Class 10 Maths: Probability – Extra Questions with Answers
Q. 1 A box contains 7 blue marbles and 9 green marbles. If one marble is selected at random, what is the probability that the marble is green?
Check Solution
Ans: D
Solution:
The probability of selecting a green marble is the number of green marbles divided by the total number of marbles.
Total marbles = 7 (blue) + 9 (green) = 16
Probability (Green) = 9/16
Answer: 9/16
Q. 2 A fair coin is tossed 200 times. What is the probability of getting a tail on any single toss?
Check Solution
Ans: B
Solution:
Answer: 1/2
Q. 3 In a single throw of two dice, what is the probability of rolling a sum of 7?
Check Solution
Ans: C
Solution:
There are 6 possible outcomes for each die, so 36 total outcomes. The combinations that sum to 7 are (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1). There are 6 favorable outcomes. The probability is 6/36.
Answer: 1/6
Q. 4 A fair six-sided die is rolled three times. What is the probability of rolling a 6 on all three rolls?
Check Solution
Ans: D
Solution:
(1/6) * (1/6) * (1/6) = 1/216
Answer: 1/216
Q. 5 If a fair six-sided die is rolled, what is the probability of rolling a 7?
Check Solution
Ans: B
Solution: A standard die has faces numbered 1 through 6. Therefore, it is impossible to roll a 7.
Answer: 0
Q. 6 What is the product of the numerator and denominator of the simplified probability of choosing a consonant randomly from the word ‘MATHEMATICS’?
Check Solution
Ans: A
Solution:
1. Count the consonants in “MATHEMATICS”: M, T, H, M, T, C, S. There are 7 consonants.
2. Count the total letters: There are 11 letters.
3. Probability of choosing a consonant: 7/11
4. The fraction is already simplified.
5. Product of numerator and denominator: 7 * 11 = 77
Answer: 77
Q. 7 A jar contains 7 green marbles and some yellow marbles. If the probability of selecting a yellow marble is three times the probability of selecting a green marble, how many yellow marbles are in the jar?
Check Solution
Ans: D
Solution:
Let $y$ be the number of yellow marbles.
The probability of selecting a green marble is $\frac{7}{7+y}$.
The probability of selecting a yellow marble is $\frac{y}{7+y}$.
The probability of selecting a yellow marble is three times the probability of selecting a green marble.
So, $\frac{y}{7+y} = 3 \times \frac{7}{7+y}$.
Multiplying both sides by $7+y$, we have $y = 3 \times 7$.
Thus, $y = 21$.
Answer: 21
Q. 8 A bag contains 35 marbles, some are red and some are yellow. If a marble is selected at random from the bag, the probability that it is red is $\frac{3}{7}$. How many yellow marbles are in the bag?
Check Solution
Ans: B
Solution:
Let $r$ be the number of red marbles and $y$ be the number of yellow marbles. We know that $r+y=35$. The probability of selecting a red marble is $\frac{r}{35} = \frac{3}{7}$. Multiplying both sides by 35, we get $r = \frac{3}{7} \times 35 = 3 \times 5 = 15$. Since $r+y=35$, we have $15+y=35$, so $y=35-15=20$.
Answer: 20
Q. 9 (i) In a single throw of a dice, the probability of getting a number less than 4 is $P$. (ii) A card is drawn from a deck of 52 cards. The probability of drawing a black card is $Q$. (iii) A bag contains 4 red and 6 blue marbles, then the probability of drawing a red marble is $R$. What is the value of $P + Q + R$?
Check Solution
Ans: A
Solution:
(i) $P = \frac{3}{6} = \frac{1}{2}$
(ii) $Q = \frac{26}{52} = \frac{1}{2}$
(iii) $R = \frac{4}{10} = \frac{2}{5}$
$P + Q + R = \frac{1}{2} + \frac{1}{2} + \frac{2}{5} = 1 + \frac{2}{5} = \frac{7}{5}$
Answer: 7/5
Q. 10 Two fair six-sided dice are rolled. What is the probability that the sum of the numbers rolled is at most 5?
Check Solution
Ans: B
Solution:
The possible outcomes with a sum at most 5 are: (1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1). There are 10 favorable outcomes. The total number of possible outcomes is 6*6 = 36.
The probability is 10/36 = 5/18.
Answer: 5/18
Next Topic: Real Numbers
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