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Mathematics [2 Mark Question Answer]

Probability · ICSE STD 10 (ICSE - English Medium). 232 questions.

Questions · Page 2 of 5

[2 Mark Question Answer]

Question 512 Marks
A bag contains 3 red balls, 4 blue balls and 1 yellow ball, all the balls being identical in shape and size. If a ball is taken out of the bag without looking into it; find the probability that the ball is: yellow
Answer
Total number of balls in the bag=3+4+1=8 balls
Number of possible outcomes=8=n(s)
Event of drawing a yellow ball={Y}
n(E) = 1
Probability of drawing a yellow ball =$\frac{n(E)}{n(s)}=\frac{1}{8}$

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Question 522 Marks
Two dice are thrown at the same time. Find the probability that the sum of the two numbers appearing on the top of the dice is : 8
Answer
The number of possible outcomes =$6 \times 6=36$
The outcomes favourble to the event the sum of the two number is 8=E={(2,6),(3,5),(4,4),(5,3),(6,2)}
The number of outcomes favourable to E= n(E) = 5
Hence,$p(E)=\frac{n(E)}{n(S)}=\frac{5}{36}$
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Question 532 Marks
In a bundle of 50 shirts, 44 are good, 4 have minor defects and 2 have major defects. What is the probability that : it is acceptable to a trader who rejects only a shirt with major defects?
Answer
Total number of shirts= 50
Total number of elementary events= 50= n(s)
Since trader rejects shirts with major defect only and number of shirts with major defect =2
Event of accepting shirts=50-2=48=n(E)
Probability of accepting shirts =$\frac{n(E)}{n(S)}=\frac{48}{50}=\frac{24}{25}$
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Question 542 Marks
In a bundle of 50 shirts, 44 are good, 4 have minor defects and 2 have major defects. What is the probability that : it is acceptable to a trader who accepts only a good shirt?
Answer
Total number of shirts= 50
Total number of elementary events= 50= n(s)
Since trader accepts only good shirts and number of good shirts= 44
Event of accepting good shirts =44= n(E)
Probability of accepting a good shirts =
$\frac{n(E)}{n(s)}=\frac{44}{50}=\frac{22}{25}$
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Question 552 Marks
A card is drawn from a well shuffled pack of 52 cards. Find the probability that the card drawn is : a red or a king
Answer
Number of possible outcomes when card is drawn from pack of 52 cards=52
n(s)=52
E= event of drawing a red oor a king =26 red cards(13h+13D)+2 black kings[since 26 red cards contain 2 red kings]
n(E)=28
Probability of drawing red or a king
$=\frac{n(E)}{n(s)}=\frac{28}{52}=\frac{7}{13}$
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Question 562 Marks
A card is drawn from a well shuffled pack of 52 cards. Find the probability that the card drawn is : ace and king
Answer
Number of possible outcomes when card is drawn from pack of 52 card=52
n(s)=52
A card cannot br both an ace as well as a king.
