Download App

Mathematics [2 Mark Question Answer]

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

Questions · Page 4 of 5

[2 Mark Question Answer]

Question 1512 Marks
A bag contains 100 identical marble stones which are numbered 1 to 100. If one stone is drawn at random from the bag, find the probability that it bears:    a number divisible by 4 and 5 
Answer
Total number of possible outcomes $=100$
Numbers which are divisible by 4 and $5=200,40,60,80,100$
Number of favourable outcomes $=5$
$p$ (number divisible by 4 and 5$)=\frac{5}{100}=\frac{1}{20}$
View full question & answer→
Question 1522 Marks
A bag contains $100$ identical marble stones which are numbered $1$ to $100.$ If one stone is drawn at random from the bag, find the probability that it bears:  a number divisible by $4$ or $5$ 
Answer
Total number of possible outcomes $=100$
Number which are divisible by $4$ or
$5=4,5,8,10,12,15,16,20,24,25,28,30,32,35,36,40,44,45,48,50,52,55,56,60,$
$64,65,68,70,72,75,76,80,84,85,88,90,92,95,96,100$
Number of faavorable outcomes $=40$
$p ($numer divisible by $4$ or $5)$
$=\frac{40}{100}$
$=\frac{2}{5}$
View full question & answer→
Question 1532 Marks
A bag contains $100$ identical marble stones which are numbered $1$ to $100.$ If one stone is drawn at random from the bag, find the probability that it bears:  a number divisible by $5$ 
Answer
Total number of possible outcomes $=100$
Number which are divisible by
$5=\{5,10, \ldots 15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95,100\}$
Number of favorable outcomes $=20$
$p ($number diviisible by $5)$
$=\frac{20}{100}$
$=\frac{1}{5}$
View full question & answer→
Question 1542 Marks
A bag contains $100$ identical marble stones which are numbered $1$ to $100.$ If one stone is drawn at random from the bag, find the probability that it bears:  a number divisible by $4$
Answer
Total number of possible outcomes $=100$
Number which are divisible by
$4=4,8,12,16,20,24,28,32,36,40,44,48,52,56,60,64,68,72,76,80,84,88,922,96,100$
Number of favorable outccomes $=25$
$p ($ number divisible by $4)$
$=\frac{25}{100}$
$=\frac{1}{4}$
View full question & answer→
Question 1552 Marks
A bag contains 100 identical marble stones which are numbered 1 to 100. If one stone is drawn at random from the bag, find the probability that it bears:   a perfect square number 
Answer
Total number of posibble outcomes $=100$
Number which are perfeect squares=
$
1,4,9,16,25,36,49,64,81,100
$
Number of favorable outcome $=10$
$
p(a \text { perfect square })=\frac{10}{100}=\frac{1}{10}
$
View full question & answer→
Question 1562 Marks
 Suppose the pen drawn in (i) is defective and is not replaced. Now one more pen is drawn at random from the rest. What is the probability that this pen is:
a) defective
b) not defective?
Answer
If defective pen drawn in first draw is not reeplaced, total possible outcomes $=20-1=19$
a) Number of defctive pens $=3$
$
p(\text { defective pens })=\frac{3}{19}
$
b) Number of not defective pens $=16$
$
p(\text { not defective pens })=\frac{16}{19}
$
View full question & answer→
Question 1572 Marks
4 defective pens are accidentally mixed with 16 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is drawn at random from the lot. What is the probability that the pen is defective?  
Answer
Total number of pens $=4+16=20$
Total possible outcomes $=20$
Number of defective pens $=4$
$p($ defective pen $)=\frac{4}{20}=\frac{1}{5}$
View full question & answer→
Question 1582 Marks
A box contains 150 bulbs out of which 15 are defective. It is not possible to just look at a bulb and tell whether or not it is defective. One bulb is taken out at random from this box. Calculate the probability that the bulb taken out is:  a good one 
Answer
Total number of possible outcome $=150$
Out of 150 bulbs, 15 are defective
Number of bulbs which are good 150-15=135
$P($ taking out a good bulb $)=\frac{135}{150}=\frac{9}{10}$
View full question & answer→
Question 1592 Marks
A dice is thrown once. What is the probability of getting a number:   greater than 2?
Answer
Number of possible outcomes when dice is throw={1,2,3,4,5,6}
n(S)=6
Event of getting a number greater than 2=E={3,4,5,6}
n(E)=4
Probability of getting a number greater than 2 $=\frac{n(E)}{n(S)}=\frac{4}{6}=\frac{2}{3}$
View full question & answer→
Question 1602 Marks
From a deck of 52 cards, all the face cards are removed and then the remaining cards are shuffled. Now one card is drawn from the remaining deck. Find the probability that the card drawn is:  a king of black color   
