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Maths 2 Marks Questions

Probability · Maharashtra Board STD 10 (MSBSHSE - English Medium). 65 questions.

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Question 12 Marks
A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is:
A heart.
Answer
Given: A card is drawn at random from a pack of 52 cardsTo Find: Probability of the following
Total number of cards = 52
A heart: There are 14 cards of heart
$\therefore$ m = 13
$\therefore\ \text{P(A)}=\frac{\text{m}}{\text{n}}=\frac{13}{52}=\frac{1}{4}$
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Question 22 Marks
A box contains 20 cards numbered from 1 to 20. A card is drawn at random from the box. Find the probability that the number on the drawn card is:
  1. Divisible by 2 or 3.
  2. A prime number.
Answer
Total number of outcomes = 20
  1. The numbers from 1 to 20 which are divisible by 2 or 3 are 2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18 and 20.
So, the favourable number of outcomes are 13.

$\therefore$ P(number on the drawn card is divisible by 2 or 3)

$=\frac{\text{Favourable number of outcomes}}{\text{Total number of outcomes}}=\frac{13}{20}$
  1. The prime numbers from 1 to 20 are 2, 3, 5, 7, 11, 13, 17 and 19.
So, the favourable number of outcomes are 8.

$\therefore$ P(number on the drawn card is a prime number)

$=\frac{\text{Favourable number of outcomes}}{\text{Total number of outcomes}}$

$=\frac{8}{20}=\frac{2}{5}$
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Question 32 Marks
A number is selected at random from first 50 natural numbers. Find the probability that it is a multiple of 3 and 4.
Answer
Total number of natural numbers = 50
Numbers which are divisible by 3 and 4
= 12, 24, 36, 48 = 4 numbers
$\therefore\ \text{Probability P}_\text{(E)}=\frac{4}{50}=\frac{2}{25}$
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Question 42 Marks
A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is:
Jack.
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
Jack : There are 4 jacks
$\therefore$ m = 4
$\therefore\ \text{P(A)}=\frac{4}{52}=\frac{1}{13}$
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Question 52 Marks
Examine each of the following statements and comment:
If a die in thrown once, there are two possible outcomes - an odd number or an even number. Therefore, the probability of obtaining an odd number is $\frac{1}{2}$ and the probability of obtaining an even number is $\frac{1}{2}$.
Answer
Correct.
When a dice is thrown, the possible outcomes are 1, 2, 3, 4, 5, and 6. Out of these 1, 3, 5 are odd and 2, 4, 6 are even numbers.
Therefore, the probability of getting an odd number is $\frac{1}{2}$.
Similarly, the probability of getting an even number is $\frac{1}{2}$.
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Question 62 Marks
A bag contains 3 red and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is not red?
Answer
Number of red balls = 3 Number of black balls = 5 Total number of balls = 3 + 5 = 8 balls No. of favourable outcomes = 5$\therefore\ \text{Probability}=\frac{5}{8}$
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Question 72 Marks
Tickets numbered from 1 to 20 are mixed up and a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 7?
Answer
No. of tickets bearing numbers from 1 to 20 = 20
Numbers which are multiple of 3 of 7 can be 3, 6, 7, 96, 12, 14, 15, 18 = which are 8
$\therefore$ m = 8
$\therefore\ \text{P(A)}=\frac{\text{m}}{\text{n}}=\frac{8}{20}=\frac{2}{5}$
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Question 82 Marks
A die is thrown once. What is the probability of getting a number greater than 4?
Answer
Numbers greater then 4 on the dice are 5 and 6
$\therefore\ \text{P(E)}=\frac{\text{m}}{\text{n}}=\frac{2}{6}=\frac{1}{3}$
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Question 92 Marks
A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is:
Neither a king nor 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
Neither a king nor a queen
Number of favourable outcomes
= 52 - (4 + 4)
= 52 - 8 = 44
$\therefore\ \text{Probability}=\frac{44}{52}=\frac{11}{13}$
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Question 102 Marks
In a lottery there are 10 prizes and 25 blanks. What is the probability of getting a prize?