E= event of drawing an ace and a king =0
n(E)=0
Probability of drawing an ace and king=0
$\frac{n(E)}{n(s)}==\frac{0}{52}=0$
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Question 572 Marks
A card is drawn from a well shuffled pack of 52 cards. Find the probability that the card drawn is : Jack or queen
Answer
Number of possible outcomes when card is drawn from pack of 52 cards=52
n(s)=52
E=event of drawing a jack or a queen= {JH,JS,JD,JC,QH,QS,QD,QC}
n(E)=8
Probability of drawinng a jack or queen
$=\frac{n(E)}{n(s)}=\frac{8}{52}=\frac{2}{13}$
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Question 582 Marks
A card is drawn from a well shuffled pack of 52 cards. Find the probability that the card drawn is : 5 of heart or diamond
Answer
Number of possible outcomes when card is drawn from pack of 52 card = 52
n(52)=52
E= event of drawing a 5 of heart or of diamond={5H,5D}
n(E)=2
$\frac{n(E)}{n(s)}=\frac{2}{52}=\frac{1}{26}$
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Question 592 Marks
A card is drawn from a well shuffled pack of 52 cards. Find the probability that the card drawn is : a face card
Answer
Number of possible outcomes when card is drawn from pack of 52 cards =52
n(s)=52
Number of face cards ( 4 king +4 queen $4 jacks )=12= E =$ event of drawing a face card
n(E)=12
Probability of drwaing a red card=
$\frac{n(E)}{n(s)}=\frac{12}{52}=\frac{3}{13}$
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Question 602 Marks
A card is drawn from a well shuffled pack of 52 cards. Find the probability that the card drawn is : a red card
Answer
Number of possible outcomes when card is drawn from pack of 52 cards =52
n(s)=52
Number of red cards (hearts+diamonds)=26=E=event of drawing a red card
n(E)=26
Probability of drwaing a red card=
$\frac{n(E)}{n(s)}=\frac{26}{52}=\frac{1}{2}$
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Question 612 Marks
A card is drawn from a well shuffled pack of 52 cards. Find the probability that the card drawn is : a spade
Answer
Number of possible outcome when card id drawn from pack of 52 cards=52
n(s)=52
Number of spade cards = 13=E event of drwing a spade
n(E)=13
Probability of drawing a sade =$\frac{n(E)}{n(s)}=\frac{13}{52}=\frac{1}{4}$
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Question 622 Marks
If two coins are tossed once, what is the probability of getting : both heads or both tails
Answer
When two coins are tossed possible number of outcomes={HH,TH,HT,TT}
n(s)=4
E=event of getting both heads or both tails = {HH,TT}
n(E)=2
Probability of getting both heads or both tails=
$\frac{n(E)}{n(s)}=\frac{2}{4}=\frac{1}{2}$
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Question 632 Marks
If two coins are tossed once, what is the probability of getting : at least one head.
Answer
When two coins are tossed together possible number of outcomes= {HH,TH,HT,TT}
n(s)= 4
E= event of getting at least one head ={HH,TH,HT}
n(E)=3
Probability of getting at least one head =
$\frac{n(E)}{n(s)}=\frac{3}{4}$
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Question 642 Marks
If two coins are tossed once, what is the probability of getting : both heads
Answer
When two coins are tossed together possible number of outcomes= {HH,TH,HT,TT}
n(s)=4
E= event of getting both heads= {HH}
n(E)=1
Probability of getting both heads=
$\frac{n(E)}{n(s)}=\frac{1}{4}$
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Question 652 Marks
A pair of dice is thrown. Find the probability of getting a sum of 10 or more, if 5 appears on the first die.
Answer
In throwing dice, total possible outcome ={1,2,3,4,5,6}
n(s)=6
for two dice, n(s) = $6 \times 6=36$
Favorable cases where the sum is 10 or more with 5 on 1st die = {(5,5),(5,6)}
Event of getting the sum is 10 or more with 5 on 1st die = n (E)=2
Probability of getting a sum of 10 or more with 5 on 1st die =
$\frac{n(E)}{n(s)}=\frac{2}{36}=\frac{1}{18}$
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Question 662 Marks
A book contains 85 pages. A page is chosen at random. What is the probability that the sum of the digits on the page is 8?
