CBSE Class 10 Maths Chapter 14 Probability Model Questions - 2 Marks - Part 1
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CBSE Class 10 Maths Chapter 14 Probability Model Questions - 2 Marks - Part 1
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SECTION B — Very Short Answer Type Questions
[2 Marks Each]
Q1. A coin is tossed twice. Find the probability of getting at least one head.
Answer: $\frac{3}{4}
Justification & Calculation:
When a coin is tossed twice, the sample space is:
$$\text{S} = \{HH, HT, TH, TT\}$$
Total number of possible outcomes = $4$.
Let $E$ be the event of getting at least one head. The favorable outcomes are $\{HH, HT, TH\}$, so the number of favorable outcomes = $3$.
$$\text{P}(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{4}$$
Q2. Two coins are tossed simultaneously. What is the probability of getting at most one tail?
Answer: $\frac{3}{4}
Justification & Calculation:
When two coins are tossed simultaneously, the sample space is:
$$\text{S} = \{HH, HT, TH, TT\}$$
Total number of possible outcomes = $4$.
Let $E$ be the event of getting at most one tail (i.e., $0$ or $1$ tail). The favorable outcomes are $\{HH, HT, TH\}$, so the number of favorable outcomes = $3$.
$$\text{P}(E) = \frac{3}{4}$$
Q3. If a fair coin is tossed 3 times, find the probability of getting exactly two heads.
Answer: $\frac{3}{8}
Justification & Calculation:
When a coin is tossed 3 times, the sample space contains $2^3 = 8$ outcomes:
$$\text{S} = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$$
Total number of possible outcomes = $8$.
Let $E$ be the event of getting exactly two heads. The favorable outcomes are $\{HHT, HTH, THH\}$, so the number of favorable outcomes = $3$.
$$\text{P}(E) = \frac{3}{8}$$
Q4. A coin is tossed 3 times. Find the probability of getting alternating outcomes (HTH or THT).
Answer: $\frac{1}{4}
Justification & Calculation:
When a coin is tossed 3 times, the total number of possible outcomes is $8$:
$$\text{S} = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$$
Let $E$ be the event of getting alternating outcomes. The favorable outcomes are $\{HTH, THT\}$, so the number of favorable outcomes = $2$.
$$\text{P}(E) = \frac{2}{8} = \frac{1}{4}$$
Q5. Two coins are tossed. Find the probability that both show the same face.
Answer: $\frac{1}{2}
Justification & Calculation:
When two coins are tossed, the sample space is:
$$\text{S} = \{HH, HT, TH, TT\}$$
Total number of possible outcomes = $4$.
Let $E$ be the event that both coins show the same face. The favorable outcomes are $\{HH, TT\}$, so the number of favorable outcomes = $2$.
$$\text{P}(E) = \frac{2}{4} = \frac{1}{2}$$
Q6. Three unbiased coins are tossed together. Find the probability of getting no head.
Answer: $\frac{1}{8}
Justification & Calculation:
When three coins are tossed together, the total number of possible outcomes is $8$:
$$\text{S} = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$$
Let $E$ be the event of getting no head (i.e., all tails). The only favorable outcome is $\{TTT\}$, so the number of favorable outcomes = $1$.
$$\text{P}(E) = \frac{1}{8}$$
Q7. A coin is biased such that a head is twice as likely to occur as a tail. If the coin is tossed once, find the probability of getting a tail.
Answer: $\frac{1}{3}
Justification & Calculation:
Let the probability of getting a tail be $P(T) = x$.
Since a head is twice as likely as a tail, the probability of getting a head is $P(H) = 2x$.
The sum of all probabilities in a sample space is equal to $1$:
$$P(H) + P(T) = 1$$
$$2x + x = 1 \implies 3x = 1 \implies x = \frac{1}{3}$$
Therefore, the probability of getting a tail is $\frac{1}{3}$.
Q8. Two coins are tossed simultaneously. Find the probability of getting at least two tails.
Answer: $\frac{1}{4}
Justification & Calculation:
When two coins are tossed simultaneously, the sample space is:
$$\text{S} = \{HH, HT, TH, TT\}$$
Total number of possible outcomes = $4$.
