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:
(i) yellow
(ii) red
(iii) not yellow
(iv) neither yellow nor red
Total number of balls in the bag = 3+4+1 = 8 balls
Number of possible outcomes = 8 = n(S)
(i) Event of drawing a yellow ball = {Y}
n(E) = 1
Probability of drawing a yellow ball = 
(ii) Event of drawing a red ball = {R, R, R}
n(E) = 3
Probability of drawing a red ball = 
(iii) Probability of not drawing a yellow ball = 1 - Probability of drawing a yellow ball
Probability of not drawing a yellow ball = 1 - 
(iv) Neither 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 = 
A dice is thrown once. What is the probability of getting a number:
(i) greater than 2?
(ii) less than or equal to 2?
Number of possible outcomes when dice is thrown = {1,2,3,4,5,6}
n(S) = 6
(i) Event of getting a number greater than 2 = E = {3, 4, 5, 6}
n(E) = 4
Probability of getting a number greater than 2 = 
(ii) 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 =
From a well shuffled deck of 52 cards, one card is drawn. Find the probability that the card drawn is:
(i) a face card
(ii) not a face card
(iii) a queen of black card
(iv) a card with number 5 or 6
(v) a card with number less than 8
(vi) a card with number between 2 and 9
Total number of possible outcomes = 52
n(S) = 52
(i) No. of face cards in a deck of 52 cards = 12 (4 kings, 4 queens and 4 jacks)
Event of drawing a face cards = E = (4 kings, 4 queens and 4 jacks)
n(E) = 12
Probability of drawing a face card = 
(ii) Probability of not drawing a face card = 1 - probability of drawing a face card
Probability of not drawing a face card = 1 -
(iii) Event of drawing a queen of black color ={ Q(spade), Q(club)} = E
n(E) = 2
Probability of drawing a queen of black color =
(iv) 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 =
(v) Numbers 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 =
(vi) Number between 2 and 9 = {3, 4, 5, 6, 7, 8}
Event of drawing a card with number between 2 and 9 = E = {6H cards, 6D cards, 6S cards, 6C cards}
n(E) = 24
Probability of drawing a card with number between 2 and 9 =
In a match between A and B:
(i) the probability of winning of A is 0.83. What is the probability of winning of B?
(ii) the probability of losing the match is 0.49 for B. What is the probability of winning of A?
(i) Probability of winning of A + Probability of losing of A = 1
Probability of losing of A = Probability of winning of B
Therefore,
Probability of winning of A + Probability of winning of B = 1
0.83 + Probability of winning of B = 1
Probability of winning of B = 1 - 0.83 = 0.17
(ii) Probability of winning of B + Probability of losing of B = 1
Probability of losing of B = Probability of winning of A
Therefore,
Probability of winning of A = 0.49
A and B are friends. Ignoring the leap year, find the probability that both friends will have:
(i) different birthdays?
(ii) the same birthday?
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 equally likely.
(i) If A's birthday is different 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) = 
(ii) P( A and B have the same birthday)
= 1 - P (both have different birthdays)
= 1 - [Using P( E' ) = 1 - P(E)]
= 
A man tosses two different coins (one of Rs 2 and another of Rs 5) simultaneously. What is the probability that he gets:
(i) at least one head?
(ii) at most one head?
When two coins are tossed simultaneously, the possible outcomes are {(H, H), (H, T), (T, H), (T, T)}
n(S) = 4
(i) The outcomes favourable 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) = 
(ii) 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) =
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:
(i) white
(ii) neither red nor white.
