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Mathematics [3 marks sum]

Probability · ICSE STD 10 (ICSE - English Medium). 89 questions.

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[3 marks sum]

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50 questions · timed · auto-graded

Question 13 Marks
In a musical chairs game, a person has been advised to stop playing the music at any time within 40 seconds after its start. What is the probability that the music will stop within the first 15 seconds?
Answer
Total result = 0 sec to 40 sec
Total possible outcomes =40
n(s)=40
Favorable results = 0 sec to 15 sec
Favourable outcomes =15
n(E)=15
Probability that the music will stop in first 15 sec = $\frac{n(E)}{n(S)}=\frac{15}{40}=\frac{3}{8}$
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Question 23 Marks
All the three face cards of spades are removed from a well shuffled pack of 52 cards. A card is then drawn at random from the remaining pack. Find the probability of getting : a black card
Answer
Total number of cards =52
3 face card of spades are removed
Remaining =52-3=49= number of possible outcomes
n(s)=49
Number of black cards left =23 cards (13 club +10 spade)
Event of drawing a black card=E=23
n(E)=23
Probability of drawing a black card=$\frac{n(E)}{n(s)}=\frac{23}{49}$
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Question 33 Marks
All the three face cards of spades are removed from a well shuffled pack of 52 cards. A card is then drawn at random from the remaining pack. Find the probability of getting : a queen.
Answer
Total number of cards =52
3 face card of spades are removed
Remaining =52-3=49= number of possible outcomes
n(s)=49
Number of queen cards left= 3
Event of drawing a black face card=E=3
n(E)=3
Probability of drwaing a queen card =$\frac{n(E)}{n(s)}=\frac{3}{49}$


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Question 43 Marks
All the three face cards of spades are removed from a well shuffled pack of 52 cards. A card is then drawn at random from the remaining pack. Find the probability of getting : a black face card
Answer
Total number of cards =52
3 face card of spades are removed
Remaining =52 - 3 = 49 = number of possible outcomes
n(s) = 49
Number of black face cards left = 3 face cards of club
Event of drawing a black face cards = E = 3
n(E) = 3
Probability of drawing a black face card =$\frac{n(E)}{n(s)}=\frac{3}{49}$
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Question 53 Marks
A box contains a certain number of balls. On each of $60\%$ balls, letter A is marked. On each of $30\%$ balls, letter B is marked and on each of remaining balls, letter C is marked. A ball is drawn from the box at random. Find the probability that the ball drawn is :
i. marked C
ii. A or B
iii. neither B nor C
Answer
A box contains,
$60 \%$ balls, letter A is marked.
$30 \%$ balls, letter $B$ is marked.
$10 \%$ balls, letter $C$ is marked.
i). Total number of all possible outcomes $=100$
Number of favourable outcomes $=10$
$\therefore \text { Required Probability }=\frac{\text { Number of favourable outcomes }}{\text { Total number of all possible outcomes }}=\frac{10}{100}=\frac{1}{10}$
ii)
The probability that the ball drawn is marked $A=$
$\frac{\text { Number of favourable outcomes }}{\text { Total number of all possible outcomes }}=\frac{60}{100}=\frac{6}{10}$ ......(1)
The probability that the ball drawn is marked $B=$
$\frac{\text { Number of favourable outcomes }}{\text { Total number of all possible outcomes }}=\frac{30}{100}=\frac{3}{10}$ ......(2)
(iii)
The probability that the ball drawn is neither $B$ nor $C$
$=1-[P(B)+P(C)]$
$=1-\left[\frac{3}{10}+\frac{1}{10}\right]$
$=1-\frac{4}{10}$
$=\frac{6}{10}$
$=\frac{3}{5}$
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Question 63 Marks
Sixteen cards are labelled as a, b, c, ……………. m, n, o, p. They are put in a box and shuffled.a boy is asked to draw a card from the box. What is the probability that the card drawn is :
(1) a vowel
(2) a consonant
(3) none of the letters of the word median
Answer
Outcomes: a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p
Total number of all possible outcomes 16
1) When the selected card has a vowel, the possible outcomes are $a, e, i, o$
Number of favourable outcomes 4
$\therefore$ Required probability $=\frac{4}{16}=\frac{1}{4}$
2) When the selected card has a consonant,
Number of favourable outcomes $16-4=12$
$\therefore$ Required probability $=\frac{12}{16}=\frac{3}{4}$
3) When the selected card has none of the letters from the word median
the possible outcomes are b, c, f, g, h, j, k, l, o, p.