Answer
Therefore are 12 face card in a deck.
Therefore, possible number of outcomes $=52-12=40$
Since all face cards are removed
Number of favorable outcomes for a king of black color $=$ 0
$p($ getting a king of black color $)=\frac{0}{40}=0$
View full question & answer→
Question 1612 Marks
From a deck of 52 cards, all the face cards are removed and then the remaining cards are shuffled. Now one card is drawn from the remaining deck. Find the probability that the card drawn is:  8 of red color 
Answer
There are 12 face cards in a deck.
Therefore, possible number of outcomes $=52-12=40$
Number of favorable outcomes for 8 of red color $=2$ $p\left(\right.$ getting a card with 8 of red color) $=\frac{2}{40}=\frac{1}{20}$
View full question & answer→
Question 1622 Marks
From a deck of 52 cards, all the face cards are removed and then the remaining cards are shuffled. Now one card is drawn from the remaining deck. Find the probability that the card drawn is:  a black card
Answer
There are 12 face cards in a deck
Therefore, possible number of outcomes $=52-12=40$
Number of favourable outcome for black $=26$ cards- 6 face cards $=20$
$
P ( a \text { black card })=\frac{20}{40}=\frac{1}{2}
$
View full question & answer→
Question 1632 Marks
One card is drawn from a well shuffled deck of 52 cards. Find the probability of getting:   a diamond or a spade 
Answer
Total possible outcomes $=52$
Number of favorable outcomes for a diamond or a spade $=13+13=26$
Number of favorable outcomes $=26$
$P ($ getting a diamond or a spade $)=\frac{26}{52}=\frac{1}{2}$
View full question & answer→
Question 1642 Marks
One card is drawn from a well shuffled deck of 52 cards. Find the probability of getting:  a diamond 
Answer
Total possible outcomes $=52$
Number of favorable outcomes for a diamond $=13$
Number of favorable outcomes $=13$
$P($ getting a diamond $)=\frac{13}{52}=\frac{1}{4}$
View full question & answer→
Question 1652 Marks
One card is drawn from a well shuffled deck of 52 cards. Find the probability of getting:  the jack or the queen of the hearts   
Answer
Total possible outcomes $=52$
Favorable outcomes for jack or queen of hearts $=1$ jack +1 queen
Number of favrable outcomes $=2$
$p\left(\right.$ jack or queen of hearts) $=\frac{2}{52}=\frac{1}{26}$
View full question & answer→
Question 1662 Marks
One card is drawn from a well shuffled deck of 52 cards. Find the probability of getting:  a black face card 
Answer
Total possible outcomes $=52$
Number of black cards $=26$
Number of black face cards $=6$
Number of favorable outcome $=6$
$P($ black face card $)=\frac{6}{52}=\frac{3}{26}$
View full question & answer→
Question 1672 Marks
One card is drawn from a well shuffled deck of 52 cards. Find the probability of getting:  a queen of red color
Answer
Total possible outcomes $=52$
Number queen of redd color $=2$
Number of favorable outcome $=2$
$p(q u e e n$ of red color $)=\frac{2}{52}=\frac{1}{26}$
View full question & answer→
Question 1682 Marks
A bag contains twenty Rs 5 coins, fifty Rs 2 coins and thirty Re 1 coins. If it is equally likely that one of the coins will fall down when the bag is turned upside down, what is the probability that the coin:
will neither be a Rs 5 coin nor be a Re 1 coin?
Answer
Total number of coins =20+50+30=100
Total possible outcomes=100=n(s)
Number of favorable outcomes for niether Re1 nor Rs5 coins= Number of favourable outcomes for Rs2 coins=50= n(E)
Probability (neither Re 1 nor Rs 5 coin)=
$\frac{n(E)}{n(s)}=\frac{50}{100}=\frac{1}{2}$
View full question & answer→
Question 1692 Marks
A bag contains twenty Rs 5 coins, fifty Rs 2 coins and thirty Re 1 coins. If it is equally likely that one of the coins will fall down when the bag is turned upside down, what is the probability that the coin:
will not be a Rs 2 coin? 
Answer
Total number of coins=20+50+30=100
Total possible outcomes=100=n(s)
Number of favorable outcomes for not a Rs 1 or Rs 5 coins=30+20+50=n(E)
Probability (not Rs 2 coin)=$=\frac{n(E)}{n(S)}=\frac{50}{100}=\frac{1}{2}$
View full question & answer→
Question 1702 Marks
A bag contains twenty Rs 5 coins, fifty Rs 2 coins and thirty Re 1 coins. If it is equally likely that one of the coins will fall down when the bag is turned upside down, what is the probability that the coin: 
will be a Re 1 coin? 
Answer
Total number of coins = 20+50+30=100
Total possible outcomes=100=n(S)
Number of favorable outcomes for Re 1 coins=30=n(s)
Probability (Re 1 coin)=$\frac{n(E)}{n(S)}=\frac{30}{100}=\frac{3}{10}$

