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Maths 2 Marks Questions

Probability · CBSE STD 10 (CBSE - English Medium). 65 questions.

Questions · Page 2 of 2

2 Marks Questions

Question 512 Marks
A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is:
The seven of clubs.
Answer
Given: A card is drawn at random from a pack of 52 cards
To Find: Probability of the following
Total number of cards = 52
Total number of 7 of club is 1
We know that $\text{Probability}=\frac{\text{Number of favourable event}}{\text{Total number of event}}$
Hence probability of getting a 7 of club is equal to $\frac{1}{52}$
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Question 522 Marks
Examine each of the following statements and comment:
If two coins are tossed at the same time, there are 3 possible outcomes-two heads, two tails, or one of each. Therefore, for each outcome, the probability of occurrence is $\frac{1}{3}$.
Answer
Incorrect.
When two coins are tossed, the possible outcomes are (H, H), (H, T), (T, H), and (T, T). It can be observed that there can be one of each in two possible ways - (H, T), (T, H).
Therefore, the probability of getting two heads is $\frac{1}{4}$, the probability of getting two tails is $\frac{1}{4}$, and the probability of getting one of each is $\frac{1}{2}$.
It can be observed that for each outcome, the probability is not $\frac{1}{3}$.
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Question 532 Marks
A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is:
A black king.
Answer
Given: A card is drawn at random from a pack of 52 cards
To Find: Probability of the following
Total number of cards = 52
Cards which are black king is 2
We know that $\text{Probability}=\frac{\text{Number of favourable event}}{\text{Total number of event}}$
Hence probability of getting a black king is equal to $\frac{2}{52}=\frac{1}{26}$
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Question 542 Marks
Harpreet tosses two different coins simultaneously (say, one is of Re 1 and other of Rs. 2). What is the probability that he gets at least one head?
Answer
Total no. of possible outcomes = 4 which are {HT, HH, TT, TH}
E → event of getting at least one head
No of favourable outcomes = 3 {HT, HH, TH}
Probability, $\text{P(E)}=\frac{\text{No. of favorable outcomes}}{\text{Total no. of possible outcomes}}$
$\text{P(E)}=\frac{3}{4}$
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Question 552 Marks
A die is thrown once. What is the probability of getting a prime number?
Answer
Total numbers on a die (n) = 6 (from 1 to 6)
Prime numbers are 2, 3, 5 i.e. 3
$\therefore\ \text{Probability}=\frac{\text{m}}{\text{n}}=\frac{3}{6}=\frac{1}{2}$
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Question 562 Marks
One card is drawn at random from a well shuffled deck of 52 cards. What is the probability of getting an ace?
Answer
Total number of cards in a deck (n) = 52
Number of aces in the deck (m) = 4
$\therefore\ \text{Probability}=\frac{\text{m}}{\text{n}}=\frac{4}{52}=\frac{1}{13}$
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Question 572 Marks
A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is:
A queen.
Answer
Given: A card is drawn at random from a pack of 52 cards
To Find: Probability of the following
Total number of cards = 52
Total number of queen is 4
We know that $\text{Probability}=\frac{\text{Number of favourable event}}{\text{Total number of event}}$
Hence probability of getting a queen $\frac{4}{52}=\frac{1}{13}$
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Question 582 Marks
A bag contains tickets numbered 11, 12, 13, ....., 30. A ticket is taken out from
the bag at random. Find the probability that the number on the drawn ticket:
  1. Is a multiple of 7.
  2. Is greater than 15 and a multiple of 5.
Answer
In a bag ticket no. are 11 to 30 i.e. 20 ticket
$\therefore$ n = 20
One ticket is drawn at random
  1. Number which are multiple of 7 are : 14, 21, 28
$\therefore$ m = 3

$\therefore$ Probability $=\frac{\text{m}}{\text{n}}=\frac{3}{20}$
  1. Number greater that 15 and a multiple of 5 are 20, 25, 30
$\therefore$ m = 3

$\therefore$ Probability $=\frac{\text{m}}{\text{n}}=\frac{3}{20}$
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Question 592 Marks
A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out:
  1. An orange flavoured candy?
  2. A lemon flavoured candy?
Answer
In a bag there are lemon flavoured candies = (n)
  1. Orange flavoured candy = Zero (0)
$\therefore$ Probability = 0
  1. Lemon candies = n
$\therefore\ \text{Probability}=\frac{\text{n}}{\text{n}}=1$
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Question 602 Marks
In a lottery of 50 tickets numbered 1 to 50, one ticket is drawn. Find the probability that the drawn ticket bears a prime number.
Answer
Number of lottery tickets (n) = 50
One ticket is drawn is a prime number
$\therefore$ Prime number upto 50 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 = 15
$\therefore$ m = 15
$\therefore\ \text{P(A)}=\frac{\text{m}}{\text{n}}=\frac{15}{50}=\frac{3}{10}$
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Question 612 Marks
A letter of English alphabet is chosen at random. Determine the probability that the chosen letter is a consonant.
Answer
Number of English alphabet = 26
Number of English outcomes = 26
Number of favourable outcomes = Consonants
= 26 - 5 = 21
$\therefore\ \text{Probability}=\frac{21}{26}$
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Question 622 Marks
Two coines are tossed simultaneously. Find the probability of getting exactly one head.
Answer
$\because$ Two coins are tossed
$\therefore$ Possible outcome will be (HH, HT, TH, TT)
Total = 4
$\therefore$ Actually outcomes will be HT, TH = 2
$\therefore\ \text{P(E)}=\frac{\text{No. of actual outcomes}}{\text{No. of possible outcomes}}=\frac{2}{4}=\frac{1}{2}$
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Question 632 Marks
If $\bar{\text{E}}$ denoted the complement or negation of an even E, what is the value of $\text{P(E)}+\text{P}(\bar{\text{E}})?$
Answer
$\bar{\text{E}}$ denotes the complement of an even E
$\therefore\ \text{P}(\bar{\text{E}})+\text{P}(\text{E})=1$
[$\therefore$ Sum of the probability of all outcomes (elementary evens) of an experiment is]
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Question 642 Marks
What is the probability that a number selected at random from the numbers 3, 4, 5, .... 9 is a multiple of 4?
Answer
Given: numbers are 3, 4, 5, 6, ....., 9
To Find: Probability of Getting multiple of 4
Total number is 9 - 3 + 1 = 7
Numbers which are multiple of 4 between 3 and 9 are 4 and 8
Total number which are multiple of 4 between 3 to 9 is 4 and 8 is 2
We know that $\text{Probability}=\frac{\text{Number of favourable event}}{\text{Total number of event}}$
Hence probability of getting a number which is a multiple of 4 is $\frac{2}{7}$
Hence probability of getting a number which is a multiple of 4 is equal to $\frac{2}{7}$
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Question 652 Marks
From a well shuffled pack of cards, a card is drawn at random. Find the probability of getting a black queen.
Answer
No. of cards in a pack of cards (n) = 52
One card is drawn at randow
No. of blank queens (n) = 2
$\therefore$ Probability of getting a black queen
$=\frac{\text{m}}{\text{n}}=\frac{2}{52}=\frac{1}{26}$
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2 Marks Questions - Page 2 - Maths STD 10 Questions - Vidyadip