Answer
Number of pages in the book =85
Number of possible outcomes=n(s)=85
out of 85 pages, pages that sum up to 8= {8,17,26,35,44,53,62,71,80}
pages that sum up to 8 = n(E) = 9
Probability of choosing a page with the sum of digits on the page equal 8= $\frac{n(E)}{n(s)}=\frac{9}{85}$
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Question 672 Marks
A die is thrown once. Find the probability of getting a number : greater than 6
Answer
In throwing a dice, total possible outcomes= {1,2,3,4,5,6,}
n(s) = 6
On a dice, numbers greater than 6 = 0
n(E) = 0
probabbility of getting a number greater than 6 = $\frac{n(E)}{n(s)}=\frac{0}{6}=0$
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Question 682 Marks
A die is thrown once. Find the probability of getting a number : less than 8
Answer
In throwing a dice, total possible outcomes = {1,2,3,,4,5,6}
n(s)=6
On a dice, number less thaan 8={1,2,3,4,5,6}
n(E)=6
Probability of getting a number less than 8 $\frac{n(E)}{n(s)}=\frac{6}{6}=1$
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Question 692 Marks
A die is thrown once. Find the probability of getting a number : greater than or equal to 4
Answer
In throwing a dice, total possible outcomes= {1,2,,3,4,5,6}
n(s)=6
On a dice, numbers greater than or equal to ={4,5,6}
n(E)=3
Probability of getting a number greater than or equal to 4=
$\frac{n(E)}{n(s)}=\frac{3}{6}=\frac{1}{2}$
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Question 702 Marks
A die is thrown once. Find the probability of getting a number : less than 3
Answer
In throwing a dice, total possible outcomes = {1,2,3,4,,5,6}
n(S) = 6
on a dice, number less than 3={1,2}
n(E) = 2
Probability of getting a number less than $3=\frac{n(E)}{n(s)}=\frac{2}{6}=\frac{1}{3}$ .
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Question 712 Marks
From 25 identical cards, numbered 1, 2, 3, 4, 5, ……, 24, 25: one card is drawn at random. Find the probability that the number on the card drawn is a multiple of : 3 or 5
Answer
There are 25 cards from which one card is drawn
Total number of elementary events=n(s)=25
From numbers 1 to 25, there are 12 numbers which are multiple of 3 or 5 i,e {3,5,6,9,10,12,15,18,20,,,21,24,25} favourable number of event n(E) =12
Probability of selecting card with a multiple of 3 or 5 = $\frac{n(E)}{n(s)}=\frac{12}{25}$
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Question 722 Marks
From 25 identical cards, numbered 1, 2, 3, 4, 5, ……, 24, 25: one card is drawn at random. Find the probability that the number on the card drawn is a multiple of : 3 and 5
Answer
There are 25 careds from which one card is drawn
Total number of elementary event = n(s) = 25
From numbers 1 to 25, there is only one number which is multiple of 3 and 5 i.e. {15} favourable number of events = n(E)=1
probabilty of selecting a card with a multiple of 3 and 5 =
$\frac{n(E)}{n(S)}=\frac{1}{25}$
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Question 732 Marks
From 25 identical cards, numbered 1, 2, 3, 4, 5, ……, 24, 25: one card is drawn at random. Find the probability that the number on the card drawn is a multiple of : 5
Answer
There are 25 cards from which one card is drawn.
Total number of elementry event ==n(s)=25
From number 1 to 25, there are 5 number which are multiple of 5 i.e. [5,10,15,20,25] Favourable number of event = n(E)=5
Probability of selecting cards with a multiple of 5= $\frac{n(E)}{n(s)}=\frac{5}{25}=\frac{1}{5}$
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Question 742 Marks
From 25 identical cards, numbered 1, 2, 3, 4, 5, ……, 24, 25: one card is drawn at random. Find the probability that the number on the card drawn is a multiple of : 3
Answer
There are 25 cared from which one card is drawn.
Total number of elementary event=n(s)=25
From numbers 1 to 25, there are 8 numbers which are multiple of i.e [3,6,9,12,15,18,21,24] Favorable number of event =n(E)=8
Probability of selecting a card with a multiple of 3=
$\frac{n(E)}{n(s)}=\frac{8}{25}$
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Question 752 Marks
Hundred identical cards are numbered from 1 to 100. The cards The cards are well shuffled and then a card is drawn. Find the probability that the number on card drawn is : less than 48
Answer
There are 100 cards from which one card is drawn.
Total number opf elementry event =n(s)=100
From number 1 to 100, there are 47 number which are less than 48 i.e. (1,2,................46,47) favorable number of event =n(E)=47
Probabilty of selecting a card with a number less than 48 = $\frac{n(E)}{n(s)}-\frac{47}{100}$
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Question 762 Marks
Hundred identical cards are numbered from 1 to 100. The cards The cards are well shuffled and then a card is drawn. Find the probability that the number on card drawn is : greater than 85
Answer
There are 100 cards from which one card is drawn.