Answer
Total no. of possible outcomes = 35 {10 prizes, 25 blanks}
E ⟶ event of getting prize
No. of favourable outcomes = 10 {10 prizes}
Probability, $\text{P(E)}=\frac{\text{No. of favorable outcomes}}{\text{Total no. of possible outcomes}}=\frac{10}{35}=\frac{2}{7}$
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Question 112 Marks
A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is:
Either a black card or a 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
Either a black card of a king
m = there are 26 black cards and 2 red kings
= 26 + 2 = 28
$\therefore\ \text{P(A)}=\frac{\text{m}}{\text{n}}=\frac{28}{52}=\frac{7}{13}$
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Question 122 Marks
A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is:
Neither an ace nor a king.
Answer
Given: A card is drawn at random from a pack of 52 cardsTo Find: Probability of the following
Total number of cards = 52
Total number of ace card are 4 and king are 4
Total number of cards that are a ace and a king is equal to 4 + 4 = 5
Hence Total number of cards that are neither an ace nor a kin is 52 - 8 = 44
We know that $\text{Probability}=\frac{\text{Number of favourable event}}{\text{Total number of event}}$
Hence probability of getting cards neither an ace nor a king $=\frac{44}{52}=\frac{11}{13}$
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Question 132 Marks
If the probability of winning a game is 0.3, what is the probability of loosing it?
Answer
Given: Probability of winning a game P(E) = 0.3
To Find: Probability of losing the game $\text{P}(\bar{\text{E}})$
Calculation: We know that sum of probability of occurrence of an event and probability of non occurrence of an event is 1.
$\text{P(E)}+\text{P}(\bar{\text{E}})=1$
$0.3+\text{P}(\bar{\text{E}})=1$
$\text{P}(\bar{\text{E}})=1-0.3$
$\text{P}(\bar{\text{E}})=0.7$
Hence the probability of losing the game is $\text{P}(\bar{\text{E}})=0.7$.
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Question 142 Marks
Cards each marked with one of the numbers 4, 5, 6, ....., 20 are placed in a box and mixed thoroughly. One card is drawn at random from the box what is the probability of getting an even number?
Answer
No. of card having marks from 4 to 20 (n) = 17
One card is drwan at random
Even numbers on the cards are 4, 6, 8, 10, 12, 14, 16, 18, 20
Total (m) = 9
$\therefore\ \text{Probability}=\frac{\text{m}}{\text{n}}=\frac{9}{17}$
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Question 152 Marks
A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is:
A ten.
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 ten is 4
We know that $\text{Probability}=\frac{\text{Number of favourable event}}{\text{Total number of event}}$
Hence probability of getting a ten is $\frac{4}{52}=\frac{1}{13}$
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Question 162 Marks
A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is:
Neither a heart nor a 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
Neither a heart nor a king:
There are 13 cards of heart and 3 kings more
i.e., 13 + 3 = 16 total
$\therefore$ m = cards other than the above
= 52 - 16 = 36
$\therefore\ \text{P(A)}=\frac{\text{m}}{\text{n}}=\frac{36}{52}=\frac{9}{13}$
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Question 172 Marks
A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is:
A black card.
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
E ⟶ event of getting a black card.
No. of favourable outcomes = 26 {13 cards of spades & 13 cards of clubs}
$\text{P(E)}=\frac{26}{52}=\frac{1}{2}$
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Question 182 Marks
A die is thrown once. What is the probability of getting a number lying between 2 and 6?
