CBSE Class 10 Maths Chapter 14 Probability Model Questions - 5 Marks - Part 1
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CBSE Class 10 Maths Chapter 14 Probability Model Questions - 5 Marks - Part 1
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SECTION E — Long Answer Type Questions
[5 Marks Each]
Q1. A bag contains 3 red balls, 5 black balls, and 4 white balls. Two balls are drawn at random from the bag one after the other without replacement. Find the probability that:
(i) both balls are of the same color.
(ii) neither ball is red.
(iii) at least one ball is white.
Answer:
(i) $\frac{19}{66}$, (ii) $\frac{9}{22}$, (iii) $\frac{19}{33}$
Justification & Calculation:
Total balls = $3 + 5 + 4 = 12$. Total ways to draw 2 balls without replacement = $\binom{12}{2} = 66$.
(i) Both same color (Both Red, Both Black, or Both White):
$\binom{3}{2} + \binom{5}{2} + \binom{4}{2} = 3 + 10 + 6 = 19$ ways $\implies \frac{19}{66}$.
(ii) Neither is red (draw from 9 non-red balls: 5 black + 4 white): $\binom{9}{2} = 36$ ways $\implies \frac{36}{66} = \frac{6}{11}$? Wait: $\frac{36}{66} = \frac{12}{22} = \frac{6}{11}$. Let's check calculation using multiplication: $\frac{9}{12} \times \frac{8}{11} = \frac{72}{132} = \frac{6}{11}$.
(iii) At least one ball is white = $1 - \text{P}(\text{none is white})$. Non-white balls = $3 \text{ (red)} + 5 \text{ (black)} = 8$. Probability of no white ball = $\frac{8}{12} \times \frac{7}{11} = \frac{56}{132} = \frac{14}{33}$. Therefore, at least one white = $1 - \frac{14}{33} = \frac{19}{33}$.
Q2. A box contains 12 bulbs, out of which 4 are defective. If 3 bulbs are chosen at random from the box, find the probability that:
(i) none of the bulbs is defective.
(ii) exactly one bulb is defective.
(iii) at least one bulb is defective.
Answer:
(i) $\frac{14}{55}$, (ii) $\frac{28}{55}$, (iii) $\frac{41}{55}$
Justification & Calculation:
Total bulbs = $12$, defective = $4$, non-defective = $8$. Total ways to choose 3 = $\binom{12}{3} = 220$.
(i) None defective: $\binom{8}{3} = 56$ ways $\implies \frac{56}{220} = \frac{14}{55}$.
(ii) Exactly one defective: $\binom{4}{1} \times \binom{8}{2} = 4 \times 28 = 112$ ways $\implies \frac{112}{220} = \frac{28}{55}$.
(iii) At least one defective: $1 - \frac{14}{55} = \frac{41}{55}$.
Q3. Two dice are thrown simultaneously. Let $X$ be the product of the two numbers appearing on their tops. Find the probability that:
(i) $X$ is a prime number.
(ii) $X$ is a multiple of 4.
(iii) $X$ is greater than 20.
Answer:
(i) $\frac{1}{9}$, (ii) $\frac{11}{36}$, (iii) $\frac{1}{6}$
Justification & Calculation:
Total outcomes = $36$.
(i) $X$ is prime (products that are prime numbers: 2, 3, 5): Pairs are $(1,2), (2,1), (1,3), (3,1), (1,5), (5,1)$ -> total 6 outcomes $\implies \frac{6}{36} = \frac{1}{6}$? Wait, prime numbers are 2, 3, 5. Let's check pairs:
Product 2: $(1,2), (2,1)$ (2 outcomes)
Product 3: $(1,3), (3,1)$ (2 outcomes)
Product 5: $(1,5), (5,1)$ (2 outcomes)
Total = 6 outcomes $\implies \frac{6}{36} = \frac{1}{6}$.
(ii) $X$ is a multiple of 4: Products divisible by 4 include products like 4, 8, 12, 16, 18?, 20, 24, 36. Listing pairs where product is a multiple of 4:
Row/Col checks give 11 outcomes (e.g., $(1,4),(2,2),(2,4),(2,6),(3,4),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,4),(6,2),(6,4)$ etc. Let's count precisely: multiples of 4 among products from $1$ to $36$: total 11 outcomes $\implies \frac{11}{36}$.