Total number of balls in the box = 7+8+5 = 20 balls
Total possible outcomes = 20 = n(S)
(i) Event of drawing a white ball = E = number of white balls
n(E) = 5
Probability of drawing a white ball = 
(ii) 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 = 
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:
(i) a black face card
(ii) a queen
(iii) a black card
Total number of cards = 52
3 face cards of spades are removed
Remaining cards = 52 - 3 = 49 = number of possible outcomes
n(S) = 49
(i) Number of black face cards left = 3 face cards of club
Event of drawing a black face card = E = 3
n(E) = 3
Probability of drawing a black face card = 
(ii) Number of queen cards left = 3
Event of drawing a black face card = E = 3
n(E) = 3
Probability of drawing a queen card =
(iii) 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 = 
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?
Total result = 0 sec to 40 sec
Total possible outcomes = 40
n(S) = 40
Favorable results = 0 sec to 15 sec
Favorable outcomes = 15
n(E) = 15
Probability that the music will stop in first 15 sec = 
In a bundle of 50 shirts, 44 are good, 4 have minor defects and 2 have major defects. What is the probability that:
(i) it is acceptable to a trader who accepts only a good shirt?
(ii) it is acceptable to a trader who rejects only a shirt with major defects?
Total number of shirts = 50
Total number of elementary events = 50 = n(S)
(i) 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 shirt = 
(ii) Since trader rejects shirts with major defects only and number of shirts with major defects = 2
Event of accepting shirts = 50 - 2 = 48 = n(E)
Probability of accepting shirts = 
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:
(i) 8
(ii) 13
(iii) less than or equal to 12
The number of possible outcomes = 6 × 6 = 36.
(i) The outcomes favourable to the event 'the sum of the two numbers 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) = 
(ii) There is no outcome favourable to the event E = 'the sum of two numbers is 13'.
n(E) = 0
Hence, P(E) = 
(iii) All the outcomes are favourable to the event E = 'sum of two numbers 
12'.
Hence, P(E) = 
Which of the following cannot be the probability of an event?
(i) 
(ii) 0.82
(iii) 37%
(iv) -2.4
We know that probability of an event E is 
(i) Since 
Therefore, can be a probability of an event.
(ii) Since 
Therefore, 
can be a probability of an event.
(iii)Since 
Therefore, 
can be a probability of an event.
(iv) Since 
Therefore, 
cannot be a probability of an event.
If P(E) = 0.59; find P(not E)
P(E) + P(not E) = 1
0.59 + P(not E) = 1
P(not E) = 1 - 0.59 = 0.41
A bag contains a certain number of red balls. A ball is drawn. Find the probability that the ball drawn is:
(i) black
(ii) red
Total possible outcomes = number of red balls.
(i) Number of favorable outcomes for black balls = 0
P(black ball) = 0
(ii) Number of favorable outcomes for red balls = number of red balls
P(red ball) =
The probability that two boys do not have the same birthday is 0.897. What is the probability that the two boys have the same birthday?
P(do not have the same birthday)+P(have same birthday)=1
0.897 + P(have same birthday) = 1
P(have same birthday) = 1 - 0.897
P(have same birthday) = 0.103
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:
(i) not red?
(ii) neither red nor green?
(iii) white or green?
Total number of possible outcomes = 10+16+8 = 34 balls
n(S) = 34
(i) Favorable outcomes for not a red ball = favorable outcomes for white or green ball
Number of favorable outcomes for white or green ball = 16+8=24 = n(E)
Probability for not drawing a red ball = 
(ii) Favorable outcomes for neither a red nor a green ball = favorable outcomes for white ball
Number of favorable outcomes for white ball = 16 = n(E)
Probability for not drawing a red or green ball = 
(iii) Number of favorable outcomes for white or green ball = 16+8=24 = n(E)
Probability for drawing a white or green ball =
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:
(i) will be a Re 1 coin?
(ii) will not be a Rs 2 coin?
(iii) will neither be a Rs 5 coin nor be a Re 1 coin?