Number of favourable outcomes 10
$\therefore$ Required probability $=\frac{10}{16}=\frac{5}{8}$
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Question 73 Marks
A game of numbers has cards marked with 11, 12, 13, … 40. A card is drawn at random. Find the probability that the number on the card drawn is : 1) A perfect square
2) Divisible by 7
Answer
Total number of outcomes $=30$
1) The perfect squares lying between 11 and 40 are 16,25 and 36 .
So the number of possible outcomes 3
$\therefore$ Probability that the number on the card drawn is a perfect square
$
=\frac{\text { Number of possible outcomes }}{\text { Total number of outcomes }}=\frac{3}{30}=\frac{1}{10}
$
2) The numbers from 11 to 40 that are divisible by 7 are $14,21,28$ and 35 .
So the number of possible outcomes
The probability that the number on the card drawn is divisible by 7
$=\frac{\text { Number of possible outcomes }}{\text { Total number of outcomes }}=\frac{4}{30}=\frac{2}{15}$
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Question 83 Marks
A bag contains $5$ white balls, $6$ red balls and $9$ green balls. A ball is drawn at random from the bag. Find the probability that the ball is drawn is :
1) a green ball
2) a white or a red ball
3) is neither a green ball nor a white ball.
Answer
Total number of balls $= 5 + 6 + 9 = 20$
1) Number of green balls 9 Number of favourable cases
$\therefore P$ (Green ball) $=$
$\frac{\text { Number of favourable cases }}{\text { Total number of balls }}=\frac{9}{20}$
2) Number of white balls 5, Number of red balls 6
Number of favourable cases $= 5 + 6 = 11$
$\therefore$ P(White ball or Red ball) $=$
$\frac{\text { Number of favourable cases }}{\text { Total number of balls }}=\frac{11}{20}$
3) P(Neither green ball nor white ball) P(Red ball)
$=\frac{\text { Number of Red balls }}{\text { Total number of balls }}$
$=\frac{6}{20}=\frac{3}{10}$
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Question 93 Marks
A die has 6 faces marked by the given numbers as shown below

The die is thrown once. What is the probability of getting
1) a positive integer
2) an integer greater than -3.
3) the smallest integer.
Answer
Given that the die has 6 faces marked by the given numbers as below :

Total number of outcomes = 6
1) For getting a positive integer, the favourable outcomes are 1, 2, 3
⇒ Number of favourable outcomes = 3
⇒ Required probability =$\frac{3}{6}=\frac{1}{2}$
2) For getting an integer greater than -3, the favourable outcomes are -2, -1, 1, 2, 3
⇒ Number of favourable outcomes = 5
⇒ Required probability = $\frac{5}{6}$
3) For getting the smallest integer, the favourable outcome is -2
⇒Number of favourable outcomes = 1
⇒ Required probability =$\frac{1}{6}$
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Question 103 Marks
From a pack of 52 playing cards, all cards whose numbers are multiples of 3 are removed. A card is now drawn at random. What is the probability that the card drawn is A face card (King, Jack or Queen).
Answer
No. of total cards $=52$ cards removed of 4 colours of multiples of 3
$
=3,6,9=4 \times 3=12
$
Remaining cards $=52-12=40$
No.of face cards $=12$ cards
$\Rightarrow$ Probability $p(E)=\frac{12}{40}=\frac{3}{10}$
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Question 113 Marks
Offices in Delhi are open for five days in a week (Monday to Friday). Two employees of an office remain absent for one day in the same particular week. Find the probability that they remain absent on : consecutive day
Answer
Total number of possible outcomes $=5 \times 5=25$
The possible outcomes are :
MM,MT,MW,MTh,MF,TM,TT,TW,TTh,TF,WM,WT,WW,WTh,W F,ThM,ThT,ThM, ThT,Thw,ThTh,THF,FM,FT,FW,FTh,FF
Favorable outcomes for two employee remaining absent on consecutive days : MT,TM,TW,WT,WTh,ThW,ThF,FTh
Number of favorable outcomes $=8$
$P ($ consecutive days $)=\frac{8}{25}$
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Question 123 Marks
Offices in Delhi are open for five days in a week (Monday to Friday). Two employees of an office remain absent for one day in the same particular week. Find the probability that they remain absent on : the same day
Answer
Total number of possible outcomes=$5 \times 5=25$
The possible outcomes are :
MM,MT,MW,MTh,MF,TM,TT,TW,TTh,TF,WM,WT,WW,WTh,WF,ThM,ThT,ThM,ThT,Thw,ThTh,THF,FM,FT,FW,FTh,FF
Favorable outcome for two employees remaining absennt on same day are: MM,TT,WW,ThTh,FF
Number of favorable outcome = 5
$P($ same day $)=\frac{5}{25}=\frac{1}{5}$
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Question 133 Marks
Two dice are thrown simultaneously. What is the probability that :
4 will not come up either time?