Answer
Total numbers on the die = 6 (from 1 to 6)
$\therefore$ Probability of number lying between 2 and 6
$(\text{i.e. }3, 4, 5)=\frac{3}{6}=\frac{1}{2}$
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Question 192 Marks
In a class, there are 18 girls and 16 boys. The class teacher wants to choose one pupil for class monitor. What she does, she writes the name of each pupil on a card and puts them into a basket and mixes thoroughly. A child is asked to pick one card from the basket. What is the probability that the name written on the card is:
  1. The name of a girl.
  2. The name of a boy.
Answer
No. of student in a class (n)
= 18 girls + 16 boys = 34
One students is to be a class monitor
  1. No. of girls (m) = 18
$\therefore$ Probability of being a girls as monitor of the

$\text{class}=\text{P(A)}=\frac{\text{m}}{\text{n}}=\frac{18}{34}=\frac{9}{17}$
  1. No. of boys (m) = 16
$\therefore$ Probability of being a boy as monitor of the

$\text{class}=\text{P(A)}=\frac{\text{m}}{\text{n}}=\frac{16}{34}=\frac{8}{17}$
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Question 202 Marks
One card is drawn from a well shuffled deck of 52 playing cards. What is the probability of getting a non-face card?
Answer
No. of cards in the deck of playing cards (n) = 52
No. of face cards = 3 × 4 = 12
Remaining non-face cards = 52 - 12 = 40
$\therefore$ Probability of non-face card $=\frac{40}{52}=\frac{10}{13}$
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Question 212 Marks
A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is:
Spade or an ace.
Answer
Given: A card is drawn at random from a pack of 52 cardsTo Find: Probability of the following
Total number of cards = 52
E ⟶ event of getting spade or an all.
No. of favourable outcomes = 13 + 3 = 16 {13 spades & 3 aces each of hearts, diamonds & clubs}
$\text{P(E)}=\frac{16}{52}=\frac{4}{13}$
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Question 222 Marks
A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is:
A jack, queen or a 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
A jack, queen or a king are 3 from each 4 suits
Total number of a jack, queen and king are 12
We know that $\text{Probability}=\frac{\text{Number of favourable event}}{\text{Total number of event}}$
Hence probability of getting a jack, queen or a king is $=\frac{12}{52}=\frac{3}{13}$
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Question 232 Marks
There are 30 cards, of same size, in a bag on which numbers 1 to 30 are written. One card is taken out of the bag at random. Find the probability that the number on the selected card is not divisible by 3.
Answer
Totan numbar of cards bearing numbers from 1 to 30 (n) = 30
Now numbers which are divisible by 3 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, which are 10
$\therefore$ Numbers which are not divisible by 3(n) = 30 - 10 = 20
Now probability $\text{P(A)}=\frac{\text{m}}{\text{n}}=\frac{20}{30}=\frac{2}{3}$
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Question 242 Marks
A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is:
A spade.
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
A spade : There are 13 cards of spade
$\therefore$ m = 13
$\text{P(A)}=\frac{\text{m}}{\text{n}}=\frac{13}{52}=\frac{1}{4}$
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Question 252 Marks
An urn contains 10 red and 8 white balls. One ball is drawn at random. Find the probability that the ball drawn is white.
Answer
Total no of possible outcomes = 18 {10 red balls, 8 white balls}
E ⟶ event of drawing white ball
No. of favourable outcomes = 8 {8 white balls}
Probability, $\text{P(E)}=\frac{\text{No. of favorable outcomes}}{\text{Total no. of possible outcomes}}$
$=\frac{8}{18}=\frac{4}{9}$
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Question 262 Marks
A number is chosen at random from the number -3, -2, -1, 0, 1, 2, 3. What will be the probability that square of this number is less then or equal to 1?
Answer
S = {-3, -2, -1, 0, 1, 2, 3}
Let E be the event of getting a number whose square is less than or equal to 1.
So, E = {-1, 1, 0}
$\text{P(E)}=\frac{3}{7}$
Hence, the probability of getting a number whose square is less than or equal to is $\frac{3}{7}$.
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Question 272 Marks
Five cards the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.