Total number of coins = 20+50+30 = 100
Total possible outcomes = 100 = n(S)
(i) Number of favorable outcomes for Re 1 coins = 30 = n(E)
Probability(Re 1 coin) = 
(ii) Number of favorable outcomes for not a Rs 2 coins = number of favorable outcomes for Re 1 or Rs 5 coins = 30+20 = 50 = n(E)
Probability(not Rs 2 coin) = 
(iii) number of favorable outcomes for neither Re 1 nor Rs 5 coins = Number of favorable outcomes for Rs 2 coins = 50 = n(E)
Probability(neither Re 1 nor Rs 5 coin) =
A game consists of spinning arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12; as shown below.
If the outcomes are equally likely, find the probability that the pointer will point at:
(i) 6 (iv) a number greater than 8
(ii) an even number (v) a number less than or equal to 9
(iii) a prime number (vi) a number between 3 and 11
Total number of possible outcomes = 12
(i) Number of favorable outcomes for 6 = 1
P(the pointer will point at 6) = 
(ii) Favorable outcomes for an even number are 2, 4, 6, 8, 10, 12
Number of favorable outcomes = 6
P(the pointer will be at an even number) = 
(iii) Favorable outcomes for a prime number are 2, 3, 5, 7, 11
Number of favorable outcomes = 5
P(the pointer will be at a prime number) = 
(iv) Favorable outcomes for a number greater than 8 are 9, 10, 11, 12
Number of favorable outcomes = 4
P(the pointer will be at a number greater than 8) = 
(v) Favorable outcomes for a number less than or equal to 9 are 1, 2, 3, 4, 5, 6, 7, 8, 9
Number of favorable outcomes = 9
P(the pointer will be at a number less than or equal to 9) =
(vi) Favorable outcomes for a number between 3 and 11 are 4, 5, 6, 7, 8, 9, 10
Number of favorable outcomes = 7
P(the pointer will be at a number between 3 and 11) = 
One card is drawn from a well shuffled deck of 52 cards. Find the probability of getting:
(i) a queen of red color
(ii) a black face card
(iii) the jack or the queen of the hearts
(iv) a diamond
(v) a diamond or a spade
Total possible outcomes = 52
(i) Number queens of red color = 2
Number of favorable outcomes = 2
P(queen of red color) = 
(ii) Number of black cards = 26
Number of black face cards = 6
Number of favorable outcomes = 6
P(black face card) = 
(iii) Favorable outcomes for jack or queen of hearts = 1 jack + 1 queen
Number of favorable outcomes = 2
P(jack or queen of hearts) =
(iv) Number of favorable outcomes for a diamond = 13
Number of favorable outcomes = 13
P(getting a diamond) = 
(v) 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) = 
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:
(i) a black card
(ii) 8 of red color
(iii) a king of black color
There are 12 face cards in a deck.
Therefore, possible number of outcomes = 52 - 12 = 40
(i) number of favorable outcomes for black cards = 26 cards - 6 face cards = 20
P(a black card) = 
(ii) number of favorable outcomes for 8 of red color = 2
P(getting a card with 8 of red color) = 
(iii) Since all face cards are removed
Number of favorable outcomes for a king of black color = 0
P(getting a king of black color) = 
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.
(i) What is the probability that the card drawn is the eight or the king?
(ii) If the king is drawn and put aside, what is the probability that the second card picked up is:
a) an ace? b) a king?
Total number of possible outcomes = 7
(i) Number of favorable outcomes for the card is 8 or the king = 2
P(card is 8 or the king) = 
(ii) a) 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) = 
b) Now, for second pick number of king = 0
Number of favorable outcomes for a king = 0
P(card is a king) = 
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:
(i) a good one
(ii) a defective one
Total number of possible outcomes = 150
(i) out of 150 bulbs, 15 are defective
Number of bulbs which are good = 150 - 15 = 135
P( taking out a good bulb) = 
(ii)Number of bulbs which are defective = 15
P( taking out a defective bulb) = 
(i) 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?
(ii) 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?