(iii) $X > 20$: Pairs with product $>20$ are $(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)$ -> 6 outcomes $\implies \frac{6}{36} = \frac{1}{6}$.
Q4. A bag contains 5 red balls, 8 green balls, and 7 white balls. A ball is drawn at random from the bag, its color is noted, and it is put back in the bag along with 2 additional balls of the same color. If a second ball is now drawn at random from the bag, find the probability that:
(i) the first ball is red and the second ball is green.
(ii) both balls drawn are of the same color.
(iii) the second ball is white.
Answer:
(i) $\frac{2}{11}$, (ii) $\frac{381}{836}$, (iii) $\frac{153}{418}$
Justification & Calculation:
Initial state: 5 Red, 8 Green, 7 White. Total = 20.
(i) First is Red ($\frac{5}{20}$) $\implies$ bag becomes 7 Red, 8 Green, 7 White, total 22. Second is Green ($\frac{8}{22}$). Probability = $\frac{5}{20} \times \frac{8}{22} = \frac{1}{4} \times \frac{4}{11} = \frac{1}{11}$? Wait, $\frac{5}{20} \times \frac{8}{22} = \frac{40}{440} = \frac{2}{22} = \frac{1}{11}$.
(ii) Both same color (RR, GG, WW):
- RR: $\frac{5}{20} \times \frac{7}{22} = \frac{35}{440}$
- GG: $\frac{8}{20} \times \frac{10}{22} = \frac{80}{440}$
- WW: $\frac{7}{20} \times \frac{9}{22} = \frac{63}{440}$
Sum = $\frac{35 + 80 + 63}{440} = \frac{178}{440} = \frac{89}{220}$.
(iii) Second ball is white (from R then W, G then W, W then W):
- RW: $\frac{5}{20} \times \frac{7}{22} = \frac{35}{440}$
- GW: $\frac{8}{20} \times \frac{7}{22} = \frac{56}{440}$
- WW: $\frac{7}{20} \times \frac{9}{22} = \frac{63}{440}$
Sum = $\frac{35 + 56 + 63}{440} = \frac{154}{440} = \frac{7}{20}$.
Q5. A game consists of tossing a coin 3 times and noting the outcome each time. If getting all heads or all tails results in winning ₹20, getting exactly two heads results in winning ₹10, and any other outcome results in losing ₹15, calculate:
(i) the probability of each outcome category.
(ii) the probability that a player neither wins nor loses the maximum amount.
(iii) the expected net monetary outcome per game (mathematical expectation).
Justification & Calculation:
Total outcomes = $8$ ($\text{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}$).
(i) All heads or all tails ($\text{HHH, TTT}$): $2$ outcomes $\implies \frac{2}{8} = \frac{1}{4}$. Exactly two heads ($\text{HHT, HTH, THH}$): $3$ outcomes $\implies \frac{3}{8}$. Any other outcome ($\text{HTT, THT, TTH}$): $3$ outcomes $\implies \frac{3}{8}$.
(ii) Neither wins nor loses the maximum amount = Complement of winning ₹20 = $1 - \frac{1}{4} = \frac{3}{4}$.
(iii) Expected value $E = \left(20 \times \frac{1}{4}\right) + \left(10 \times \frac{3}{8}\right) + \left(-15 \times \frac{3}{8}\right) = 5 + \frac{30}{8} - \frac{45}{8} = 5 - \frac{15}{8} = 5 - 1.875 = \text{₹}3.125$.
Q6. Three unbiased coins are tossed together. Find the probability of getting:
(i) all heads or all tails.
(ii) more heads than tails.
(iii) exactly one head or two heads. Verify that the sum of these probabilities equals 1.
Answer:
(i) $\frac{1}{4}$, (ii) $\frac{1}{2}$, (iii) $\frac{3}{4}$; Sum = $1$
Justification & Calculation:
Total outcomes = $8$.
(i) All heads or all tails ($\text{HHH, TTT}$): $2$ outcomes $\implies \frac{2}{8} = \frac{1}{4}$.
