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Maths 2 Marks Questions

Probability · Gujarat Board (GSEB) STD 10 (GSEB - 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