View full question & answer→
Question 1712 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:
white or green? 
Answer
Total number of possible outcomes=10+16+8=34
n(s)=34
Number of favorable outcomes for white or green ball =16+8=24 =n(E)
Probability for drawing s white or green ball=
$\frac{n(E)}{n(S)=} \frac{24}{34}=\frac{12}{17}$
View full question & answer→
Question 1722 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:
neither red nor green?
Answer
Total number of possible outcomes = 10+16+8=34 balls
n(s)=34
Favorable outcomes for neither a red nor a green ball=favourable outcomes for white ball number of favorable outcomes for white ball=16=n(E)
Probability for not drawing a red or green ball=
$\frac{n(E)}{n(s)}=\frac{16}{34}=\frac{8}{17}$
View full question & answer→
Question 1732 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:  neither yellow nor red 
Answer
Total number of balls in the bag=3+4+1=8
Number of possible outcomes=8=n(s)
Neithher yellow ball nor red ball means a blue ball
Event of not drawing a yellow or red ball=E=4
n(E)=4
Probability of not drawing a yellow or red ball=$\frac{n(E)}{n(S)}=\frac{4}{8}=\frac{1}{2}$
View full question & answer→
Question 1742 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:   not yellow
Answer
Total number of balls in the bag=3+4+1=8 balls
Number of possible outcomes =8=n(s)
Probability of not drrawing a yellow ball=1- Probability of drawing a yellow ball Probability of not drawing a yellow ball$=1-\frac{1}{8}=\frac{8-1}{8}=\frac{7}{8}$
View full question & answer→
Question 1752 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:
 red 
Answer
Total number of balls in the bag=3+4+1=8 balls
Number of possible outcomes=8=n(s)
Event of drawing a red ball ={R,R,R}
n(E)=3
Probability of drawing a reddd ball=$\frac{n(E)}{n(S)}=\frac{3}{8}$
View full question & answer→
Question 1762 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 =$\frac{n(E)}{n(s)}=\frac{1}{8}$