  1. What is the probability that the card is the queen?
  2. If the queen is drawn and put a side, what is the probability that the second card picked up is:
  1. An ace?
  2. A queen?
Answer
5 cards are the ten, jack, queen, king ace (n) = 5
One card is picked up at random:
  1. Probability of a card being a queen $=\frac{\text{m}}{\text{n}}=\frac{1}{5}$
  2. Queen is drawn and put aside, then number of cards will be = 4
  1. Probability of card being a ace $=\frac{\text{m}}{\text{n}}=\frac{1}{4}$
  2. Probability of card being a queen $=\frac{\text{m}}{\text{n}}=\frac{0}{4}=0$
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Question 282 Marks
Two unbiased dice are thrown. Find the probability that the total of the numbers on the dice is greater than 10.
Answer
When a pair of dice are thrown, then total no. of possible outcomes = 6 × 6 = 36
let E ⟶ event of getting sum on dice greater than 10
Then no of favourable outcomes = 3 {(5, 6) (6, 5) (6, 6)}
We know that, $\text{P(E)}=\frac{\text{No. of favorable outcomes}}{\text{Total no. of possible outcomes}}$
i.e., $\text{P(E)}=\frac{3}{36}=\frac{1}{12}$
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Question 292 Marks
A bag contains 5 red, 8 green and 7 white balls, One ball is drawn at random from the bag. What is the probability of getting a white ball or a green ball?
Answer
Given: A bag contains 5 red, 8 green and 7 white balls
To Find: Probability of getting a white ball or a green ball.
Total number of balls 5 + 8 + 7 = 20
Total number of green or white balls 8 + 7 = 15
We know that $\text{Probability}=\frac{\text{Number of favourable event}}{\text{Total number of event}}$
Hence probability of getting a green or a white ball $=\frac{15}{20}=\frac{3}{4}$
Hence probability of getting an green or white ball $=\frac{3}{4}$
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Question 302 Marks
Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn bears a number which is a multiple of 3?
Answer
Total number of tickets (from 1 to 20) = 20
One ticket is drawb at random
Number which are nultiple of 3 are : 3, 6, 9, 12, 15, 18
Total numbers (m) = 6
$\therefore\ \text{Probability}=\frac{\text{m}}{\text{n}}=\frac{6}{20}=\frac{3}{10}$
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Question 312 Marks
What is the probability that a number selected at random from the number 1, 2, 2, 3, 3, 3, 4, 4, 4, 4 will be their average?
Answer
Given: A number is selected from the numbers 1, 2, 2, 3, 3, 3, 4, 4, 4, 4
To Find: Probability that the selected number is the average of the numbers
Total numbers are 10
Average of numbers is
$=\frac{1+2+2+3+3+3+4+4+4+4}{10}$
$=\frac{30}{10}$
$=3$
Total numbers of numbers which are average of these numbers are 3
We know that $\text{Probability}=\frac{\text{Number of favourable event}}{\text{Total number of event}}$
Hence Probability that the selected number is the average of the numbers $=\frac{3}{10}$
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Question 322 Marks
All kings and queens are removed from a pack of 52 cards. The remaining cards are well-shuffled and then a card is randomly drawn from it. Find the probability that this card is:
  1. A red face card.
  2. A black card.
Answer
All king and all Queens are removed then the remaining cards = 52 - 8 = 44
  1. A red face cards are = 2
$\therefore\ \text{Pribability}=\frac{2}{44}=\frac{1}{22}$
  1. A black cards are = 26 - 4 = 22
$\therefore$ Probability of a black cards $=\frac{22}{44}=\frac{1}{2}$
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Question 332 Marks
Die is thrown once. Find the probability of getting a number less than 3.
Answer
Given: A dice is thrown once
To Find: Probability of getting a number less than 3
Total number on a dice is 6.