4 will come up at least once?
Answer
When two dice are thrown , total possible outccomes=36
Favorable outcomes for 4 will not come up either time:
{(1,1)(1,2)(1,3)(1,5)(1,6)}
{(2,1)(2,2)(2,3)(2,5)(2,6)}
{(3,1)(3,2)(3,3)(3,5)(3,6)}
{(5,1)(5,2)(5,3)(5,5)(5,6)}
{(6,1)(6,2)(6,3)(6,5)(6,6)}
Number of favorable outcomes = 25
p(4 will not come up) = 25/36
P(4 will come up ones) = 1- p(4 will not come up either time)
p(4 will come up once) =$1-\frac{25}{36}$
p(4 will come up once) = (36-25)/36=11/36
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Question 143 Marks
A circle with diameter $20\ cm$ is drawn somewhere on a rectangular piece of paper with length $40\ cm$ and width $30\ cm.$ This paper is kept horizontal on table top and a die, very small in size, is dropped on the rectangular paper without seeing towards it. If the die falls and lands on paper only, find the probability that it will fall and land : outside the circle
Answer
Diameter of the circle $= 20$
Radius $= 10\ cm$
Area of circle $=\pi r^2=\frac{22}{7} \times 10 \times=\frac{2200}{7} cm ^2$
Length of paper $= 40\ cm$
Width of paper $= 30\ cm$
Area of paper $=1200\ cm ^2$
Total possible outcomes $=$ area of rectangular paper
P(outside the circle) $= 1 - P$ (inside the circle)
$=1-\frac{11}{42}$
$ =\frac{31}{42}$
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Question 153 Marks
A circle with diameter 20 cm is drawn somewhere on a rectangular piece of paper with length 40 cm and width 30 cm. This paper is kept horizontal on table top and a die, very small in size, is dropped on the rectangular paper without seeing towards it. If the die falls and lands on paper only, find the probability that it will fall and land : inside the circle
Answer
Diameter of the circle= 20
Radius= 10 cm
Area of circle $=\pi r^2=\frac{22}{7} \times 10 \times 10 \times=\frac{2200}{7} cm ^2$
Length of paper =40 cm
Width of paper =30 cm
Area of paper $=1200 cm ^2$
|Total possible outccomes=arrea of rectangular paper
Since paper is kept on table top and die falls and land on paper.
Number of favorable outcomes= are of circle.
P(inside thhe circle)=$\frac{\text { Area of arde }}{\text { Area of rectangular paper }}=\frac{\frac{2200}{7}}{1200}=\frac{11}{42}$
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Question 163 Marks
Seven cards:- the eight, the nine, the ten, jack, queen, king and ace of diamonds are well shuffled. One card is then picked up at random. If the king is drawn and put aside, what is the probability that the second card picked up is :
an ace?
a king?
Answer
Total number of possible outcomes = 7
If a king is drawn and put aside, then total possible outcomes = 6
Number of favorable outcomes for an ace =1
p(card is an ace) = $\frac{1}{6}$
Now, for second pick number of king = 0
Number of favorable outcomes for a king = 0
p(card is a king) = $\frac{0}{6}=0$
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Question 173 Marks
Two dice are rolled together. Find the probability of getting : a multiple of 2 on one die and an odd number on the other die.
