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Mathematics [3 marks sum]

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

Questions · Page 2 of 2

[3 marks sum]

Question 513 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 523 Marks
A bag contains 10 red balls, 16 white balls and 8 green balls. A ball is drawn out of the bag at random. What is the probability that the ball drawn will be:   not red? 
Answer
Total number of possible outcomes = 10+16+8=34 balls
n(s)=34
Favorable outcomes for not a red ball= favourable outcomes for white or green ball
Number of favorable outcome for white or green ball=16+8=24=n(E)
Probability for not drawing a red ball =
$\frac{n(E)}{n(s)}=\frac{24}{34}=\frac{12}{17}$
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Question 533 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 543 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 553 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 563 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 573 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 583 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 593 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 603 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 613 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 623 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 633 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 643 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 653 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 663 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 673 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 683 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 693 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 703 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 713 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 723 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 733 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 743 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 753 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 763 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 773 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 783 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 793 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 803 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 813 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 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 823 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 833 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 843 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 ≤p(E) ≤1.
Proof : Let's be the sample space and 'E' be the event
Then
0 ≤ n(E) ≤ n(S)
0/n(E) ≤ / n(E)/ ≤ n(S)/ n(S)
or 0 ≤P(E) ≤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 853 Marks
In a race between Mahesh and John, the probability that John will lose the race is 0.54. Find the probability of:
(i) winning of Mahesh
(ii) winning of John
Answer
(i) But if John looses, Mahesh wins
Hence, probability of John losing the race = Probability of Mahesh winning the race since it is a race between these two only
Therefore, P(winning of Mahesh) = 0.54
(ii) P(winning of Mahesh) + P(winning of John) = 1
0.54 + P(winning of John) = 1
P(winning of John) = 1 – 0.54
P(winning of John) = 0.46
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Question 863 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 873 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 883 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 893 Marks
From a well shuffled deck of 52 cards, one card is drawn. Find the probability that the card drawn will:   be a black 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
probability of drawing a black card $=\frac{ p ( E )}{ P ( s )}=\frac{26}{52}=\frac{1}{2}$
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[3 marks sum] - Page 2 - Mathematics STD 10 Questions - Vidyadip