Number less than 3 are 1 and 2
Total number of numbers less than 3 is 2
We know that $\text{Probability}=\frac{\text{Number of favourable event}}{\text{Total number of event}}$
Hence probability of getting a number less than 3 is equal to $\frac{2}{6}=\frac{1}{3}$
Hence probability of getting a number less than $3=\frac{1}{3}$
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Question 342 Marks
All jacks, queens and kings are removed from a pack of 52 cards. The remaining cards are well-shuffled and then a card is randomly drawn from it. Find the probability that this cards is:
  1. A black face card.
  2. A red card.
Answer
All jacks, queen and kings are removed
$\therefore$ Remaining cards = 52 - 4 × 3 = 52 - 12 = 40
  1. A black face cards = 0
$\therefore$ Probability = 0
  1. A red car = 20
$\therefore\ \text{probability}=\frac{20}{40}=\frac{1}{2}$
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Question 352 Marks
A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is:
Other than an ace.
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
E ⟶ event of getting card other than an ace.
No. of favourable outcomes = 52 - 4 = 48 {Since we have 4 ace cards}
$\text{P(E)}=\frac{48}{52}=\frac{12}{13}$
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Question 362 Marks
A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is:
The ace of spades.
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
E ⟶ event of getting the ace of spades.
No. of favourable outcomes = 1{ace of spades}
$\text{P(E)}=\frac{1}{52}$
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Question 372 Marks
Two coins are tossed simultaneously. What is the probability of getting at least one head?
Answer
Given: Two coins are tossed simultaneously.
To Find: Probability of getting at least one head.
When two coins are tossed then the outcome will be
TT, HT, TH, HH.
Hence total number of outcome is 4.
At least one head means 1H or 2H.
Hence total number of favorable outcome i.e. at least one head is 3
We know that $\text{Probability}=\frac{\text{Number of favourable event}}{\text{Total number of event}}$
Hence probability of getting at least one head $=\frac{3}{4}$
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Question 382 Marks
What is the probability that a leap year has 53 Sundays and 53 Mondays?
Answer
A leap year has 366 days i.e. 52 weeks and 2 days
When a leap year has 53 Sundays and 53 Mondays
$\therefore$ m = 1 and n = 7
$\therefore\ \text{P(A)}=\frac{1}{7}$
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Question 392 Marks
Find the probability that a number selected from the number 1 to 25 is not a prime number when each of the given numbers is equally likely to be selected.
Answer
Givne: A number is selected from numbers 1 to 25
To Find: Probability of getting a number which is not a prime.
Total number of cards is 25.
Total number of elementary events = 25
Cards bearing non prime numbers are 1, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25
Total number of cards bearing non-prime numbers = 16
Number of favourable elementary events = 16
We know that, $\text{Probability}=\frac{\text{number of favourable elementary events}}{\text{Total number of elementary events}}$
So, P(getting a card bearing a non prime number) $=\frac{16}{25}$
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Question 402 Marks
It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?
Answer
Let E ⟶ event of 2 students having same birthday P(E) is given as 0.992
Let $(\bar{\text{E}})\rightarrow$ event of 2 students not having same birthday.
We know that, $\text{P}(\text{E})+\text{P}(\bar{\text{E}})=1$
$\text{P}(\bar{\text{E}})=1-\text{P(E)}$
$=1-0.992$
$=0.008$
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Question 412 Marks
A bag contains 5 white and 7 red balls. One ball is drawn at random. What is the probability that ball drawn is white?
Answer
Total no. of possible outcomes = 12 {5 white, 7 red}
E ⟶ event of drawing white ball.
No. of favorable outcomes = 5 {white balls are 5}
Probability, $\text{P(E)}=\frac{\text{No. of favorable outcomes}}{\text{Total no. of possible outcomes}}$
$\text{P(E)}=\frac{5}{12}$
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Question 422 Marks
Why is tossing a coin considered to be a fair way of deciding which team should choose ends in a game of cricket?