Answer
In throwing a dice, total possible outcomes= {1,2,3,4,5,6}
n(s)
for two dice,
$n(s)=6 \times 6=36$
E= event of getting a multiple of 2 on one die and an odd number on the othher = (2,1),(2,3),(2,5),(4,1),(4,3),(4,5),(6,1),(6,3),(6,5),(1,2),(3,2),(5,2),(1,4),(3,4),(5,4),(1,6),(3,6),(5,6)
n(E)=18
Probability of getting a multiple of 2 one die and an odd number on the other= $\frac{n(E)}{n(s)}=\frac{18}{36}=\frac{1}{2}$
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Question 183 Marks
Two dice are rolled together. Find the probability of getting : a total of at least 10
Answer
n(s) = 36 i.e.
(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)
(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)
(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)
(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)
(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)
(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)
In throwing a dice, total possible outcomes={1,2,3,4,5,6}
n(s)=6
for two dice, n(s)= 6 x 6=36
E= event of getting a total of at least 10={(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)}
n(E)=6
Probability of getting a total of at least 10=
$\frac{n(E)}{n(s)}=\frac{6}{36}=\frac{1}{6}$
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Question 193 Marks
In a single throw of two dice, find the probability of : an odd number on one dice and a number less than or equal to 4 on the other dice.
Answer
The number of possible outcomes is $6 \times 6=36$, We write them as given below :
1,11,21,31,41,51,6
2,12,22,32,42,52,6
3,13,23,33,43,53,6
4,14,24,34,44,54,6
5,15,25,35,45,55,6
6,16,26,36,46,56,6
n(s)=36
E= Event of getting an odd number on dice 1 and a number less than or equal to 4 on dice 2=
{(1,1)(1,2)(1,3)(1,,4)(1,5)}
{(2,1)(2,3)(2,5)}
{(3,1)(3,2)(3,3)(3,4)(3,5)}
{(4,1)(4,3)(4,5)}
{(5,1)(5,2)(5,3)(5,4)}
Therefore total number of favorable ways = 20 = n(E)
Probability of getting an odd number on dice 1 and a number less than or equal to 4 on dice 2 = $\frac{n(E)}{n(s)}=\frac{20}{36}=\frac{5}{9}$
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Question 203 Marks
In a single throw of two dice, find the probability of : a total of at most 10
Answer
The number of possible outcomes is $6 \times 6=36$
We write them as given below :
1,11,21,,31,41,51,6
2,12,22,32,42,52,6
3,13,23,33,43,53,6
4,14,24,34,44,54,6
5,15,25,35,45,55,6
6,16,26,36,46,56,6
n(s)=36
E= Event of getting a total of at most 10=
{(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)}
{(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)}
{(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)}
{(5,1)(5,2)(5,3)(5,4)(5,5)}
{(6,1)(6,2)(6,3)(6,4)}
Thereefore total number of favorable ways=33=n(E)
Probability of getting a total of at most 10=
$\frac{n(E)}{n(s)}=\frac{33}{36}=\frac{11}{12}$
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Question 213 Marks
In a single throw of two dice, find the probability of : an odd number as a sum
Answer
The number of possible outcomes is $6 \times 6=36$ We write them as given below :
1,11,21,31,41,51,6
2,12,22,32,42,52,6
3,13,23,33,43,53,6
4,14,24,34,44,54,6
5,15,25,35,45,55,6
6,16,26,36,46,56,6
n(s)=36
E=Event of getting an odd number as a sum ={(1,2)(1,4)(1,6)(2,1)(2,3)(2,5)(3,2)(3,4)(3,6)(4,1)(4,3)(4,5)(5,2)(5,6)(6,1)(6,3)(6,5)}
n(E)=18
Probability of getting an odd number as sum
$\frac{n(E)}{n(s)}=\frac{18}{36}=\frac{1}{2}$
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Question 223 Marks
In a single throw of two dice, find the probability of : a number less than 3 on each dice
Answer
The number of possible outcomes is$6 \times 6=36$,We write them as given below :
1,11,21,31,41,51,6
2,12,22,32,42,52,6
3,13,23,33,43,53,6
4,14,24,34,44,54,6
5,15,25,35,45,55,6
6,16,26,36,46,56,6
n(s)=36
E= Event of getting a doublet={(1,1)(2,2)(3,3)(4,4)(5,5)(6,6)}
n(E)=6
Probability of getting a doublet =$\frac{n(E)}{n(S)}=\frac{4}{36}=\frac{1}{6}$
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Question 233 Marks