Total number of elementary event =n(s0=100
From numbers 1 to 100 there are 15 numbers which are greater than 85 i.e {86,87,.......98,99,100} favorable number of event =n(E) =15
Probabilty of selecting card with a number greater than 85=
$\frac{n(E)}{n(s)}-\frac{15}{100}-\frac{3}{20}$
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Question 772 Marks
Hundred identical cards are numbered from 1 to 100. The cards The cards are well shuffled and then a card is drawn. Find the probability that the number on card drawn is : between 40 and 60
Answer
There are 100 cards from which one card is drawn.
Total number of elementary event = n(s)=100
From numbers 1 to 100, there are 19 number which are between 40 and 60 i.e. {41,42,43,44,45,46,4,748,49,50,51,52,53,54,55,56,57,58,59} favourable number of event = n(E)=19
Probability of selecting a cards between 40 and 60
$=\frac{n(E)}{n(S)}-\frac{19}{100}$
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Question 782 Marks
Hundred identical cards are numbered from 1 to 100. The cards The cards are well shuffled and then a card is drawn. Find the probability that the number on card drawn is : a multiple of 6
Answer
There are 100 cards from which one card is drawn.
Total number of elementary event -n(s)-100
From number 1 to 100, there are 16 numbers which are multiple of 6 i.e.{6,12,18,24,30,36,42,48,54,60,66,72,78,84,90,96}
favorable number of event =n(E)=16
Probability of seleting card with a multiple of 6=
$\frac{n(E)}{n(s)}-\frac{16}{100}-\frac{4}{25}$
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Question 792 Marks
Hundred identical cards are numbered from 1 to 100. The cards The cards are well shuffled and then a card is drawn. Find the probability that the number on card drawn is : a multiple of 5
Answer
There are 100 cards from which one card is drawn.
Total number of elementary events - n(s)-100
From numbers 1 to 100, threr are 20 numbers which are multiple of i.e {5,10,15,20,25,30,,35,40,45,50,55,60,65,70,75,80,85,90,95,100} favorable number of event =n(E)=20
Probability of selecting card with a multiple of 5 =
$\frac{n(E)}{n(s)}-\frac{20}{100}-\frac{1}{5}$
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Question 802 Marks
Nine cards (identical in all respects) are numbered 2 to 10. A card is selected from them at random. Find the probability that the card selected will be : an even number or a multiple of 3
Answer
There are 9 cards from which one card is drawn.
Total number of elementry events = n(E)=9
From numbers 2 to 10, there are 7 numbers which are even numbers or a multiple of 3 i.e. 2, 3, 4, 6, 8, 9, 10
Favorable number of events = n(E) = 7
Probability of selecting a card with a number which is an even number or a multiple of 3 = $\frac{n(E)}{n(s)}=\frac{7}{9}$
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Question 812 Marks
Nine cards (identical in all respects) are numbered 2 to 10. A card is selected from them at random. Find the probability that the card selected will be : an even number and a multiple of 3
Answer
There are 9 cards from which one card is drawn.
favorable number of event =n(s)=9
From numbers 2 to 10, there is one number which is an even number as well as multiples of 3 i.e.6
favourable number of event = n(E)=1
probability of selecting card with a multiple of 3
$=\frac{n(E)}{n(s)}=\frac{1}{9}$
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Question 822 Marks
Thirty identical cards are marked with numbers 1 to 30. If one card is drawn at random, find the probability that it is : a multiple of 3 or 5
Answer
There are 30 aaards from which one card is drawn.
Total number of elementary events=n(s)=30
From numbers 1 to 30, there are 14 numbers which is multiple of 3 or 5 i.e {3,5,6,9,10,12,15,18,20,21,24,25,27,30} favorable number of event =n(E)=14 Probability of selecting a card with a multiple of 3 or 5 =$\frac{n(E)}{n(s)}=\frac{14}{30}=\frac{7}{15}$
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Question 832 Marks
Thirty identical cards are marked with numbers 1 to 30. If one card is drawn at random, find the probability that it is : a multiple of 3 and 5
Answer
There are 30 cards from which one card is drawn.
