Page 1
Q u e s t i o n : 1
The probability that it will rain tomorrow is 0.85. What is the probability that it will not rain tomorrow?
S o l u t i o n :
Let A be the event of raining tomorrow. The probability that it will rain tomorrow, P(A), is 0. 85. Since the event of raining tomorrow and not raining tomorrow are complementary to each other
Q u e s t i o n : 2
A die is thrown. Find the probability of getting:
i
a prime number
i i
2 or 4
i i i
a multiple of 2 or 3
S o l u t i o n :
When a die is thrown, the possible outcomes are 1, 2, 3, 4, 5 and 6. Thus, the sample space will be as follows: S = {1, 2, 3, 4, 5, 6} i Let A be the event of getting a prime number. There
Q u e s t i o n : 3
In a simultaneous throw of a pair of dice, find the probability of getting:
i
8 as the sum
i i
a doublet
i i i
a doublet of prime numbers
i v
a doublet of odd numbers
v
a sum greater than 9
v i
an even number on first
v i i
an even number on one and a multiple of 3 on the other
v i i i
neither 9 nor 11 as the sum of the numbers on the faces
i x
a sum less than 6
x
a sum less than 7
x i
a sum more than 7
x i i
at least once
x i i i
a number other than 5 on any dice.
S o l u t i o n :
When a pair of dice is thrown simultaneously, the sample space will be as follows: S = {(1, 1), (1, 2), (1, 3), (1, 4), ?(6, 5), (6, 6)}Hence, the total number of outcomes is 36. i Let A be the
Q u e s t i o n : 4
Three coins are tossed together. Find the probability of getting:
i
exactly two heads
i i
at least two heads
i i i
at least one head and one tail
i v
no tails
S o l u t i o n :
When 3 coins are tossed together, the outcomes are as follows: S = {(h, h, h), (h, h, t), (h, t, h), (h, t, t), (t, h, h), (t, h, t), (t, t, h), (t, t, t)}Therefore, the total number of outcomes is 8. i Let A be
Q u e s t i o n : 5
A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is:
i
a black king
i i
either a black card or a king
i i i
black and a king
i v
a jack, queen or a king
v
neither a heart nor a king
v i
spade or an ace
v i i
neither an ace nor a king
v i i i
neither a red card nor a queen.
i x
other than an ace
x
a ten
( )
( )
( )
Page 2
Q u e s t i o n : 1
The probability that it will rain tomorrow is 0.85. What is the probability that it will not rain tomorrow?
S o l u t i o n :
Let A be the event of raining tomorrow. The probability that it will rain tomorrow, P(A), is 0. 85. Since the event of raining tomorrow and not raining tomorrow are complementary to each other
Q u e s t i o n : 2
A die is thrown. Find the probability of getting:
i
a prime number
i i
2 or 4
i i i
a multiple of 2 or 3
S o l u t i o n :
When a die is thrown, the possible outcomes are 1, 2, 3, 4, 5 and 6. Thus, the sample space will be as follows: S = {1, 2, 3, 4, 5, 6} i Let A be the event of getting a prime number. There
Q u e s t i o n : 3
In a simultaneous throw of a pair of dice, find the probability of getting:
i
8 as the sum
i i
a doublet
i i i
a doublet of prime numbers
i v
a doublet of odd numbers
v
a sum greater than 9
v i
an even number on first
v i i
an even number on one and a multiple of 3 on the other
v i i i
neither 9 nor 11 as the sum of the numbers on the faces
i x
a sum less than 6
x
a sum less than 7
x i
a sum more than 7
x i i
at least once
x i i i
a number other than 5 on any dice.