In a single throw of two dice, find the probability of : a doublet
Answer
The number of possible outcomes is$6 \times 6=36$,We write them as given below :
1,11,21,31,41,51,6
2,12,22,32,42,52,6
3,13,23,33,43,53,6
4,14,24,34,44,54,6
5,15,25,35,45,55,6
6,16,26,36,46,56,6
n(s)=36
E= Event of getting a doublet={(1,1)(2,2)(3,3)(4,4)(5,5)(6,6)}
n(E)=6
Probability of getting a doublet =$\frac{n(E)}{n(S)}=\frac{6}{36}=\frac{1}{6}$
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Question 243 Marks
A ball is drawn at random from a box containing 12 white, 16 red and 20 green balls. Determine the probability that the ball drawn is : red or white
Answer
Total number of balls in the box $=48$
Total possible outcomes on drawing $g$ ball $=48$
$
n ( s )=48
$
red or a white ball $=12+16=28$ balls
Event of drrawing a red or white ball $= E =28$
$
n ( E )=28
$
Probability of not drawing a green ball $=\frac{n(E)}{n(s)}=\frac{28}{488}=\frac{7}{12}$
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Question 253 Marks
A ball is drawn at random from a box containing 12 white, 16 red and 20 green balls. Determine the probability that the ball drawn is : not green
Answer
Total number of balls in the box=48
Total possible outcomes on drawing a ball=48
n(s)=48
Event of drawing a green ball=E=2
n(E)=20
Probability of drawing a green ball $=\frac{n(E)}{n(s)}=\frac{20}{48}=\frac{5}{12}$
Probability of not drawing a green ball  = $1-\frac{5}{12}=\frac{12-5}{12}=\frac{7}{12}$
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Question 263 Marks
For an event E, write a relation representing the range of values of P(E)
Answer
The probability of an event lies between 0 and ' 1 '.
i.e $0 \leq p ( E ) \leq 1$.
Proof : Let's be the sample space and 'E' be the event
Then
$
0 \leq n ( E ) \leq n ( S )
$
$
0 / n(E) \leq / n(E) / \leq n(S) / n(S)
$
or $0 \leq P ( E ) \leq 1$
The number of elements in ' $E$ ' can't be less than ' 0 ' i.e. negative and greater than the number of elements in $S$
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Question 273 Marks
From a well shuffled deck of 52 cards, one card is drawn. Find the probability that the card drawn will : be a face card.
Answer
Total number of cards = 52
Total number of outcomes = P(s) = 52
There are 13 cards of each type. The cards of heart and diamond are red in colour. Spade and diamond are black. So,
There are 26 red cards and 26 black cards.
There are 52 cards in a deck of cards, and 12 of these cards are face cards (4 kings, 4 queens, and 4 jacks).
P(E) = 12
Probability of drwaing a face card $=\frac{ P ( E )}{ P ( s )}=\frac{12}{52}=\frac{3}{13}$
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Question 283 Marks
From a well shuffled deck of 52 cards, one card is drawn. Find the probability that the card drawn will : be a red card.
Answer
Total number of cards = 52
Total number of outcomes = P(s) = 52
There are 13 cards of each type. The cards of heart and diamond are red in colour. Spade and diamond are black. So, there are 26 red cards and 26 black cards.
(i) Number of black cards in a deck = 26
P(E) = favourable outcomes for the event of drawing a black card = 26
Number of red card in a deck =26
P(E)=favourable outcomes for the event of drwaing a
red card $=\frac{P(E)}{P(s)}=\frac{26}{52}=\frac{1}{2}$
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Question 293 Marks
From a well shuffled deck of 52 cards, one card is drawn. Find the probability that the card drawn will : not be a red card.
Answer
Total number of cards = 52
Total number of outcomes = P(s) = 52
There are 13 cards of each type. The cards of heart and diamond are red in colour. Spade and diamond are black. So, there are 26 red cards and 26 black cards.
Number of black card in a deak= 26
Therefore, number of non-red =52-26 = 26
p(E) =favourable outcomes for the event of not drwaing a red card= 26
probability of not drawing a red card =
$\frac{p(E)}{p(s)}=\frac{26}{52}=\frac{1}{2}$
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Question 303 Marks
Ramesh chooses a date at random in january for a party.