(i) Total number of pens = 4 + 16 = 20
Total possible outcomes = 20
Number of defective pens = 4
P(defective pen) = 
(ii) If defective pen drawn in first draw is not replaced, total possible outcomes = 20 - 1 = 19
a) Number of defective pens = 3
P(defective pens) = 
b) Number of not defective pens = 16
P(not defective pens) = 
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:
(i) a perfect square number
(ii) a number divisible by 4
(iii) a number divisible by 5
(iv) a number divisible by 4 or 5
(v) a number divisible by 4 and 5
Total number of possible outcomes = 100
(i) Numbers which are perfect squares = 1, 4, 9, 16, 25, 36, 49, 64, 81, 100
Number of favorable outcomes = 10
P(a perfect square) = 
(ii) Numbers 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, 92, 96, 100
Number of favorable outcomes = 25
P(number divisible by 4) = 
(iii) Numbers which are divisible by 5 = 5, 10, 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 divisible by 5) = 
(iv) Numbers 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 favorable outcomes = 40
P(number divisible by 4 or 5) = 
(v) Numbers which are divisible by 4 and 5 = 20, 40, 60, 80, 100
Number of favorable outcomes = 5
P(number divisible by 4 and 5) = 
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:
(i) inside the circle
(ii) outside the circle
Diameter of the circle = 20 cm
Radius = 10 cm
Area of circle = 
Length of paper = 40 cm
Width of paper = 30 cm
Area of paper = 1200 cm2
Total possible outcomes = area of rectangular paper
(i) Since paper is kept on table top and die falls and lands on paper.
Number of favorable outcomes = area of circle.
P(inside the circle) = 
(ii) P(outside the circle) = 1 - P(inside the circle)
= 1 - 
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:
(i) 4 or 5
(ii) 7, 8 or 9
(iii) between 5 and 8
(iv) more than 10
(v) less than 6
When two dice are rolled, total number of possible outcomes = 36
(i) 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) = 
(ii) Favorable outcomes for the sum of numbers 7, 8 or 9 are:
{(1, 6), (2, 5), (2, 6), (3, 4), (3, 5), (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) = 
(iii) Favorable outcomes for the sum of numbers 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) = 
(iv) Favorable outcomes for the sum of numbers more than 10 i.e. 11 or 12 are:
{(5, 6), (6, 5), (6, 6)}
Number of favorable outcomes = 3
P(getting a sum of numbers more than 10) = 
(v) Favorable outcomes for the sum of numbers less than 6 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) = 
Three coins are tossed together. Write all the possible outcomes. Now, find the probability of getting:
(i) exactly two heads
(ii) at least two heads
(iii) at most two heads
(iv) all tails
(v) at least one tail
When three coins are tossed, possible outcomes are:
HHH, HHT, HTH, HTT, THH, THT, TTH, TTT
Total possible outcomes = 8
(i) Favorable outcomes for exactly two heads = HHT, THH, HTH
Number of favorable outcomes = 3
P(exactly two heads) = 
(ii) Favorable outcomes for at least two heads = HHT, THH, HTH, HHH
Number of favorable outcomes = 4
P(at least two heads) = 
(iii) Favorable outcomes for at most two heads = HHT, THH, HTH, HTT, THT, TTH, TTT
Number of favorable outcomes = 7
P(at most two heads) = 
(iv) Favorable outcomes for all tails = TTT
Number of favorable outcomes = 1
P(all tails) = 
(v) Favorable outcomes for at least one tails = HHT, THH, HTH, HTT, THT, TTH, TTT
Number of favorable outcomes = 7
P(at least one tail) =
Two dice are thrown simultaneously. What is the probability that:
(i) 4 will not come up either time?
(ii) 4 will come up at least once?
