option (C).

   Concept and Intuition

   The device's report can be wrong. To find the true state given an observed (possibly wrong) report, we must weigh both ways the report could have arisen: a genuinely non-defective item correctly reported as such, or a genuinely defective item incorrectly reported as non-defective. Bayes' theorem combines these.

   Step-by-Step Solution

   1. Let $D$: item defective, $P(D)=0.3$; $D'$: item not defective, $P(D')=0.7$.
   2. Device accuracy $=0.8$, so it errs with probability $0.2$.
   3. "Reports not defective" while the item is defective means the device made an error: $P(\text{report ND}\mid D)=0.2$.
   4. "Reports not defective" while the item is not defective means the device was accurate: $P(\text{report ND}\mid D')=0.8$.
   5. By the total probability rule: $P(\text{report ND}) = 0.3(0.2)+0.7(0.8) = 0.06+0.56=0.62$.
   6. By Bayes' theorem: $P(D\mid\text{report ND})=\dfrac{0.3\times0.2}{0.62}=\dfrac{0.06}{0.62}=\dfrac{6}{62}=\dfrac{3}{31}$.

   Common Mistakes

   • Swapping which conditional (accurate vs. inaccurate) applies to which true state.
   • Forgetting to normalize by the total probability of the observed report (just using the numerator).

   ✓Final answer

   The correct option is (C) — $\dfrac{3}{31}$.

   ANSWER: C

6. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.In a bolt factory, machines A, B, C manufacture 25%, 35%, 40% of the total output respectively. There is a chance of having 5%, 4%, 2% defective bolts manufactured by A, B, C respectively. If a bolt is drawn at random from the output, then the probability that it is defective is (A) $\dfrac{69}{2000}$ (B) $\dfrac{59}{2000}$ (C) $\dfrac{79}{2000}$ (D) $\dfrac{89}{2000}$

   ›Reveal solutionSolution

   Classic total-probability ("law of total probability") setup: weight each machine's defect rate by its share of output and sum. The answer is (A).

   Concept and Intuition

   When an item can come from several mutually exclusive, exhaustive sources (here, machines A, B, C), and each source has its own conditional probability of producing a defect, the overall probability of a defect is the weighted average:

   $P(D) = \sum_i P(\text{source}_i)\,P(D\mid \text{source}_i).$

   This is the law of total probability — it's the natural way to combine "how much each machine contributes" with "how likely each machine's own output is defective."

   Step-by-Step Solution

   1. Let $A, B, C$ denote drawing a bolt from each machine: $P(A)=0.25$, $P(B)=0.35$, $P(C)=0.40$ (these sum to 1, as they must).
   2. Conditional defect probabilities: $P(D\mid A)=0.05$, $P(D\mid B)=0.04$, $P(D\mid C)=0.02$.
   3. By the law of total probability:

   $P(D) = P(A)P(D\mid A) + P(B)P(D\mid B) + P(C)P(D\mid C).$

   4. Compute each term: $0.25\times0.05=0.0125$; $0.35\times0.04=0.014$; $0.40\times0.02=0.008$.
   5. Sum: $0.0125+0.014+0.008 = 0.0345$.
   6. Express as a fraction: $0.0345 = \dfrac{345}{10000} = \dfrac{69}{2000}$ (dividing numerator and denominator by 5).

   Common Mistakes

   • Averaging the three defect rates directly ($\frac{0.05+0.04+0.02}{3}$) instead of weighting by each machine's output share — that ignores that the machines don't contribute equally.
   • Arithmetic slips when reducing the decimal to the given fractional form.

   ✓Final answer

   The correct option is (A) — $\dfrac{69}{2000}$.