January
Mon 6132027
Tue 7142128
Wed 18152229
Thurs 29162330
Fri 310172431
sat 4111825
Sun 5121926
Find the probability that the choose :
(1) a Wednesday
(2) a friday
(3) a tuesday or a saturday.
Answer
Number of possible outcomes= number of day in the month=31
n(s)=31
(1) E= event of selection of a Tuesday or a saturday ={1,8,15,22,29}
n(E)=5
Probability of selection of a wednesday=
$P(s)=\frac{n(E)}{n(s)}=\frac{5}{31}$
E=event of selection of a friday ={3,10,17,24,31}
n(E)=5
probability of selection of a friday =
$P(s)=\frac{n(E)}{n(s)}=\frac{5}{31}$
E=event of seletionn of a tuesday or a saturday ={4,7,11,14,18,21,25,28}
n(E)=8
Probability of selection of a Tuesday or a saturday
$=P(s)=\frac{n(E)}{n(S)}=\frac{8}{31}$
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Question 313 Marks
In a musical chairs game, a person has been advised to stop playing the music at any time within 40 seconds after its start. What is the probability that the music will stop within the first 15 seconds? 
Answer
Total result = 0 sec to 40 sec
Total possible outcomes =40
n(s)=40
Favorable results = 0 sec to 15 sec
Favourable outcomes =15
n(E)=15
Probability that the music will stop in first 15 sec=
$\frac{n(E)}{n(S)}=\frac{15}{40}=\frac{3}{8}$
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Question 323 Marks
All the three face cards of spades are removed from a well shuffled pack of 52 cards. A card is then drawn at random from the remaining pack. Find the probability of getting:  a black card   
Answer
Total number of cards =52
3 face card of spades are removed
Remaining =52-3=49= number of possible outcomes
n(s)=49
Number of black cards left =23 cards (13 club +10 spade)
Event of drawing a black card=E=23
n(E)=23
Probability of drawing a black card=$\frac{n(E)}{n(s)}=\frac{23}{49}$
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Question 333 Marks
All the three face cards of spades are removed from a well shuffled pack of 52 cards. A card is then drawn at random from the remaining pack. Find the probability of getting:  a queen 
Answer
Total number of cards =52
3 face card of spades are removed
Remaining =52-3=49= number of possible outcomes
n(s)=49
Number of queen cards left= 3
Event of drawing a black face card=E=3
n(E)=3
Probability of drwaing a queen card =$\frac{n(E)}{n(s)}=\frac{3}{49}$


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Question 343 Marks
All the three face cards of spades are removed from a well shuffled pack of 52 cards. A card is then drawn at random from the remaining pack. Find the probability of getting:   a black face card 
Answer
Total number of cards =52
3 face card of spades are removed
Remaining =52-3=49= number of possible outcomes
n(s)=49
Number of black face cards left = 3 face cards of club
Event of drawing a black face cards =E=3
n(E)=3
Probability of drawing a black face card =$\frac{n(E)}{n(s)}=\frac{3}{49}$
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Question 353 Marks
A box contains 7 red balls, 8 green balls and 5 white balls. A ball is drawn at random from the box. Find the probability that the ball is:  neither red nor white.
Answer
Total number of balls in the box =7+8+5=20
Total possible outcomes = 20= n(s)
Neither red ball nor white ball = green ball
Event of not drawing a red or white ball =E= number of green ball n(E)=8
Probability of drawing a white ball = $\frac{n(E)}{n(S)}=\frac{8}{20}=\frac{2}{5}$

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Question 363 Marks
A man tosses two different coins (one of Rs 2 and another of Rs 5) simultaneously. What is the probability that he gets:   at most one head? 
Answer
When two coins are tosses simultaneously , the possiuble outcomes are {(H,H)(H,T)(T,H)(T,T)}
n(s)=4
The outcomes favourable to the event E,' at most one head are {(T,H)(H,T)(T,T)}
So, the number of outcomes favourable to E is 3=n(E)
Therefore, p(E)=
$\frac{n(E)}{n(s)}=\frac{3}{4}$
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Question 373 Marks
A man tosses two different coins (one of Rs 2 and another of Rs 5) simultaneously. What is the probability that he gets:  at least one head?  