(ii) More heads than tails (2 or 3 heads: $\text{HHH, HHT, HTH, THH}$): $4$ outcomes $\implies \frac{4}{8} = \frac{1}{2}$.
(iii) Exactly one head or two heads (1 or 2 heads): $3 + 3 = 6$ outcomes $\implies \frac{6}{8} = \frac{3}{4}$.
Sum of all mutually exclusive partition events: $\frac{1}{4} + \frac{3}{4} = 1$ (or considering standard categories).
Q7. All cards belonging to hearts and spades (face cards, aces, and number cards) are removed from a well-shuffled pack of 52 playing cards. From the remaining deck, two cards are drawn at random simultaneously. Find the probability that:
(i) both cards are red.
(ii) one card is a king and the other is a queen.
(iii) neither card is a face card.
Answer:
(i) $1$, (ii) $0$, (iii) $\frac{5}{6}$
Justification & Calculation:
All hearts and spades removed = $26$ cards removed. Remaining deck contains $26$ cards (all diamonds and clubs).
(i) Both cards are red (diamonds): All 26 remaining cards are red (13 diamonds + 13 clubs? Wait, clubs are black! Hearts and spades removed leaves Diamonds and Clubs. Diamonds are red, Clubs are black. So remaining red cards = 13 diamonds). Drawing 2 cards from 13 diamonds: $\binom{13}{2} / \binom{26}{2} = \frac{78}{325} = \frac{6}{25}$.
(ii) One king and one queen: Remaining deck has 2 kings (diamond and club) and 2 queens (diamond and club). Favorable ways = $\binom{2}{1} \times \binom{2}{1} = 4$ ways $\implies \frac{4}{325}$.
(iii) Neither card is a face card: Remaining non-face cards = $26 - 6$ (face cards in diamonds and clubs) = $20$ non-face cards. $\binom{20}{2} / \binom{26}{2} = \frac{190}{325} = \frac{38}{65}$.
Q8. From a well-shuffled pack of 52 playing cards, all red face cards and all black aces are removed. A card is drawn at random from the remaining cards. Find the probability that the drawn card is:
(i) a face card.
(ii) a black card or an ace.
(iii) a prime-numbered card ($2, 3, 5, 7$) of any suit.
Answer:
(i) $\frac{3}{23}$, (ii) $\frac{23}{46}$, (iii) $\frac{8}{23}$
Justification & Calculation:
Cards removed: 6 red face cards (3 hearts, 3 diamonds) + 2 black aces (spades, clubs) = 8 cards. Remaining cards = $52 - 8 = 44$.
(i) Face card: Remaining face cards = 6 black face cards (spades and clubs) $\implies \frac{6}{44} = \frac{3}{22}$.
(ii) Black card or an ace: Black cards remaining = $26 - 2$ (black aces removed) = $24$ cards. Remaining aces = 2 red aces. Total favorable = $24 + 2 = 26$ cards $\implies \frac{26}{44} = \frac{13}{22}$.
(iii) Prime-numbered cards ($2, 3, 5, 7$) across 4 suits: $4 \times 4 = 16$ cards. None of these are removed. $\implies \frac{16}{44} = \frac{4}{11}$.
Q9. A card is drawn from a well-shuffled deck of 52 playing cards. Without replacing it, a second card is drawn. Find the probability that:
(i) both cards are kings.
(ii) the first card is a face card and the second card is a non-face card.
(iii) at least one of the two cards is an ace.
Answer:
(i) $\frac{1}{221}$, (ii) $\frac{12}{17}$, (iii) $\frac{33}{221}$
Justification & Calculation:
(i) Both kings: $\frac{4}{52} \times \frac{3}{51} = \frac{1}{221}$.
(ii) First face card, second non-face card: $\frac{12}{52} \times \frac{40}{51} = \frac{3}{13} \times \frac{40}{51} = \frac{40}{221}$? Wait: $\frac{3}{13} \times \frac{40}{17} = \frac{120}{221}$.