   ANSWER: A

7. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.Two cards are drawn at random from a pack of 52 playing cards. If both the cards drawn are found to be black in colour, then the probability that atleast one of them is a face card is (A) $\dfrac{3}{13}$ (B) $\dfrac{3}{5}$ (C) $\dfrac{9}{65}$ (D) $\dfrac{27}{65}$

   ›Reveal solutionSolution

   This is a conditional probability restricted to the 26 black cards. Using the complement (no face card among the two) is the fastest route, giving $27/65$.

   Concept and Intuition

   Once we're told both drawn cards are black, the sample space shrinks to just the 26 black cards (13 spades + 13 clubs). Among these, 6 are face cards (J, Q, K of spades and clubs) and 20 are non-face cards. "At least one face card" is easiest via the complement: $1-P(\text{no face card})$.

   Step-by-Step Solution

   1. Black cards $=26$; black face cards $=6$ (J,Q,K $\times$ 2 suits); black non-face cards $=20$.
   2. Total ways to pick 2 from the 26 black cards: $\binom{26}{2}=325$.
   3. Ways with no face card (both from the 20 non-face black cards): $\binom{20}{2}=190$.
   4. $P(\text{no face card})=\dfrac{190}{325}=\dfrac{38}{65}$.
   5. $P(\text{at least one face card})=1-\dfrac{38}{65}=\dfrac{27}{65}$.

   Common Mistakes

   • Forgetting to restrict the sample space to black cards only (using all 52 cards instead of 26) — the conditioning on "both black" changes the base.
   • Miscounting black face cards as 12 (all face cards) instead of 6 (only the black-suited ones).

   ✓Final answer

   The correct option is (D) — $\dfrac{27}{65}$.

   ANSWER: D

8. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.A, B, C are mutually exclusive and exhaustive events of a random experiment and E is an event that occurs in conjunction with one of the events A, B, C. The conditional Probabilities of E given the happening of A, B, C are respectively 0.6, 0.3 and 0.1. If $P(A) = 0.30$ and $P(B) = 0.50$, then $P(C|E) =$ (A) $\dfrac{2}{35}$ (B) $\dfrac{15}{35}$ (C) $\dfrac{18}{35}$ (D) $\dfrac{17}{35}$

   ›Reveal solutionSolution

   This is a direct Bayes' theorem application over three mutually exclusive, exhaustive causes; $P(C|E)=2/35$.

   Concept and Intuition

   When an event $E$ can occur alongside any of several mutually exclusive, exhaustive causes $A,B,C$, Bayes' theorem lets us find the probability of a particular cause given that $E$ has occurred, by weighing each cause's prior probability by how likely $E$ is under it, then normalizing.

   Step-by-Step Solution

   1. Since $A,B,C$ are mutually exclusive and exhaustive, $P(C)=1-P(A)-P(B)=1-0.30-0.50=0.20$.
   2. Total probability of $E$: $P(E)=P(E|A)P(A)+P(E|B)P(B)+P(E|C)P(C)$

   $=0.6(0.30)+0.3(0.50)+0.1(0.20)=0.18+0.15+0.02=0.35$

   3. By Bayes' theorem: $P(C|E)=\dfrac{P(E|C)\,P(C)}{P(E)}=\dfrac{0.1\times0.2}{0.35}=\dfrac{0.02}{0.35}$.
   4. Simplify: $\dfrac{0.02}{0.35}=\dfrac{2}{35}$.

   Common Mistakes

   • Forgetting to first compute $P(C)$ from the exhaustiveness condition.
   • Using $P(E)$ instead of $P(E|C)P(C)$ in the numerator (or vice versa).

   ✓Final answer

   The correct option is (A) — $\dfrac2{35}$.

   ANSWER: A

9. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.A box P contains 3 white and 7 red balls. A bag Q contains 4 green and 5 blue balls. Two balls are randomly drawn from box P. If both are of same color, one ball is drawn from bag Q and if the two balls are of different color, 2 balls are drawn from the bag Q. If it is known that there is exactly one green ball among the ball or balls drawn from bag Q, then the probability that the two balls drawn from box P are of different colors is (A) $\dfrac{35}{67}$ (B) $\dfrac{21}{62}$ (C) $\dfrac{20}{43}$ (D) $\dfrac{32}{67}$

   ›Reveal solutionSolution

   A two-stage Bayes' theorem problem; conditioning on "exactly one green ball from Q" gives $P(\text{different colors from P}) = \frac{35}{67}$.