Answer
When two coins are tossed simultaneossly, the possible outcomes are {(H,H)(H,T)(T,H)(T,T)}
n(s)=4
The outcomes favouraable to the event E,' at least one head' are
{(H,H)(H,T)(T,H)}
So, the number of outcomes favourable to E is 3=n(E)
Therefore, p(E)=$\frac{n(E)}{n(S)}=\frac{3}{4}$
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Question 383 Marks
A and B are friends. Ignoring the leap year, find the probability that both friends will have:   the same birthday? 
Answer
Out of the two friends, A's birthday can be any day of the year. Now, B's birthday can also be any day of 365 days in the year.
We assume that these 365 outcomes are qually likely.
P(A and B have the same birthday)
=1-p(both have different birthday)
(=1-\frac{364}{365}\left[\right.$ Using $\left.p\left(E^{\prime}\right)=1-p(E)\right]$
$\frac{1}{365}$
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Question 393 Marks
A box contains a certain number of balls. Some of these balls are marked $A,$ some are marked $B$ and the remaining are marked $C$. When a ball is drawn at random from the box $P(A)=\frac{1}{3}$ and $P(B)=\frac{1}{4}$. If there are $40$ balls in the box which are marked $C$, find the number of balls in the box.
Answer
$P(C)=1-[P(A)+P(B)]= 1-\left[\frac{1}{3}+\frac{1}{4}\right]=1-\frac{7}{12}=\frac{5}{12}$
$\text{Probability} = \text{ Number of favorable outcomes} / \text{Total $\mu$ mberofallpossib $\leq$ outcomes}$
Given that $40$ balls in the box are marked $C$.
$\Rightarrow \frac{5}{12}=\frac{40}{\text { Total number of all possible outcomes }}$
$\Rightarrow \text { Total number of all possible outcomes }=\frac{40 \times 12}{5}=96$
$\therefore$ the number of balls in the box is $96 .$
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Question 403 Marks
From a well shuffled deck of 52 cards, one card is drawn. Find the probability that the card drawn is:  a card with number between 2 and 9 
Answer
Total number of possible outcomes=52
n(s)=52
Number between 2 and 9 = {3,4,5,6,7,8}
Event of drawing card with number between 2 and 9=E= {6H cards,6D cards,6Ss cards,6c cards}
n(E)=24
Probability of drawing a card with number between 2 and 9= $\frac{n(E)}{n(S)}=\frac{24}{52}=\frac{6}{13}$
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Question 413 Marks
From a well shuffled deck of 52 cards, one card is drawn. Find the probability that the card drawn is   a card with number less than 8
Answer
Total number of possible outcomes=52
n(s) = 52
Number less than 8={2,3,4,5,6,7}
Event of drawing a card with number less than 8 = E = [ 6H cards , 6D cards ,6s cards ,6c cards]
n(E) = 24
Probability of drawing a card with number less than 8 =$\frac{n(E)}{n(s)}=\frac{24}{52}=\frac{6}{13}$
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Question 423 Marks
From a pack of 52 playing cards, all cards whose numbers are multiples of 3 are removed. A card is now drawn at random. What is the probability that the card drawn is An even-numbered red card? 
Answer
No. of total cards $=52$
cards removed of 4 colours of multiples of 3
$
=3,6,9=4 \times 3=12
$
Remaining cards $=52-12=40$
An even number red cards $=2,4,8,10$ card $=2 x$
$
4=8
$
$\Rightarrow$ Probability $P(E)=\frac{8}{40}=\frac{1}{5}$
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Question 433 Marks
From a pack of 52 playing cards, all cards whose numbers are multiples of 3 are removed. A card is now drawn at random. What is the probability that the card drawn is A face card (King, Jack or Queen). 