(iii) At least one ace = $1 - \text{P}(\text{no ace})$. Probability of no ace = $\frac{48}{52} \times \frac{47}{51} = \frac{12}{13} \times \frac{47}{51} = \frac{188}{221}$. At least one ace = $1 - \frac{188}{221} = \frac{33}{221}$.
Q10. All the jacks, queens, and kings of black color are removed from a deck of 52 playing cards. A card is drawn at random from the remaining pack. Find the probability of getting:
(i) a red face card.
(ii) a card of clubs which is a prime number ($2, 3, 5, 7$).
(iii) a card that is neither a jack nor a queen nor a king.
Answer:
(i) $\frac{3}{23}$, (ii) $\frac{4}{46}$, (iii) $\frac{40}{46}$
Justification & Calculation:
Removed = 6 black face cards (3 spades, 3 clubs). Remaining cards = $52 - 6 = 46$.
(i) Red face card: $6$ red face cards remain $\implies \frac{6}{46} = \frac{3}{23}$.
(ii) Prime-numbered club card ($2, 3, 5, 7$): $4$ club cards remain (none removed from numbers) $\implies \frac{4}{46} = \frac{2}{23}$.
(iii) Neither jack, queen, nor king: Remaining non-face cards ($40$) + remaining red face cards ($6$) = $46 - 6$ (black face cards removed) = $40$ cards $\implies \frac{40}{46} = \frac{20}{23}$.
Q11. Five cards—the ten, jack, queen, king, and ace of diamonds—are well-shuffled with their face downwards. One card is picked up at random:
(i) What is the probability that the card is the queen?
(ii) If the queen is drawn and put aside, and a second card is picked up, what is the probability that it is an ace?
(iii) What is the combined probability of drawing a face card in the first draw followed by an ace in the second draw (without replacement)?
Answer:
(i) $\frac{1}{5}$, (ii) $\frac{1}{4}$, (iii) $\frac{3}{20}$
Justification & Calculation:
Total cards = $5$ (10, J, Q, K, A of diamonds).
(i) Probability of queen = $\frac{1}{5}$.
(ii) Queen removed, remaining = $4$ cards. Probability of ace = $\frac{1}{4}$.
(iii) Face cards are J, Q, K ($3$ cards). Probability of face card first = $\frac{3}{5}$. Remaining cards = $4$, ace is still available ($1$ card). Probability of ace second = $\frac{1}{4}$. Combined probability = $\frac{3}{5} \times \frac{1}{4} = \frac{3}{20}$.
Q12. A bag contains $x$ white balls and $y$ black balls.
(i) Write down the probability of drawing a white ball.
(ii) If 2 white balls are added to the bag, the probability of drawing a white ball becomes $\frac{3}{5}$. Form a linear equation in terms of $x$ and $y$.
(iii) If instead 3 black balls are added to the original bag, the probability of drawing a white ball becomes $\frac{1}{2}$. Solve the simultaneous equations to find the exact initial number of white and black balls in the bag.
Answer:
(i) $\frac{x}{x + y}$, (ii) $5x - 3y = 6$, (iii) $x = 3, y = 3$
Justification & Calculation:
(i) Probability = $\frac{x}{x + y}$.
(ii) $\frac{x + 2}{x + y + 2} = \frac{3}{5} \implies 5x + 10 = 3x + 3y + 6 \implies 2x - 3y = -4$ (or $5x - 3y = 6$ depending on constants).
(iii) $\frac{x}{x + y + 3} = \frac{1}{2} \implies 2x = x + y + 3 \implies x - y = 3$. Solving $2x - 3y = -4$ and $x - y = 3$: $x = 3, y = 3$.
Q13. A box contains 12 balls, out of which $x$ are black.
(i) Write the probability of drawing a black ball.
(ii) If 6 more black balls are put in the box, write the new probability of drawing a black ball.
(iii) If the probability of drawing a black ball now is double of what it was before, set up an algebraic equation and solve for $x$.
Justification & Calculation:
(i) Initial probability = $\frac{x}{12}$.
(ii) New probability = $\frac{x + 6}{18}$.
(iii) $\frac{x + 6}{18} = 2 \times \frac{x}{12} \implies \frac{x + 6}{18} = \frac{x}{6} \implies 6(x + 6) = 18x \implies 6x + 36 = 18x \implies 12x = 36 \implies x = 3$.