Let $E$ be the event of getting at least two tails. The only favorable outcome is $\{TT\}$, so the number of favorable outcomes = $1$.
$$\text{P}(E) = \frac{1}{4}$$
Q9. A coin and a die are thrown simultaneously. Find the probability of getting a head on the coin and an even number on the die.
Answer: $\frac{1}{4}
Justification & Calculation:
Total number of possible outcomes when a coin and a die are thrown = $2 \times 6 = 12$.
The sample space consists of:
$$\text{S} = \{(H,1), (H,2), (H,3), (H,4), (H,5), (H,6), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)\}$$
Let $E$ be the event of getting a head and an even number. The favorable outcomes are $\{(H,2), (H,4), (H,6)\}$, so the number of favorable outcomes = $3$.
$$\text{P}(E) = \frac{3}{12} = \frac{1}{4}$$
Q10. In a single toss of two coins, find the probability of getting more heads than tails.
Answer: $\frac{1}{4}
Justification & Calculation:
When two coins are tossed, the sample space is:
$$\text{S} = \{HH, HT, TH, TT\}$$
Total number of possible outcomes = $4$.
Let $E$ be the event of getting more heads than tails (i.e., $2$ heads and $0$ tails). The only favorable outcome is $\{HH\}$, so the number of favorable outcomes = $1$.
$$\text{P}(E) = \frac{1}{4}$$
Q11. A die is thrown once. Find the probability of getting a prime number.
Answer: $\frac{1}{2}$
Justification & Calculation:
When a die is thrown once, the sample space is:
$$\text{S} = \{1, 2, 3, 4, 5, 6\}$$
Total number of possible outcomes = $6$.
Let $E$ be the event of getting a prime number. The prime numbers on a die are $\{2, 3, 5\}$, so the number of favorable outcomes = $3$.
$$\text{P}(E) = \frac{3}{6} = \frac{1}{2}$$
Q12. Two dice are thrown simultaneously. Find the probability that the sum of the two numbers appearing is 7.
Answer: $\frac{1}{6}$
Justification & Calculation:
Total number of outcomes when two dice are thrown = $6 \times 6 = 36$.
Let $E$ be the event that the sum of the numbers is $7$. The favorable outcomes are:
$$\{(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\}$$
Number of favorable outcomes = $6$.
$$\text{P}(E) = \frac{6}{36} = \frac{1}{6}$$
Q13. A die is thrown once. What is the probability of getting a number lying between 2 and 6?
Answer: $\frac{1}{2}$
Justification & Calculation:
Sample space = $\{1, 2, 3, 4, 5, 6\}$, Total outcomes = $6$.
Numbers lying between $2$ and $6$ are $\{3, 4, 5\}$. Number of favorable outcomes = $3$.
$$\text{P}(E) = \frac{3}{6} = \frac{1}{2}$$
Q14. Two dice are rolled once. Find the probability that the product of the two numbers appearing is 12.
Answer: $\frac{1}{9}$
Justification & Calculation:
Total number of outcomes = $36$.
Favorable outcomes where the product is $12$:
$$\{(2,6), (3,4), (4,3), (6,2)\}$$
Number of favorable outcomes = $4$.
$$\text{P}(E) = \frac{4}{36} = \frac{1}{9}$$
Q15. A die is thrown twice. What is the probability that the number 5 will not come up either time?
Answer: $\frac{25}{36}$
Justification & Calculation:
Total number of outcomes = $36$.
Outcomes where $5$ comes up at least once include any outcome where the first or second die is $5$ ($6 + 6 - 1 = 11$ outcomes).
Therefore, outcomes where $5$ does not come up either time = $36 - 11 = 25$.
Alternatively, probability of not getting $5$ in a single throw is $\frac{5}{6}$. Since the throws are independent:
$$\text{P}(E) = \frac{5}{6} \times \frac{5}{6} = \frac{25}{36}$$
Q16. Two dice are thrown simultaneously. Find the probability that the sum of the numbers is greater than 8.
Answer: $\frac{5}{18}$
Justification & Calculation:
Total outcomes = $36$.