Total number of elementary events= n(s)=30
From numbers 1 to 30, there are 2 numbers which are multiple of 3 and 5 i.e. {15,30} favourable number of events =n(E)=2
Probability of selecting card with a multiple of 3 and 5=$\frac{n(E)}{n(s)}=\frac{2}{30}=\frac{1}{15}$
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Question 842 Marks
Thirty identical cards are marked with numbers 1 to 30. If one card is drawn at random, find the probability that it is : a multiple of 4 or 6
Answer
There are 30 cards from which one card is drawn.
Total number of elementary events =n(s)=30
From numbers 1 to 30, there are 10 numbers which are multiple of 4 or 6 i.e {4,6,8,12,16,18,20,24,28,30} Favourable number of event =10
Probability os selecting a card with a multiple of 4 or 6
$\frac{n(E)}{n(s)}=\frac{10}{30}=\frac{1}{3}$
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Question 852 Marks
Nine cards (identical in all respects) are numbered 2 to 10. A card is selected from them at random. Find the probability that the card selected will be : a multiple of 3
Answer
There are 9 cards from which one card is drawn.
Total number of elements events = n(s)=9
From numbers 2 to 10, there are 3 numbers which are multiples of 3 i.e 3,6,9
favorable number of events =n(E)=3
Probability of selecting card with a multiple of
$3=\frac{n(E)}{n(s)}=\frac{3}{9}=\frac{1}{3}$
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Question 862 Marks
A card is drawn from a pack of 52 cards. Find the probability that the card drawn is : a queen or a jack
Answer
Number of possible outcomes when card is drawn from pack of 52 ards =52
n(s)=52
E= event of drawing a jack or queen = {JH,JS,JD,JC,QH,QS,QD,QC}
n(E)=8
Pprobability of drawing a jack or a queen =$\frac{n(E)}{n(s)}=\frac{8}{52}=\frac{2}{13}$
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Question 872 Marks
A card is drawn from a pack of 52 cards. Find the probability that the card drawn is : not a club
Answer
Number of possible outcomes when card is drawn from pack of 52 card =52
n(s)=52
Number of club cards=13=E= event of drawing a club card
n(E)=13
Probabilty of not drawing a club card = $\frac{n(E)}{n(s)}=\frac{13}{52}=\frac{1}{4}$
Probability of not drawing a club card$=1-\frac{1}{4}=\frac{4-1}{4}=\frac{3}{4}$
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Question 882 Marks
A card is drawn from a pack of 52 cards. Find the probability that the card drawn is : ace of diamonds
Answer
Number of possible outcomes when card is drawn from pack of 52 cards=52
n(s)=52
There is only one ace of diamonds.
E= event of drawing an ace of diamond
n(E)=1
Probability of drawing an ace of diamonds= $\frac{n(E)}{n(s)}=\frac{1}{52}$

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Question 892 Marks
A card is drawn from a pack of 52 cards. Find the probability that the card drawn is : a black ace
Answer
Number of possible outcomes when card is drawn from pack of 52 cards=52
n(s)=52
Number of black ace cards=2 =E= event of drawing an ace
n(E)=2
Probability of drawing a black ace= $\frac{n(E)}{n(s)}=\frac{2}{52}=\frac{1}{26}$
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Question 902 Marks
A card is drawn from a pack of 52 cards. Find the probability that the card drawn is : an ace
Answer
Number of possible outcomes when card is drawn from pack of 52 cards=52
n(s)=52
Numbber of ace cardd=4=E=event of drawing an ace
n(E)=4
Probability of drawing an ace=
$=\frac{n(E)}{n(s)}=\frac{4}{52}=\frac{1}{13}$
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Question 912 Marks
A card is drawn from a pack of 52 cards. Find the probability that the card drawn is : a spade
Answer
Number of possible outcomes wwhen card is drawn from packk of 52 card=52
n(s)=52
Number of ace cards=13=E=event of drawing a spade
n(E)=13
Probability of drawing spade=
$\frac{n(E)}{n(s)}=\frac{13}{52}=\frac{1}{4}$
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Question 922 Marks
A card is drawn from a pack of 52 cards. Find the probability that the card drawn is : a black card
Answer
Number of possible outcomes when card is drawn from pack of 52 cards=52
n(E)=52
Number of black cards (spade+clubs)=26=E
n(E)=26
Probability of drawinng a black card=
$\frac{n(E)}{n(s)}=\frac{26}{52}=\frac{1}{2}$
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Question 932 Marks
A card is drawn from a pack of 52 cards. Find the probability that the card drawn is : a red card
Answer
Number of possible outcomes when card is drawn from pack of 52 cards=52
n(s)=52
Number of red card (hearts+diamonds)=26=E
n(E)=26
Probabilty of drawing a red card=$\frac{n(E)}{n(S)}=\frac{26}{52}=\frac{1}{2}$
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Question 942 Marks
Nine cards (identical in all respects) are numbered 2 to 10. A card is selected from them at random. Find the probability that the card selected will be : an even number
Answer
There are 9 cards from which one card is drawn.