Answer
No. of total cards $=52$ cards removed of 4 colours of multiples of 3
$
=3,6,9=4 \times 3=12
$
Remaining cards $=52-12=40$
No.of face cards $=12$ cards
$\Rightarrow$ Probability $p(E)=\frac{12}{40}=\frac{3}{10}$
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Question 443 Marks
Offices in Delhi are open for five days in a week (Monday to Friday). Two employees of an office remain absent for one day in the same particular week. Find the probability that they remain absent on:    different days 
Answer
Total number of possible outcomes $=5 \times 5=25$
The possible outcomes are :
MM,MT,MW,MTh,MF,TM,TT,TW,TTh,TF,WM,WT,WW,WTh,W F,ThM,ThT,ThM, ThT,Thw,ThTh,THF,FM,FT,FW,FTh,FF
$p$ (absent on diff days) $=1- p$ (absent on same days)
$
=1-\frac{1}{5}
$
$
=\frac{4}{5}
$
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Question 453 Marks
Offices in Delhi are open for five days in a week (Monday to Friday). Two employees of an office remain absent for one day in the same particular week. Find the probability that they remain absent on:   consecutive day 
Answer
Total number of possible outcomes $=5 \times 5=25$
The possible outcomes are :
MM,MT,MW,MTh,MF,TM,TT,TW,TTh,TF,WM,WT,WW,WTh,W F,ThM,ThT,ThM, ThT,Thw,ThTh,THF,FM,FT,FW,FTh,FF
Favorable outcomes for two employee remaining absent on consecutive days : MT,TM,TW,WT,WTh,ThW,ThF,FTh
Number of favorable outcomes $=8$
$P ($ consecutive days $)=\frac{8}{25}$
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Question 463 Marks
Offices in Delhi are open for five days in a week (Monday to Friday). Two employees of an office remain absent for one day in the same particular week. Find the probability that they remain absent on:  the same day
Answer
Total number of possible outcomes=$5 \times 5=25$
The possible outcomes are :
MM,MT,MW,MTh,MF,TM,TT,TW,TTh,TF,WM,WT,WW,WTh,WF,ThM,ThT,ThM,ThT,Thw,ThTh,THF,FM,FT,FW,FTh,FF
Favorable outcome for two employees remaining absennt on same day are: MM,TT,WW,ThTh,FF
Number of favorable outcome = 5
$P($ same day $)=\frac{5}{25}=\frac{1}{5}$
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Question 473 Marks
Three coins are tossed together. Write all the possible outcomes. Now, find the probability of getting:  at least one tail 
Answer
When three coins are tossed, possible outcomes are:
HHH, HHT, HTH,HTT,THH, THT,TTH,TTT
Total possible outcomes $=8$
Favorable outcomes for at least one tails = HHT,THH,HTH,HTT,THT,TTH,TTT
Number of favorable outcomes $=7$
$
P (\text { at least one tail })=\frac{7}{8}
$
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Question 483 Marks
Two dice (each bearing numbers 1 to 6) are rolled together. Find the probability that the sum of the numbers on the upper-most faces of two dice is:  less than 6 
Answer
When two dice are rolled, total number of possible outcomes $=36$
Favorable outcomes for the sum of numbers less than i.e. $2,3,4$, or 5 are :
$
\{(1,1)(1,2)(1,3)(1,4)(2,1)(2,2)(2,3)(3,1)(3,2)(4,1)\}
$
Number of favorable outcomes $=10$
$P($ getting a sum of less than 6$)=\frac{10}{36}=\frac{5}{18}$
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Question 493 Marks
A circle with diameter $20 \ cm$ is drawn somewhere on a rectangular piece of paper with length $40 \ cm$ and width $30 \ cm.$ This paper is kept horizontal on table top and a die, very small in size, is dropped on the rectangular paper without seeing towards it. If the die falls and lands on paper only, find the probability that it will fall and land: outside the circle
Answer
Diameter of the circle $= 20$
Radius $= 10 \ cm$
Area of circle $=\pi r^2=\frac{22}{7} \times 10 \times=\frac{2200}{7} \ cm ^2$
Length of paper $= 40 \ cm$
Width of paper $= 30 \ cm$
Area of paper $=1200 \ cm ^2$
Total possible outcomes $=$ area of rectangular paper
$P($outside the circle$) = 1 - P($inside the circle$)$
$=1-\frac{11}{42}$
$=\frac{31}{42}$
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Question 503 Marks
A dice is thrown once. What is the probability of getting a number: less than or equal to 2?  
Answer
Number of possible outcomes when dice is thrown ={1,2,3,4,5,6}
n(S)=6
Event of getting a number less than or equal to 2=E{1,2} n(E)=2
Probability of getting a number less than or equal to 2=
$\frac{n(E)}{n(s)}=\frac{2}{6}=\frac{1}{3}$
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