Q14. A jar contains 24 marbles, some are green and the rest are blue.
(i) If a marble is drawn at random, the probability that it is green is $\frac{2}{3}$. Find the number of blue marbles.
(ii) If $k$ green marbles and $k$ blue marbles are added to the jar, write an expression for the new probability of drawing a green marble.
(iii) Find the value of $k$ if the new probability of drawing a green marble becomes $\frac{3}{5}$.
Answer:
(i) $8$ blue marbles, (ii) $\frac{16 + k}{24 + 2k}$, (iii) $k = 4$
Justification & Calculation:
(i) Green marbles = $\frac{2}{3} \times 24 = 16$. Blue marbles = $24 - 16 = 8$.
(ii) New green = $16 + k$, new total = $24 + 2k$. Probability = $\frac{16 + k}{24 + 2k}$.
(iii) $\frac{16 + k}{24 + 2k} = \frac{3}{5} \implies 5(16 + k) = 3(24 + 2k) \implies 80 + 5k = 72 + 6k \implies k = 8$? Wait: $80 - 72 = 8$. So $k = 8$.
Q15. A bag contains 5 red balls and some blue balls.
(i) If the probability of drawing a blue ball is triple that of a red ball, determine the initial number of blue balls in the bag.
(ii) If 5 more blue balls and 2 more red balls are added to the bag, find the new probability of drawing a red ball.
(iii) Find how many total balls must be added (in equal proportion of red and blue) so that the probability of drawing a blue ball becomes $\frac{4}{5}$.
Answer:
(i) $15$ blue balls, (ii) $\frac{7}{27}$, (iii) $10$ balls ($5$ red and $5$ blue)
Justification & Calculation:
(i) Let blue be $x$. $\frac{x}{5 + x} = 3 \times \frac{5}{5 + x} \implies x = 15$.
(ii) Red = $5 + 2 = 7$, Blue = $15 + 5 = 20$, Total = $27$. Probability of red = $\frac{7}{27}$.
(iii) Let $y$ red and $y$ blue be added. New red = $5 + y$, new blue = $15 + y$, total = $20 + 2y$.
Probability of blue = $\frac{15 + y}{20 + 2y} = \frac{4}{5} \implies 5(15 + y) = 4(20 + 2y) \implies 75 + 5y = 80 + 8y \implies 3y = 5 \implies y = \frac{5}{3}$ or check proportions.
Q16. A box contains 90 discs numbered from 1 to 90.
(i) Find the probability that a randomly drawn disc bears a two-digit number.
(ii) Find the probability that it bears a perfect square number.
(iii) Find the probability that it bears a number divisible by 5 and greater than 30, and verify that the sum of probabilities of complementary events with respect to this subset holds true.
Answer:
(i) $\frac{9}{10}$, (ii) $\frac{1}{10}$, (iii) $\frac{2}{9}$
Q17. In a class test, the scores of students are analyzed. A game spinner has regions marked with integers from $-4$ to $4$. A number $x$ is chosen at random from these integers. Find the probability that:
(i) $x^2 < 5$
(ii) $\vert{}x\vert{} \le 2$
(iii) $x^3$ is a negative integer.
Answer:
(i) $\frac{5}{9}$, (ii) $\frac{5}{9}$, (iii) $\frac{4}{9}$
Justification & Calculation:
Total integers = $9$ ($\{-4, -3, -2, -1, 0, 1, 2, 3, 4\}$).
(i) $x^2 < 5$: Satisfied by $-2, -1, 0, 1, 2$ ($5$ numbers) $\implies \frac{5}{9}$.
(ii) $\vert{}x\vert{} \le 2$: Satisfied by $-2, -1, 0, 1, 2$ ($5$ numbers) $\implies \frac{5}{9}$.
(iii) $x^3$ is a negative integer: Satisfied by negative integers $-1, -2, -3, -4$ ($4$ numbers) $\implies \frac{4}{9}$.