View full question & answer→
Question 1772 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:  less than or equal to 12
Answer
The number of possible outcomes =$6 \times 6=36$
All the outcomes are favourable to the event E=sum of two number $\leq 12$ :
Hence, p(E)=(n(E))/(n(S))=36/36=1
View full question & answer→
Question 1782 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:  13 
Answer
The number of possible outcomes =$6 \times 6=36$
There is no outcome favourable to the event E=the sum of two number is 13 .
n(E)=0
Hence,$p(E)=\frac{n(E)}{n(s)}=\frac{0}{36}$
View full question & answer→
Question 1792 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,$p(E)=\frac{n(E)}{n(S)}=\frac{5}{36}$
View full question & answer→
Question 1802 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 =$\frac{n(E)}{n(S)}=\frac{48}{50}=\frac{24}{25}$
View full question & answer→
Question 1812 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 =
$\frac{n(E)}{n(s)}=\frac{44}{50}=\frac{22}{25}$
View full question & answer→
Question 1822 Marks
A card is drawn from a well shuffled pack of 52 cards. Find the probability that the card drawn is:  a red and a king 
Answer
Number of possible outcomes when card is drawn from pack of 52 cards=52
n(E) = 2
event of drawing a red and a king=$\frac{n(E)}{n(s)}=\frac{2}{52}=\frac{1}{26}$
View full question & answer→
Question 1832 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
$=\frac{n(E)}{n(s)}=\frac{8}{52}=\frac{2}{13}$
View full question & answer→
Question 1842 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
$\frac{n(E)}{n(s)}=\frac{2}{52}=\frac{1}{26}$
View full question & answer→
Question 1852 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 red cards (hearts+diamonds)=26=E=event of drawing a red card
n(E)=26
Probability of drwaing a red card=
$\frac{n(E)}{n(s)}=\frac{26}{52}=\frac{1}{2}$
View full question & answer→
Question 1862 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=
$\frac{n(E)}{n(s)}=\frac{26}{52}=\frac{1}{2}$
View full question & answer→
Question 1872 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 =$\frac{n(E)}{n(s)}=\frac{13}{52}=\frac{1}{4}$
View full question & answer→
Question 1882 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=
$\frac{n(E)}{n(s)}=\frac{2}{4}=\frac{1}{2}$
View full question & answer→
Question 1892 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 =
$\frac{n(E)}{n(s)}=\frac{3}{4}$
View full question & answer→
Question 1902 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=
$\frac{n(E)}{n(s)}=\frac{1}{4}$
View full question & answer→
Question 1912 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=
$\frac{n(E)}{n(s)}=\frac{0}{6}=0$

View full question & answer→
Question 1922 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
$\frac{n(E)}{n(s)}=\frac{6}{6}=1$
View full question & answer→
Question 1932 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= $\frac{n(E)}{n(s)}=\frac{2}{52}=\frac{1}{26}$
View full question & answer→
Question 1942 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=
$=\frac{n(E)}{n(s)}=\frac{4}{52}=\frac{1}{13}$
View full question & answer→
Question 1952 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=
$\frac{n(E)}{n(s)}=\frac{13}{52}=\frac{1}{4}$
View full question & answer→
Question 1962 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=
$\frac{n(E)}{n(s)}=\frac{26}{52}=\frac{1}{2}$
View full question & answer→
Question 1972 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=$\frac{n(E)}{n(S)}=\frac{26}{52}=\frac{1}{2}$
View full question & answer→
Question 1982 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 = $\frac{n(E)}{n(s)}=\frac{16}{48}=\frac{1}{3}$
View full question & answer→
Question 1992 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
$=\frac{n(E)}{n(s)}=\frac{12}{48}=\frac{1}{4}$

View full question & answer→
Question 2002 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=$\frac{n(E)}{n(s)}=\frac{9}{16}$
View full question & answer→
Generate Paper FreeAll Probability topics
[2 Mark Question Answer] - Page 4 - Mathematics STD 10 Questions - Vidyadip