Favorable outcomes (sum $> 8$, i.e., sum of $9, 10, 11, 12$):
* Sum 9: $(3,6), (4,5), (5,4), (6,3)$ -> $4$ outcomes
* Sum 10: $(4,6), (5,5), (6,4)$ -> $3$ outcomes
* Sum 11: $(5,6), (6,5)$ -> $2$ outcomes
* Sum 12: $(6,6)$ -> $1$ outcome
Total favorable outcomes = $4 + 3 + 2 + 1 = 10$.
$$\text{P}(E) = \frac{10}{36} = \frac{5}{18}$$
Q17. A die is thrown. Find the probability of getting a multiple of 3.
Answer: $\frac{1}{3}$
Justification & Calculation:
Sample space = $\{1, 2, 3, 4, 5, 6\}$, Total outcomes = $6$.
Multiples of $3$ are $\{3, 6\}$. Number of favorable outcomes = $2$.
$$\text{P}(E) = \frac{2}{6} = \frac{1}{3}$$
Q18. Two dice are thrown at the same time. Find the probability of getting the same number on both dice (doublets).
Answer: $\frac{1}{6}$
Justification & Calculation:
Total outcomes = $36$.
Q19. A die is thrown twice. Find the probability that the difference of the numbers appearing is 2.
Answer: $\frac{2}{9}$
Justification & Calculation:
Total outcomes = $36$.
Favorable outcomes where difference is $2$:
$$\{(1,3), (2,4), (3,5), (4,6), (3,1), (4,2), (5,3), (6,4)\}$$
Number of favorable outcomes = $8$.
$$\text{P}(E) = \frac{8}{36} = \frac{2}{9}$$
Q20. Two dice are thrown simultaneously. Find the probability that the sum of the numbers is an odd number.
Answer: $\frac{1}{2}$
Justification & Calculation:
Total outcomes = $36$.
A sum is odd when one number is odd and the other is even. Exactly half of the $36$ outcomes result in an odd sum ($18$ outcomes).
$$\text{P}(E) = \frac{18}{36} = \frac{1}{2}$$
Q21. A die is numbered $1, 2, 2, 3, 3, 6$. If it is thrown once, find the probability of getting an even number.
Answer: $\frac{1}{2}$
Justification & Calculation:
Total outcomes = $6$.
Even numbers on this die are $\{2, 2, 6\}$. Number of favorable outcomes = $3$.
$$\text{P}(E) = \frac{3}{6} = \frac{1}{2}$$
Q22. Two dice are rolled. Find the probability that the sum of the numbers is less than or equal to 4.
Answer: $\frac{1}{6}$
Justification & Calculation:
Total outcomes = $36$.
Favorable outcomes (sums of $2, 3, 4$):
$$\{(1,1), (1,2), (2,1), (1,3), (2,2), (3,1)\}$$
Number of favorable outcomes = $6$.
$$\text{P}(E) = \frac{6}{36} = \frac{1}{6}$$
Q23. A card is drawn from a well-shuffled deck of 52 playing cards. Find the probability of getting a red face card.
Answer: $\frac{3}{26}$
Justification & Calculation:
Total number of cards = $52$.
Total face cards = $12$ (4 Jacks, 4 Queens, 4 Kings). Red face cards = $6$ (3 in hearts, 3 in diamonds).
$$\text{P}(E) = \frac{6}{52} = \frac{3}{26}$$
Q24. From a well-shuffled pack of 52 cards, a card is drawn at random. Find the probability of getting neither a red card nor a queen.
Answer: $\frac{6}{13}$
Justification & Calculation:
Total cards = $52$.
Red cards = $26$. Remaining non-red cards = $26$ (13 spades, 13 clubs). Among these non-red cards, two are queens (Queen of spades, Queen of clubs). Removing those two leaves $26 - 2 = 24$ cards.
$$\text{P}(E) = \frac{24}{52} = \frac{6}{13}$$
Q25. All kings, queens, and aces are removed from a pack of 52 cards. A card is drawn from the remaining cards. Find the probability of getting a black card.
Removed black cards = $2$ black kings + $2$ black queens + $2$ black aces = $6$ black cards.
Remaining black cards = $26$ (total original black) $- 6 = 20$.
$$\text{P}(E) = \frac{20}{40} = \frac{1}{2}$$
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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.
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