S o l u t i o n :
When a pair of dice is thrown simultaneously, the sample space will be as follows: S = {(1, 1), (1, 2), (1, 3), (1, 4), ?(6, 5), (6, 6)}Hence, the total number of outcomes is 36. i Let A be the
Q u e s t i o n : 4
Three coins are tossed together. Find the probability of getting:
i
exactly two heads
i i
at least two heads
i i i
at least one head and one tail
i v
no tails
S o l u t i o n :
When 3 coins are tossed together, the outcomes are as follows: S = {(h, h, h), (h, h, t), (h, t, h), (h, t, t), (t, h, h), (t, h, t), (t, t, h), (t, t, t)}Therefore, the total number of outcomes is 8. i Let A be
Q u e s t i o n : 5
A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is:
i
a black king
i i
either a black card or a king
i i i
black and a king
i v
a jack, queen or a king
v
neither a heart nor a king
v i
spade or an ace
v i i
neither an ace nor a king
v i i i
neither a red card nor a queen.
i x
other than an ace
x
a ten
( )
( )
( )
x i
a spade
x i i
a black card
x i i i
the seven of clubs
x i v
jack
x v
the ace of spades
x v i
a queen
x v i i
a heart
x v i i i
a red card
S o l u t i o n :
i
There are two black kings, spade and clover. Hence, the probability that the drawn card is a black king is: 2/52 = 1/26
i i
There are 26 black cards and 4 kings, but two kings are already black. Hence, we only need to count the red kings. Thus, the probability is: 26 +2
/52 = 7/13
i i i
This question is exactly the same as part i
. Hence, the probability is: 2/52 = 1/26
i v
There are 4 jacks, 4 queens and 4 kings in a deck. Hence, the probability of drawing either of them is: 4 +4 +4
/52 = 3/13
v
This means that we have to leave the hearts and the kings out. There are 13 hearts and 3 kings o t h e r t h a n t h a t o f h e a r t s
. Hence, the probability of drawing neither a heart nor a king is: 52 -13 -3
/52 = 9/13
v i
There are 13 spades and 3 aces o t h e r t h a n t h a t o f s p a d e s
. Hence the probability is: 13 +3
/52 = 4/13
v i i
This means that we have to leave the aces and the kings out. There are 4 aces and 4 kings. Hence, the probability of drawing neither an ace nor a king is: (52 -
4 -
4)/52 = 11/13.
v i i i
This means that we have to leave the red cards and the queens out. There are 26 red cards and 2 queens o n l y b l a c k q u e e n s a r e c o u n t e d s i n c e t h e r e d s a r e a l r e a d y c o u n t e d a m o n g t h e r e d c a r d s
. Hence, the probability of drawing neither a red card nor a queen is: 52 -26 -2
/52 = 6/13
i x
It means that we have to leave out the aces. Since there are 4 aces, then the probability is (52 -
4)/52 = 12/13
x
Since there are four 10s, the probability is: 4/52 = 1/13
x i
Since there are 13 spades, the probability is: 13/52 = 1/4
x i i
Since there are 26 black cards, the probability is: 26/52 = 1/2
x i i i
There is only one card named seven of clubs. Hence, the probability is 1/52.
x i v
Since there are 4 jacks, the probability is: 4/52 = 1/13
x v
There is only 1 card named ace of spade. Hence, the probability is 1/52.
x v i
Since there are 4 queens, the probability is: 4/52 = 1/13
x v i i
Since there are 13 hearts, the probability is: 13/52 = 1/4
x v i i i
Since there are 26 red cards, the probability is 26/52 = 1/2
Q u e s t i o n : 6
An urn contains 10 red and 8 white balls. One ball is drawn at random. Find the probability that the ball drawn is white.
S o l u t i o n :
Number of red balls = 10Number of white balls = 8Total number of balls in the urn = 10 + 8 = 18Therefore, the total number of cases is 18 and the number of favourable cases is 8. ?
Q u e s t i o n : 7
A bag contains 3 red balls, 5 black balls and 4 white balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is:
i
white?
i i
red?
i i i
black?
i v
not red?
S o l u t i o n :
Number of red balls = 3Number of black balls = 5 Number of white balls = 4Total number of balls = 3 + 5 + 4 = 12Therefore, the total number of cases is 12. i Since there are 4 white
Q u e s t i o n : 8
What is the probability that a number selected from the numbers 1, 2, 3, ..., 15 is a multiple of 4?
S o l u t i o n :
There are 15 numbers from 1, 2, ?, 15. Hence, the total number of cases is 15. Again, the multiples of 4 are 4, 8 and 12. Therefore, the total number of favourable cases is 3. ? P(the number
( )
Page 3
Q u e s t i o n : 1
The probability that it will rain tomorrow is 0.85. What is the probability that it will not rain tomorrow?