Answer
No. of possible outcomes while tossing a coin = 2 {1 head & 1 tail}
$\text{P(E)}=\frac{\text{No. of favorable outcomes}}{\text{Total no. of possible outcomes}}$
$\text{P(getting head)}=\frac{1}{2}$
$\text{P(getting tail)}=\frac{1}{2}$
Since probability of two events are equal, these are called equally like events.
Hence, tossing a coin is considered to be a fair way of deciding which team should choose ends in a game of cricket.
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Question 432 Marks
What is the probability that a number selected from the numbers 1, 2, 3, ....., 15 is a multiple of 4?
Answer
Numbers are 1, 2, 3, ....., 15
$\therefore$ n = 15
Now multiple of 4 are 4, 8, 19
$\therefore$ m = 3
$\therefore\ \text{P(A)}=\frac{\text{m}}{\text{n}}=\frac{3}{15}=\frac{1}{5}$
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Question 442 Marks
A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is:
Black and a 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
E ⟶ event of getting black & a king.
No. of favourable outcomes = 2 {king of spades & clubs}
$\text{P(E)}=\frac{2}{52}=\frac{1}{26}$
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Question 452 Marks
The faces of a red cube and a yellow cube are numbered from 1 to 6. Both cubes are rolled. What is the probability that the top face of each cube will have the same number?
Answer
Two cubes bearing numbers 1 to 6 each, are rolled
$\therefore$ Number of total outcomes = 6 × 6 = 36
Now probability of comming of two same number i.e. (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) can be
$\text{P(A)}=\frac{\text{m}}{\text{n}}=\frac{6}{36}=\frac{1}{6}$
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Question 462 Marks
A die is thrown once. What is the probability of getting an odd number?
Answer
Given: A dice is thrown once
To Find: Probability of getting an odd number.
Total number on a dice is 6.
Odd numbers on a dice are 1, 3 and 5
Total number of odd numbers on dice is 3
We know that $\text{Probability}=\frac{\text{Number of favourable event}}{\text{Total number of event}}$
Hence probability of an odd number $=\frac{3}{6}=\frac{1}{2}$
Hence probability of getting an odd number $=\frac{1}{2}$
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Question 472 Marks
The probability that it will rain tomorrow is $0.85$. What is the probability that it will not rain tomorrow?
Answer
Let $E$ be the event of happening of rain
$P ( E )$ is given as $0.85$
$E^{-} \longrightarrow ? ? ? h ? ? ? ? ? ? ? ? ? ? ? ? ? ? \\
P\left(E^{-}\right)=1-P(E)=1-0.85=0.15$
$\therefore P$ (not happening of rain) $=0.15$
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Question 482 Marks
A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is:
Neither a red card nor 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
Neither a red card nor a queen
There are 26 red card and two black queens more
$\therefore$ m = 52 - (26 + 2) = 52 - 28 = 24
$\therefore\ \text{P(A)}=\frac{\text{m}}{\text{n}}=\frac{24}{52}=\frac{6}{13}$
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Question 492 Marks
A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is:
A red card.
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
E → event of getting a red card.
No. of favourable outcomes = 26 {13 hearts, 13 diamonds}
$\text{P(E)}=\frac{26}{52}=\frac{1}{2}$
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Question 502 Marks
What is the probability that an ordinary year has 53 Sundays?
Answer
Given: An ordinary year
To Find: Probability that a non leap year has 53 Sundays.
Total number of days in an ordinary year is 365 days
Hence number of weeks in an ordinary year is $\frac{365}{7}=52\text{ weeks and 1 day}$
In an ordinary year we have 52 complete weeks and 1 day which can be any day of the week i.e. SUNDAY, MONDAY, TUESDAY, WEDNESDAY, THURSDAY, FRIDAY and SATURDAY
To make 53 Sundays the additional day should be Sunday
Hence total number of days is 7
Favorable day i.e. Sunday is 1
We know that $\text{Probability}=\frac{\text{Number of favourable event}}{\text{Total number of event}}$
Hence probability that an ordinary year has 53 Sundays is equal to $\frac{1}{7}$
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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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