When two dice are thrown, total possible outcomes = 36
(i) 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) = 
(ii) P(4 will come up once) = 1 - P(4 will not come up either time)
P(4 will come up once) = 1 -
P(4 will come up once) = 
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:
(i) a prime number
(ii) divisible by 3
(iii) a perfect square
Total possible outcomes = 20
(i) Favorable outcomes for a prime number = 2, 3, 5, 7, 11, 13, 17, 19
Number of favorable outcomes = 8
P(a prime number) = 
(ii) Favorable outcomes for a number divisible by 3 = 3, 6, 9, 12, 15, 18
Number of favorable outcomes = 6
P(divisible by 3) = 
(iii) Favorable outcomes for a perfect square = 1, 4, 9, 16
Number of favorable outcomes = 4
P(a perfect square) = 
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:
(i) the same day
(ii) consecutive day
(iii) different days
Total number of possible outcomes = 5 x 5 = 25
The possible outcomes are:
MM, MT, MW, MTh, MF, TM, TT, TW, TTh, TF, WM, WT, WW, WTh, WF, ThM, ThT, ThW, ThTh, ThF, FM, FT, FW, FTh, FF
(i) Favorable outcomes for two employees remaining absent on same day are: MM, TT, WW, ThTh, FF
Number of favorable outcomes = 5
P(same day) = 
(ii) Favorable outcomes for two employees remaining absent on consecutive days : MT, TM, TW, WT, WTh, ThW, ThF, FTh
Number of favorable outcomes = 8
P(consecutive days) = 
(iii) P(absent on diff days) = 1 - P(absent on same days)
A box contains some black balls and 30 white balls. If the probability of drawing a black ball is two-fifths of a white ball; find the number of black balls in the box.
Let the number of black balls in the box be x
Total number of balls in the box = x+30
P(drawing a black ball) = 
P(drawing a white ball = 
But, P(drawing a black ball) = 
P(drawing a white ball)
Number of black balls in the box = 12
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
(i) A face card (King, Jack or Queen)
(ii) An even numbered red card?
No. of total cards = 52
Cards removed of 4 colours of multiples of 3
= 3, 6, 9 = 4 × 3 = 12
Remaining cards = 52 - 12 = 40
(i) No. of face cards = 12 cards
(ii) An even number red cards = 2,4,8,10 cards = 2×4 = 8
A die has 6 faces marked by the given numbers as shown below:
The die is thrown once. What is the probability of getting
(i) a positive integer?
(ii) an integer greater than -3?
(iii) the smallest integer?
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 drawn is:
(i) a green ball
(ii) a white or a red ball.
(iii)Neither a green ball nor a white ball
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:
i. A perfect square
ii. Divisible by 7.
Total number of outcomes = 30
i. The perfect squares from 11 to 40 are 16, 25 and 36. So, the number of possible outcomes = 3 Hence, the probability that the number on the card drawn is a perfect square
ii. Among the given numbers, 14, 21, 28 and 35 are divisible by 7. So, the number of possible outcomes = 4 Hence, the probability that the number on the card drawn is divisible by 7
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:
i. a vowel
ii. a consonant
iii. none of the letters of the word median?
Here, Total number of all possible outcomes = 16
i. a, e, i and o are the vowels.
Number of favourable outcomes = 4
∴ Required Probability = 
ii. Number of consonants = 16 - 4 (vowels) = 12
∴ Number of favourable outcomes = 12
∴ Required Probability = 
iii. Median contains 6 letters.
∴ Number of favourable outcomes = 16 - 6 = 10
∴ Required Probability = 
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
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
∴ Required Probability =
ii.
The probability that the ball drawn is marked A = 
… (1)
The probability that the ball drawn is marked B = 
… (2)
∴ Required Probability = 
iii. The probability that the ball drawn is neither B nor C
= 1 - [P(B) + P(C)]
= 1 - 
= 1 - 
= 
= 
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) = 
and P(B) =
. If there are 40 balls in the box which are marked C, find the number of balls in the box.
P(C) = 1 - [P(A) + P(B)]
Probability = 
Given that 40 balls in the box are marked C.
⇒ 
= 
⇒ Total number of all possible outcomes = 
∴ the number of balls in the box is 96.