S o l u t i o n :
Let A be the event of raining tomorrow. The probability that it will rain tomorrow, P(A), is 0. 85. Since the event of raining tomorrow and not raining tomorrow are complementary to each other
Q u e s t i o n : 2
A die is thrown. Find the probability of getting:
i
a prime number
i i
2 or 4
i i i
a multiple of 2 or 3
S o l u t i o n :
When a die is thrown, the possible outcomes are 1, 2, 3, 4, 5 and 6. Thus, the sample space will be as follows: S = {1, 2, 3, 4, 5, 6} i Let A be the event of getting a prime number. There
Q u e s t i o n : 3
In a simultaneous throw of a pair of dice, find the probability of getting:
i
8 as the sum
i i
a doublet
i i i
a doublet of prime numbers
i v
a doublet of odd numbers
v
a sum greater than 9
v i
an even number on first
v i i
an even number on one and a multiple of 3 on the other
v i i i
neither 9 nor 11 as the sum of the numbers on the faces
i x
a sum less than 6
x
a sum less than 7
x i
a sum more than 7
x i i
at least once
x i i i
a number other than 5 on any dice.
S o l u t i o n :
When a pair of dice is thrown simultaneously, the sample space will be as follows: S = {(1, 1), (1, 2), (1, 3), (1, 4), ?(6, 5), (6, 6)}Hence, the total number of outcomes is 36. i Let A be the
Q u e s t i o n : 4
Three coins are tossed together. Find the probability of getting:
i
exactly two heads
i i
at least two heads
i i i
at least one head and one tail
i v
no tails
S o l u t i o n :
When 3 coins are tossed together, the outcomes are as follows: S = {(h, h, h), (h, h, t), (h, t, h), (h, t, t), (t, h, h), (t, h, t), (t, t, h), (t, t, t)}Therefore, the total number of outcomes is 8. i Let A be
Q u e s t i o n : 5
A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is:
i
a black king
i i
either a black card or a king
i i i
black and a king
i v
a jack, queen or a king
v
neither a heart nor a king
v i
spade or an ace
v i i
neither an ace nor a king
v i i i
neither a red card nor a queen.
i x
other than an ace
x
a ten
( )
( )
( )
x i
a spade
x i i
a black card
x i i i
the seven of clubs
x i v
jack
x v
the ace of spades
x v i
a queen
x v i i
a heart
x v i i i
a red card
S o l u t i o n :
i
There are two black kings, spade and clover. Hence, the probability that the drawn card is a black king is: 2/52 = 1/26
i i
There are 26 black cards and 4 kings, but two kings are already black. Hence, we only need to count the red kings. Thus, the probability is: 26 +2
/52 = 7/13
i i i
This question is exactly the same as part i
. Hence, the probability is: 2/52 = 1/26
i v
There are 4 jacks, 4 queens and 4 kings in a deck. Hence, the probability of drawing either of them is: 4 +4 +4
/52 = 3/13
v
This means that we have to leave the hearts and the kings out. There are 13 hearts and 3 kings o t h e r t h a n t h a t o f h e a r t s
. Hence, the probability of drawing neither a heart nor a king is: 52 -13 -3
/52 = 9/13
v i
There are 13 spades and 3 aces o t h e r t h a n t h a t o f s p a d e s
. Hence the probability is: 13 +3
/52 = 4/13
v i i
This means that we have to leave the aces and the kings out. There are 4 aces and 4 kings. Hence, the probability of drawing neither an ace nor a king is: (52 -
4 -
4)/52 = 11/13.
v i i i
This means that we have to leave the red cards and the queens out. There are 26 red cards and 2 queens o n l y b l a c k q u e e n s a r e c o u n t e d s i n c e t h e r e d s a r e a l r e a d y c o u n t e d a m o n g t h e r e d c a r d s
. Hence, the probability of drawing neither a red card nor a queen is: 52 -26 -2
/52 = 6/13
i x
It means that we have to leave out the aces. Since there are 4 aces, then the probability is (52 -
4)/52 = 12/13
x
Since there are four 10s, the probability is: 4/52 = 1/13
x i
Since there are 13 spades, the probability is: 13/52 = 1/4
x i i
Since there are 26 black cards, the probability is: 26/52 = 1/2
x i i i
There is only one card named seven of clubs. Hence, the probability is 1/52.
