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Mathematics [2 Mark Question Answer]

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

Questions · Page 3 of 5

[2 Mark Question Answer]

Question 1012 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
Answer
Balls in the bag =16= Number of balls that could be drawn
n(s)=16
E= Event of drawing a white ball =number of white balls=3
n(E)=3
Probability of drawing white ball=$\frac{n(E)}{n(s)}=\frac{3}{16}$
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Question 1022 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 red
Answer
Balls in the bag=16= Number of balls that could be drawn
n(s)=16
Not a red ball=16- number of red balls=16=7=9=E
n(E)=9
Probability of not drawing a red ball $=\frac{n(E)}{n(s)}=\frac{9}{16}$
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Question 1032 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 : red
Answer
Balls in the bag=16= Number of balls that could be drawn
n(s)=16
E=Event of drawing a red ball= number of red balls=7
n(E)=7
Probability of drawing a red ball $=\frac{n(E)}{n(S)}=\frac{7}{16}$
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Question 1042 Marks
Write the probability of an event when impossible.
Answer
The probability of an impossible event is 0 i.e p(s)=0
Proof : Since E has no element, n(E) =0
from definition of probability :
p(s)=n(E) /n(S)=0/n(s)
p(s)=0
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Question 1052 Marks
Write the probability of a sure event.
Answer
The probability of a sure event is $1$ i.e.p(s) $=1$ where $s$ is the sure event.
Proof: In a sure event $n(E)=n(s)$
[Since number of elements in Event ' $E$ ' will be equal to the number of element in samplespace.]
By definitions of probability :
$P(s)=n(E) / n(s)=1$
$p(s)=1$
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Question 1062 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 1072 Marks
In a T.T. match between Geeta and Ritu, the probability of the winning of Ritu is 0.73. Find the probability of : not winning of Ritu
Answer
Not winning of Ritu is a complementary event to winning of Ritu
Therefore,
P(winning of Ritu) + P(not winning of Ritu) = 1
P(not winning of Ritu) = 1 – P(winning of Ritu)
P(not winning of Ritu) = 1 – 0.73
P(not winning of Ritu) = 0.27
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Question 1082 Marks
In a T.T. match between Geeta and Ritu, the probability of the winning of Ritu is 0.73. Find the probability of: winning of Geeta
Answer
Winning of Geeta is a complementary event to winning of Ritu
Therefore,
P(winning of Ritu) + P(winning of Geeta) = 1
P(winning of Geeta) = 1 – P(winning of Ritu)
P(winning of Geeta) = 1 – 0.73
P(winning of Geeta) = 0.27
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Question 1092 Marks
If the probability of happening an event A is 0.46. What will be the probability of not happening of the event A?
Answer
P(A) = 0.46
Let P(B) be the probability of not happening of event A
We know,
P(A) + P(B) = 1
P(B) = 1 – P(A)
P(B) = 1 – 0.46
P(B) = 0.54
Hence the probability of not happening of event A is 0.54
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Question 1102 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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Question 1112 Marks
In a single throw of a die, find the probability that the number:will be an odd number.
Answer
sample space={1,2,3,4,5,6}
n(s) =6
E=event of not getting an even number ={1,3,5}
n(E) =3
probability of a not getting an even number=
$\frac{ n ( E )}{ n ( s )}=\frac{3}{6}=\frac{1}{2}$
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Question 1122 Marks
In a single throw of a die, find the probability that the number : will not be an even number.
Answer
sample space={1,2,3,4,5,6}
n(s) =6
E=event of not getting an even number ={1,3,5}
n(E) =3
probability of a not getting an even number=
$\frac{ n ( E )}{ n ( s )}=\frac{3}{6}=\frac{1}{2}$
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Question 1132 Marks
In a single throw of a die, find the probability that the number : will be an even number.
Answer
sample space ={1,2,3,4,5,6}
n(s) =6
E=event of getting an even number ={2,4,6}
n(E) =3
probability of a getting an event number =
$\frac{ n ( E )}{ n ( s )}=\frac{3}{6}=\frac{1}{2}$
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Question 1142 Marks
In a single throw of a die, find the probability of getting a number : not greater than 4.