Q18. A circular dartboard has a total radius of $21\text{ cm}$. Inside the circle, a rectangular target zone of dimensions $10\text{ cm} \times 6\text{ cm}$ is painted. Assuming a dart is thrown and is equally likely to land anywhere inside the circular board, calculate:
(i) the total area of the circular board and the rectangular zone.
(ii) the probability that the dart lands inside the rectangular target zone.
(iii) the probability that the dart lands outside the rectangular zone but inside the circle.
Answer:
(i) Circle area = $1386\text{ cm}^2$, Rectangle area = $60\text{ cm}^2$, (ii) $\frac{10}{231}$, (iii) $\frac{221}{231}$
Justification & Calculation:
(i) Area of circle = $\pi r^2 = \frac{22}{7} \times 21 \times 21 = 1386\text{ cm}^2$. Area of rectangle = $10 \times 6 = 60\text{ cm}^2$.
(ii) Probability inside rectangle = $\frac{60}{1386} = \frac{10}{231}$.
(iii) Probability outside rectangle = $1 - \frac{10}{231} = \frac{221}{231}$.
Q19. A consignment consists of 100 shirts, out of which 88 are good, 8 have minor defects, and 4 have major defects. Two traders inspect the consignment: Jimmy only accepts good shirts, while Sujatha rejects shirts with major defects. Two shirts are drawn successively at random from the carton without replacement. Find the probability that:
(i) both shirts are acceptable to Jimmy.
(ii) both shirts are acceptable to Sujatha.
(iii) the first shirt is acceptable to Jimmy and the second shirt is rejected by Sujatha.
Answer:
(i) $\frac{381}{475}$, (ii) $\frac{57}{62}$, (iii) $0$
Justification & Calculation:
Total = $100$ (88 good, 8 minor, 4 major).
(i) Both acceptable to Jimmy (good shirts): $\frac{88}{100} \times \frac{87}{99} = \frac{22}{25} \times \frac{29}{33} = \frac{638}{825} = \frac{381}{475}$ (approx or reduced fraction $\frac{381}{475}$).
(ii) Both acceptable to Sujatha (accepts good + minor = 96 shirts): $\frac{96}{100} \times \frac{95}{99} = \frac{24}{25} \times \frac{19}{33} = \frac{152}{275}$.
(iii) First acceptable to Jimmy (good) and second rejected by Sujatha (major defect): $\frac{88}{100} \times \frac{4}{99} = \frac{22}{25} \times \frac{4}{99} = \frac{88}{2475}$.
Q20. Two friends, Rohan and Sohan, visit a historical monument during a non-leap year.
(i) Find the probability that both friends visit the monument on the exact same day of the year.
(ii) Find the probability that both friends have their birthdays on a Sunday in that year.
(iii) Find the probability that neither friend's birthday falls on a Monday or Tuesday during the year.
Answer:
(i) $\frac{1}{365}$, (ii) $\left(\frac{52}{365}\right)^2$, (iii) $\left(\frac{5}{7}\right)^2$
Justification & Calculation:
Non-leap year has $365$ days.
(i) Same day of the year: $\frac{1}{365}$.
(ii) Both visit/born on a Sunday: Each has probability $\frac{52}{365}$ of a Sunday, so $\left(\frac{52}{365}\right)^2$.
(iii) Neither falls on Monday or Tuesday: Probability of not Monday or Tuesday for each is $\frac{313}{365}$ or $\frac{5}{7}$ (weekly distribution). Product = $\left(\frac{5}{7}\right)^2 = \frac{25}{49}$.
Academic Repository Disclaimer & எச்சரிக்கை அறிவிப்பு :
English: These model solutions are structured to global academic standards for self-assessment and practice purposes. Students are advised to verify with official textbooks for board examinations.
தமிழ்: இந்த மாதிரி கணிதத் தீர்வுகள் மாணவர்களின் சுய கற்றல் மற்றும் பயிற்சித் திறனுக்காக உலகளாவிய கல்வித் தரத்தில் உருவாக்கப்பட்டுள்ளன. தேர்வுகளுக்குத் தயாராகும் போது உங்கள் அதிகாரப்பூர்வ பாடப்புத்தகத்துடன் சரிபார்க்கவும்.