Total number of elementary events =n(s)=9
From numbers 2 to 10, there are 5 even numbers i.e 2,4,6,8,10
Favorable number of event =n(E)=5
Probability of selecting card with an even number = $\frac{n(E)}{n(S)}=\frac{5}{9}$
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Question 952 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
Answer
Total number of balls in the box=48
Total possible outcomes on drawinng a ball =48
n(s)=48
Event of drawing a red ball=E=16
n(E)=16
Probability of drawing a red ball = $\frac{n(E)}{n(s)}=\frac{16}{48}=\frac{1}{3}$
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Question 962 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 : white
Answer
Total number of balls in the box =48
Total possible outcomes on drawing a ball=48/
n(s)=48
Event of drawing a white ball =E=12
n(E)=12
Probability of drawing a white ball
$=\frac{n(E)}{n(s)}=\frac{12}{48}=\frac{1}{4}$

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Question 972 Marks
A bag contains 16 colored balls. Six are green, 7 are red and 3 are white. A ball is chosen, without looking into the bag. Find the probability that the ball chosen is : green or red or white
Answer
Balls in the bag=16= Number of balls that could be drawn
n(s)=16
E= Event of drawing a green or a white or a red ball= number of green balls++number of white balls+number of red balls=6+3+7=16
n(E)=16
Probability of drawing a green or a white or a red ball= $\frac{n(E)}{n(s)}-\frac{16}{16}-1$
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Question 982 Marks
A bag contains 16 colored balls. Six are green, 7 are red and 3 are white. A ball is chosen, without looking into the bag. Find the probability that the ball chosen is : white or green
Answer
Balls in the bag =16= Number of balls that could be drawn
n(s)=16
E= Event of drawing a green or a white ball=number of green balls+number of white balls=6+3=9
n(E)=9
Probability of drawing a green or a white ball=$\frac{n(E)}{n(s)}=\frac{9}{16}$
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Question 992 Marks
A bag contains 16 colored balls. Six are green, 7 are red and 3 are white. A ball is chosen, without looking into the bag. Find the probability that the ball chosen is : green or red
Answer
Balls in the bag=16= Number of balls that could be drawn
n(s)=16
E=Event of drawwing a green or a red ball= number of green balls+number of red balls=6+7=13
n(E)=13
Probability of drawing a green or a red ball $=\frac{n(E)}{n(s)}=13 / 16$
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Question 1002 Marks
A bag contains 16 colored balls. Six are green, 7 are red and 3 are white. A ball is chosen, without looking into the bag. Find the probability that the ball chosen is : not white
Answer
Balls in the bag =16= Number of balls that could be drawn
n(s)=16
Not a white ball =16-3=13=E n(E)=3
n(E)=13
Probability of not drawing a white ball $=\frac{n(E)}{n(s)}=\frac{13}{16}$
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[2 Mark Question Answer] - Page 2 - Mathematics STD 10 Questions - Vidyadip