Answer
E=event of getting number not greater than
4={1,2,3,4}
n(E)=4
probability of a number not grearer than 4
$\frac{ n ( E )}{ n ( s )}=\frac{4}{6}=\frac{2}{3}$
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Question 1152 Marks
In a single throw of a die, find the probability of getting a number : less than or equal to 4
Answer
Sample space ={1,2,3,4,5,6}
n(s)=6
(2)E=event of getting a number less than or equal to 4{1,2,3,4}
n(E)=4
probabitity of a number less than or equal to 4=$\frac{ n ( E )}{ n ( s )}=\frac{4}{6}=\frac{2}{3}$
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Question 1162 Marks
In a single throw of a die, find the probability of getting a number : greater than 4
Answer
Sample space ={1,2,3,4,5,6}
n(s) =6
(1) E= event of getting number greater than 4={5,6}
n(E) =2
probability of a numbber greater than
$4=\frac{ n ( E )}{ n ( s )} \frac{2}{6}=\frac{1}{3}$
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Question 1172 Marks
A bag contains 3 white, 5 black and 2 red balls, all of the same shape and size. A ball is drawn from the bag without looking into it, find the probability that the ball drawn not a black ball.
Answer
Total number of balls=3+5+2=10
Total number of events = p(n) =10
There are 3 + 2 = 5 balls which are not black
favourable number of event = P(A) = 5
Hence, P (not getting a black ball) = $\frac{ P ( A )}{ P ( n )}=\frac{5}{10}=\frac{1}{2}$
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Question 1182 Marks
A bag contains 3 white, 5 black and 2 red balls, all of the same shape and size. A ball is drawn from the bag without looking into it, find the probability that the ball drawn is : not a red ball.
Answer
There are 3+5=8 balls which are not red
Favourable number of events= P(A)=8
Hence, P (not getting a red ball) =$\frac{P(A)}{P(n)}=\frac{8}{10}=\frac{4}{5}$
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Question 1192 Marks
A bag contains 3 white, 5 black and 2 red balls, all of the same shape and size. A ball is drawn from the bag without looking into it, find the probability that the ball drawn White ball.
Answer
Total number of balls=3+5+2
Total number of events =p(n)=10
There are 3 white balls
favourable number of events=p(A)=3
Hence, p(getting a red ball)=" $P(A) " / " P(n) "=3 / 10$
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Question 1202 Marks
A bag contains 3 white, 5 black and 2 red balls, all of the same shape and size. A ball is drawn from the bag without looking into it, find the probability that the ball drawn a red ball.
Answer
Total number of balls=3+5+2
Total number of events =p(n)=10
There are 2 red balls
favourable number of events=p(A)=2
Hence,p(getting a red ball) =$\frac{ P ( A )}{ P ( n )}=\frac{2}{10}=\frac{1}{5}$
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Question 1212 Marks
A bag contains 3 white, 5 black and 2 red balls, all of the same shape and size. A ball is drawn from the bag without looking into it, find the probability that the ball drawn a black ball.
Answer
Total number of ball3+5+2=10
Total number of events =p(n)=10
(1) There are 5 black balls
favourable number of events =p(A) =5
Hence, p(getting a black ball)$=\frac{P(A)}{P(n)}=\frac{5}{10}=\frac{1}{2}$
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Question 1222 Marks
A single letter is selected at random from the word ‘Probability’. Find the probability that it is a vowel.
Answer
Possible outcomes= s={'p','r','o','b','a','b','i','l','i','t,''y'}
n(s)=11
Event of selection of vowels =E={'o','a','i','i'}
n(E)=4
Probability of selection of a vowel =
$p(s)=\frac{n(E)}{n(s)}=\frac{4}{11}$
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Question 1232 Marks
A bag contains six identical black balls. A child withdraws one ball from the bag without looking into it. What is the probability that he takes out : a white ball
Answer
Possible number of outcomes=6= number of balls in the bag
n(s)=6
E=event of drawing a white ball = number of white balls in the bag = 0
n(E)=0
Probability of drwing a white ball =
$P(s)=\frac{n(E)}{n(s)}=\frac{0}{6}=0$
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Question 1242 Marks
A coin is tossed once. Find the probability of : not getting a tail
Answer
ample space ={H,T}
n(s)=2
Not getting a tail means getting a heads.
Event of getting a heads={H}
n(A) =1
Therefor, the probability of tail=
$\frac{n( A )}{ n ( s )}=\frac{1}{2}$
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Question 1252 Marks
Which of the following cannot be the probability of an event? 0.35
Answer
0.35
$0 \leq 0.35 \leq 1$
Hence, it can be the probability of an event.
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Question 1262 Marks
A coin is tossed once. Find the probability of : getting a tail.