x i v
Since there are 4 jacks, the probability is: 4/52 = 1/13
x v
There is only 1 card named ace of spade. Hence, the probability is 1/52.
x v i
Since there are 4 queens, the probability is: 4/52 = 1/13
x v i i
Since there are 13 hearts, the probability is: 13/52 = 1/4
x v i i i
Since there are 26 red cards, the probability is 26/52 = 1/2
Q u e s t i o n : 6
An urn contains 10 red and 8 white balls. One ball is drawn at random. Find the probability that the ball drawn is white.
S o l u t i o n :
Number of red balls = 10Number of white balls = 8Total number of balls in the urn = 10 + 8 = 18Therefore, the total number of cases is 18 and the number of favourable cases is 8. ?
Q u e s t i o n : 7
A bag contains 3 red balls, 5 black balls and 4 white balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is:
i
white?
i i
red?
i i i
black?
i v
not red?
S o l u t i o n :
Number of red balls = 3Number of black balls = 5 Number of white balls = 4Total number of balls = 3 + 5 + 4 = 12Therefore, the total number of cases is 12. i Since there are 4 white
Q u e s t i o n : 8
What is the probability that a number selected from the numbers 1, 2, 3, ..., 15 is a multiple of 4?
S o l u t i o n :
There are 15 numbers from 1, 2, ?, 15. Hence, the total number of cases is 15. Again, the multiples of 4 are 4, 8 and 12. Therefore, the total number of favourable cases is 3. ? P(the number
( )
Q u e s t i o n : 9
A bag contains 6 red, 8 black and 4 white balls. A ball is drawn at random. What is the probability that ball drawn is not black?
S o l u t i o n :
Number of red balls = 6Number of black balls = 8 Number of white balls = 4Total number of balls = 6 + 8 + 4 = 18 ? Total number of cases = 18Again, number of balls that are not
Q u e s t i o n : 1 0
A bag contains 5 white and 7 red balls. One ball is drawn at random. What is the probability that ball drawn is white?
S o l u t i o n :
Number of white balls = 5Number of red balls = 7Total number of balls = 5 + 7 = 12 ? The total number of cases = 12Again, there are 5 white balls. Therefore, the number of favourable
Q u e s t i o n : 1 1
A bag contains 4 red, 5 black and 6 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is:
i
white
i i
red
i i i
not black
i v
red or white
S o l u t i o n :
Number of red balls = 4Number of black balls = 5Number of white balls = 6Total number of balls in the bag = 4 + 5 + 6 = 15Therefore, the total number of cases is 15. i Let A denote
Q u e s t i o n : 1 2
A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is:
i
red
i i
black
S o l u t i o n :
Number of red balls = 3Number of black balls = 5Total number of balls = 3 + 5 = 8 i Let A be the event of drawing a red ball. ? P(A) =
Number of favourable outcomes
Total number of outcomes
=
3
8
ii Let B be the event
Q u e s t i o n : 1 3
A bag contains 5 red marbles, 8 white marbles, 4 green marbles. What is the probability that if one marble is taken out of the bag at random, it will be
i
red
i i
white
i i i
not green
S o l u t i o n :
Number of red marbles = 5Number of white marbles = 8Number of green marbles = 4Total number of marbles in the bag = 5 + 8 + 4 = 17 ? Total number outcomes = 17 i Let A be
Q u e s t i o n : 1 4
If you put 21 consonants and 5 vowels in a bag. What would carry greater probability? Getting a consonant or a vowel? Find each probability.
S o l u t i o n :
Number of consonants = 21Number of vowels = 5Total number of possible outcomes = 21 + 5 = 26Let C be the event of getting a consonant and V be the event of getting a vowel. ?
Q u e s t i o n : 1 5
If we have 15 boys and 5 girls in a class which carries a higher probability? Getting a copy belonging to a boy or a girl. Can you give it a value?