   Concept and Intuition

   This is a compound experiment: the outcome in box P (same-color vs different-color) determines how many balls are drawn from bag Q, and hence changes the probability model for "exactly one green." We must compute, for each P-outcome, the probability of observing exactly one green from Q, then combine via Bayes' theorem using the P-outcome's prior probability.

   Step-by-Step Solution

   1. Box P has 3 white + 7 red = 10 balls. $P(\text{same color}) = \dfrac{{}^3C_2+{}^7C_2}{{}^{10}C_2} = \dfrac{3+21}{45}=\dfrac{24}{45}=\dfrac{8}{15}$. $P(\text{different colors}) = 1-\dfrac{8}{15}=\dfrac{7}{15}$ (check: $\dfrac{{}^3C_1{}^7C_1}{{}^{10}C_2}=\dfrac{21}{45}=\dfrac{7}{15}$ ✓).
   2. If same color (S): draw 1 ball from Q (4 green, 5 blue, 9 total). "Exactly one green" among 1 ball drawn just means that ball is green: $P(\text{1 green}\mid S)=\dfrac{4}{9}$.
   3. If different colors (D): draw 2 balls from Q. $P(\text{exactly one green}\mid D) = \dfrac{{}^4C_1\cdot{}^5C_1}{{}^9C_2} = \dfrac{20}{36}=\dfrac{5}{9}$.
   4. By Bayes' theorem:

   $P(D\mid \text{one green}) = \dfrac{P(D)P(\text{one green}\mid D)}{P(S)P(\text{one green}\mid S)+P(D)P(\text{one green}\mid D)} = \dfrac{\frac{7}{15}\cdot\frac{5}{9}}{\frac{8}{15}\cdot\frac{4}{9}+\frac{7}{15}\cdot\frac{5}{9}}$

   5. Numerator $=\dfrac{35}{135}$; denominator $=\dfrac{32}{135}+\dfrac{35}{135}=\dfrac{67}{135}$. So $P(D\mid\text{one green}) = \dfrac{35}{67}$.

   Common Mistakes

   • Using the same probability model for "one green" regardless of whether 1 or 2 balls were drawn from Q — the two cases have genuinely different sample spaces.
   • Forgetting to weight each conditional probability by the correct prior $P(S)$ or $P(D)$ before combining.

   ✓Final answer

   The correct option is (A) — $\dfrac{35}{67}$.

   ANSWER: A

10. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
    Q.Bag $B_1$ contains 4 white and 2 black balls. Bag $B_2$ contains 3 white and 4 black balls. A bag is chosen at random and a ball is drawn from it at random, then the probability that the ball drawn is white, is (A) $\dfrac{1}{42}$ (B) $\dfrac{42}{32}$ (C) $\dfrac{33}{42}$ (D) $\dfrac{23}{42}$

    ›Reveal solutionSolution

    Total-probability rule over the two equally likely bags gives $P(\text{white})=\frac12\cdot\frac46+\frac12\cdot\frac37=\frac{23}{42}$.

    Concept and Intuition

    The ball drawn depends on which bag was chosen first. Since the bag choice is random with $P(B_1)=P(B_2)=\frac12$, and the draw is conditionally independent given the bag, the Law of Total Probability adds the two conditional probabilities weighted by how likely each bag is.