Answer
sample space={H,T}
n(s)=2
A= Event of getting a tail = {T}
n (A)=1
Therefore, the probability of getting a tail =
$\frac{ n ( A )}{n(s)}=\frac{1}{2}$
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Question 1272 Marks
A die is thrown once. Find the probability of getting : an even number or a multiple of 3
Answer
E= the event of getting an even number or a multiple of 3{2,3,,,4,6}
n(E) = 4
probability of getting an even number or a multiple
$3=P(s)=\frac{n(E)}{n(s)}=\frac{4}{6}=\frac{2}{3}$
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Question 1282 Marks
A die is thrown once. Find the probability of getting : an even number
Answer
Sample space={1,2,3,4,5,6}
n(S)=6
E=the possible even numbers{2,4,6}
n(E)=3
probability of getting an even number=
$p(s)=\frac{n(E)}{n(s)}=\frac{3}{6}=\frac{1}{2}$
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Question 1292 Marks
In a single throw of die, find the probability of getting : a prime number
Answer
E=event of getting a prime number ={2,35}
n(E) =3
Probability of getting a prime number =
$P(s)=\frac{n(E)}{n(s)}=\frac{3}{6}=\frac{1}{2}$

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Question 1302 Marks
In a single throw of die, find the probability of getting : a number less than $8$
Answer
If we consider to find the probability of number less than 8, then all six cases are favourable $n(E) = 6$
probability of getting a number less than $8 = P(S)$
$=\frac{n(E)}{n(S)} $
$ =\frac{6}{6} $
$ =1$
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Question 1312 Marks
In a single throw of die, find the probability of getting : 8
Answer
There are only six possible outcomes in a single throw of a die. If we want to find probability of 8 to come up, then in that case number of possible or favourable outcome is 0 (zero)
n(E) =0
probability of getting a 8=$P(s)=\frac{n(E)}{n(s)}=\frac{0}{6}=0$
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Question 1322 Marks
In a single throw of die, find the probability of getting: 5
Answer
Sample space={1,2,3,4,5,6}
n(s)=6
E= event of getting a 5on a throw of die ={5}
n(E) =1
Probability of getting a 5 =$\frac{n(E)}{n(s)}=\frac{1}{6}$
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Question 1332 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:
white
Answer
Total number of balls in the box =7+8+5=20 balls
Total possible outcomes=20=n(s)
Event of drawing a white ball=E= number of white balls
Probability of drawing a white ball= $\frac{n(E)}{n(s)}=\frac{5}{20}=\frac{1}{4}$
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Question 1342 Marks
A and B are friends. Ignoring the leap year, find the probability that both friends will have:
different birthdays?
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.
If A's birthday from B's the number of favourable outcomes for his birthday is 365-1=364
So, p(A's birthday is different from B,s birthday)=$\frac{364}{365}$
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Question 1352 Marks
In a match between A and B:
 the probability of winning of A is 0.83. What is the probability of winning of B?
Answer
Probability of winning of A+ Probability of losing of A=1 Probability of losing of A=Probability of winning of B Therfore,
 Probability of winning of A+ Probability of winning of B=1 0.83+Probability of winning of B =1-1.0.83=0.17 
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Question 1362 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 5 or 6
Answer
Total number of possible outcomes=52
n(s)=52
Event of drawing a card with number 5 or 6 =E={5H,5D,5S,5C,6H,6D,6S,6C}
n(E)=8
Probability of drawing a card with number 5 or 6
$\frac{n(E)}{n(S)}=\frac{8}{52}=\frac{2}{13}$
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Question 1372 Marks
From a well shuffled deck of 52 cards, one card is drawn. Find the probability that the card drawn is:
a queen of black card 
Answer
Event of drawing a queen of black colour={Q (spande), Q(club)=E}
n(E)=2
Probability of drawing a queen of black colour=
$\frac{n(E)}{n(s)}=\frac{2}{52}=\frac{1}{26}$
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Question 1382 Marks
From a well shuffled deck of 52 cards, one card is drawn. Find the probability that the card drawn is:
 not a face card
Answer
Total number of possible outcomes=52
n(s)=52
Probability of not drawing a face card =1-Probability of drawing a face card probability of not drawing face card =$1-\frac{3}{13}=\frac{13-3}{13}=\frac{10}{13}$
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Question 1392 Marks
From a well shuffled deck of 52 cards, one card is drawn. Find the probability that the card drawn is:
a face card
Answer
Total number of possible outcomes=52
n(s)=52
No.of face cards in a deck of52 cards=12(4kings,4queens and 4 jacks)
Event of drawing a face cards=E=(4kings,4queens and 4 jacks) n(E)=12