    Step-by-Step Solution

    1. $B_1$: 4 white, 2 black out of 6 $\Rightarrow P(\text{white}\mid B_1)=\frac{4}{6}=\frac23$.
    2. $B_2$: 3 white, 4 black out of 7 $\Rightarrow P(\text{white}\mid B_2)=\frac37$.
    3. $P(\text{white})=\frac12\cdot\frac23+\frac12\cdot\frac37=\frac13+\frac{3}{14}$.
    4. Common denominator 42: $\frac13=\frac{14}{42}$, $\frac{3}{14}=\frac{9}{42}$, sum $=\frac{23}{42}$.

    Common Mistakes

    • Simplifying $\frac46$ incorrectly or forgetting to weight by $\frac12$ for each bag.
    • Adding numerators/denominators directly instead of finding a common denominator (a classic fraction-addition slip).

    ✓Final answer

    The correct option is (D) — $\dfrac{23}{42}$.

    ANSWER: D

11. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
    Q.In a class consisting of 40 boys and 30 girls, 30% of the boys and 40% of the girls are good at Mathematics. If a student selected at random from that class is found to be a girl, then the probability that she is not good at Mathematics is (A) $\dfrac{3}{5}$ (B) $\dfrac{2}{5}$ (C) $\dfrac{3}{10}$ (D) $\dfrac{7}{10}$

    ›Reveal solutionSolution

    Conditioning on "the student is a girl" restricts the sample space to the 30 girls only; the answer is $\frac35$.

    Concept and Intuition

    This is conditional probability with a twist: the condition ("selected student is a girl") is given as a fact, not something to be computed via Bayes' theorem. Once we know the student is a girl, the boys' statistics become irrelevant — we simply work within the group of girls.

    Step-by-Step Solution

    1. Total girls $=30$.
    2. Girls good at Mathematics $= 40\%$ of $30 = 12$.
    3. Girls not good at Mathematics $= 30-12=18$.
    4. Since we are told the selected student is a girl, the relevant sample space is just these 30 girls.
    5. $P(\text{not good at maths} \mid \text{girl}) = \frac{18}{30} = \frac{3}{5}.$

    Common Mistakes

    • Trying to apply Bayes' theorem to find $P(\text{girl}\mid\text{not good})$ instead of the (simpler) quantity actually asked, $P(\text{not good}\mid\text{girl})$.
    • Mixing in the boys' 30% figure, which plays no role once we're told the student is a girl.

    ✓Final answer

    The correct option is (A) — $\dfrac{3}{5}$.

    ANSWER: A

12. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
    Q.A bag contains 4 red and 3 black balls. A second bag contains 2 red and 3 black balls. One bag is selected at random. If from the selected bag, one ball is drawn at random, then the probability that the ball drawn is red is (A) $\frac{39}{70}$ (B) $\frac{41}{70}$ (C) $\frac{29}{70}$ (D) $\frac{17}{35}$

    ›Reveal solutionSolution

    Total probability theorem over the two equally-likely bags gives $\frac{17}{35}$.

    Concept and Intuition

    Since the bag is chosen at random (each with probability $\frac12$), the overall probability of drawing red is the weighted average of the conditional probabilities of drawing red from each bag.

    Step-by-Step Solution

    1. Bag 1: 4 red, 3 black (7 total) → $P(\text{red}|\text{Bag1})=\dfrac{4}{7}$.
    2. Bag 2: 2 red, 3 black (5 total) → $P(\text{red}|\text{Bag2})=\dfrac{2}{5}$.
    3. $P(\text{red}) = \dfrac12\cdot\dfrac47+\dfrac12\cdot\dfrac25 = \dfrac{4}{14}+\dfrac{2}{10}=\dfrac{2}{7}+\dfrac15$.
    4. Common denominator 35: $\dfrac{10}{35}+\dfrac{7}{35}=\dfrac{17}{35}$.

    Common Mistakes

    • Averaging the counts of red balls across bags instead of averaging the conditional probabilities.

    ✓Final answer

    The correct option is (D) — $\dfrac{17}{35}$.

    ANSWER: D