Probability of drawing a face card=$\frac{n(E)}{n(S)}=\frac{12}{52}=\frac{3}{13}$

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Question 1402 Marks
Cards marked with numbers 1, 2, 3 ……… 20 are well shuffled and a card is drawn at random. What is the probability that the number on the card is:
 a perfect square
Answer
Total possible outcomes $=20$
Favorable outcomes for a perfect square $=1,4,9,16$
Number of favorable outcomes $=4$
$
P(\text { a perfect square })=\frac{4}{20}=\frac{1}{5}
$
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Question 1412 Marks
Cards marked with numbers 1, 2, 3 ……… 20 are well shuffled and a card is drawn at random. What is the probability that the number on the card is:
divisible by 3
Answer
Total possible outcomes $=20$
Favorable outcomes for a number divisible by $3=3,6,9,12,15,18$
Number of favorable outcomes $=6$
$
P (\text { divisible by } 3)=\frac{6}{20}=\frac{3}{10}
$
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Question 1422 Marks
Cards marked with numbers 1, 2, 3 ……… 20 are well shuffled and a card is drawn at random. What is the probability that the number on the card is:
 a prime number
Answer
Total possible outcomes=20
Favorable outcomes for a prime number= 2,3,5,7,11,13,17,19
Number of favorable outcomes=8
 P(a prime number)= 8/20=2/5
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Question 1432 Marks
Three coins are tossed together. Write all the possible outcomes. Now, find the probability of getting  all tails
Answer
When three coins are tossed, possible outcomes are:
HHH, HHT,HTH, HTT, THH,THT,TTH,TTT
Total possible outcomes $=8$
Favorable outcomes for all tails= TTT
Number of favorable outcomes $=1$
$
P(\text { all tails })=\frac{1}{8}
$
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Question 1442 Marks
Three coins are tossed together. Write all the possible outcomes. Now, find the probability of getting:
 at most two heads
Answer
When three coins are tossed, possible outcomes are:
HHH,HHT,HTH,HTT,THH,THT,TTH,TTT
Total possible outcomes=8
Favourable outcomes for at most two heads=HHT,THH,HTH,HTT,THT,TTH,TTT
Number of favoraable outcomes=7
p(at most two heads)=$\frac{7}{8}$
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Question 1452 Marks
Three coins are tossed together. Write all the possible outcomes. Now, find the probability of getting:  at least two heads 
Answer
When three coins are tossed, possible outcomes are:
HHH,HHT,HTH,HTT,THH,THT,TTH,TTT
Total possible outcomes $=8$
Faavvorable outcomes for at least two heads= $HHT , THH , HTH , HHH$
Number of favorable outcomes $=4$
$P ($ at least two heads $)=\frac{4}{8}=\frac{1}{2}$
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Question 1462 Marks
Three coins are tossed together. Write all the possible outcomes. Now, find the probability of getting:  exactly two heads 
Answer
When three coins are tossed, possible outcomes are:
HHH, HHT,HTH,HTT,THH,THT,TTH,TTT
Total possible outcomes $=8$
Favorable outcomes for exactly two heads $= HHT , THH , HTH$
Number of favorable outcomes $=3$
$p ($ exactly two heads $)=\frac{3}{8}$
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Question 1472 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:  more than 10
Answer
When two dice are rolled, total number of possible outcomes $=36$
Favorable outcome for the sum of number more than 10.i.e. 11 or 12 are :
$
\{(5,6)(6,5)(6,6)\}
$
Number of favourable outcomes $=3$
$p($ getting a sum of number more than 10$)=\frac{3}{36}=\frac{1}{12}$
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Question 1482 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:  between 5 and 8 
Answer
When two dice rolled, total number of possible outcomes $=36$
Favorable outcomes for the sum of number between 5 and 8 i.e, 6 or 7 are:
$
\{(1,5)(1,6)(2,4)(2,5)(3,3)(3,4)(4,2)(4,3)(5,1)(5,2)(6,1)\}
$
Number of favorable outcomes $=11$
p(getting a sum of 6 or $7=\frac{11}{36}$
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Question 1492 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:  $7, 8$ or $9$
Answer
When two dice are rolled, total number of possible outcomes $=36$
Favorable outcome for the sum of number $7,8$ or $9$ are:
$\{(1,6)(2,5)(2,6)(3,4)(3,6)(4,3)(4,4)(4,5)(5,2)(5,3)(5,4)(6,1)(6,2)(6,3)\}$
Number of favorable outcomes $=15$
$p($ getting a sum of $7,8$ or $ 9)$
$=\frac{15}{36}$
$=\frac{5}{12}$
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Question 1502 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:  4 or 5
Answer
When two dice are rolled, total number of possible outcomes $=36$
Favorable outcomes for the sum of numbers 4 or 5 are :
$
\{(1,3),(1,4),(2,2),(2,3)(3,1)(3,2)(4,1)\}
$
Number of favorable outcomes $=7$
$p($ getting a sum of 4 or 5$)=\frac{7}{36}$
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