S o l u t i o n :
Number of boys in the class = 15Number of girls in the class = 5Total number of students in the class = 15 + 5 = 20 ? Number of possible outcomes = 20Since the number of boys
Q u e s t i o n : 1 6
If you have a collection of 6 pairs of white socks and 3 pairs of black socks. What is the probability that a pair you pick without looking is
i
white?
i i
black?
S o l u t i o n :
Number of pairs of white socks = 6 Number of pairs of black socks = 3 Total number of pairs of socks = 6 + 3 = 9 ? Number of possible outcomes = 9 i Let A be the event of getting
Q u e s t i o n : 1 7
If you have a spinning wheel with 3-green sectors, 1-blue sector and 1-red sector. What is the probability of getting a green sector? Is it the maximum?
S o l u t i o n :
Number of green sectors in the wheel = 3Number of blue sectors in the wheel = 1 Number of red sectors in the wheel = 1Total number of sectors in the wheel = 3 + 1 + 1 = 5 ? Number
Q u e s t i o n : 1 8
When two dice are rolled:
i
List the outcomes for the event that the total is odd.
i i
Find probability of getting an odd total.
i i i
( )
( ) ( )
( )
( )
Page 4
Q u e s t i o n : 1
The probability that it will rain tomorrow is 0.85. What is the probability that it will not rain tomorrow?
S o l u t i o n :
Let A be the event of raining tomorrow. The probability that it will rain tomorrow, P(A), is 0. 85. Since the event of raining tomorrow and not raining tomorrow are complementary to each other
Q u e s t i o n : 2
A die is thrown. Find the probability of getting:
i
a prime number
i i
2 or 4
i i i
a multiple of 2 or 3
S o l u t i o n :
When a die is thrown, the possible outcomes are 1, 2, 3, 4, 5 and 6. Thus, the sample space will be as follows: S = {1, 2, 3, 4, 5, 6} i Let A be the event of getting a prime number. There
Q u e s t i o n : 3
In a simultaneous throw of a pair of dice, find the probability of getting:
i
8 as the sum
i i
a doublet
i i i
a doublet of prime numbers
i v
a doublet of odd numbers
v
a sum greater than 9
v i
an even number on first
v i i
an even number on one and a multiple of 3 on the other
v i i i
neither 9 nor 11 as the sum of the numbers on the faces
i x
a sum less than 6
x
a sum less than 7
x i
a sum more than 7
x i i
at least once
x i i i
a number other than 5 on any dice.
S o l u t i o n :
When a pair of dice is thrown simultaneously, the sample space will be as follows: S = {(1, 1), (1, 2), (1, 3), (1, 4), ?(6, 5), (6, 6)}Hence, the total number of outcomes is 36. i Let A be the
Q u e s t i o n : 4
Three coins are tossed together. Find the probability of getting:
i
exactly two heads
i i
at least two heads
i i i
at least one head and one tail
i v
no tails
S o l u t i o n :
When 3 coins are tossed together, the outcomes are as follows: S = {(h, h, h), (h, h, t), (h, t, h), (h, t, t), (t, h, h), (t, h, t), (t, t, h), (t, t, t)}Therefore, the total number of outcomes is 8. i Let A be
Q u e s t i o n : 5
A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is:
i
a black king
i i
either a black card or a king
i i i
black and a king
i v
a jack, queen or a king
v
neither a heart nor a king
v i
spade or an ace
v i i
neither an ace nor a king
v i i i
neither a red card nor a queen.
i x
other than an ace
x
a ten
( )
( )
( )
x i
a spade
x i i
a black card
x i i i
the seven of clubs
x i v
jack
x v
the ace of spades
x v i
a queen
x v i i
a heart
x v i i i
a red card
S o l u t i o n :
i
There are two black kings, spade and clover. Hence, the probability that the drawn card is a black king is: 2/52 = 1/26
i i
There are 26 black cards and 4 kings, but two kings are already black. Hence, we only need to count the red kings. Thus, the probability is: 26 +2
/52 = 7/13
i i i
This question is exactly the same as part i
. Hence, the probability is: 2/52 = 1/26
i v
There are 4 jacks, 4 queens and 4 kings in a deck. Hence, the probability of drawing either of them is: 4 +4 +4
/52 = 3/13
v
This means that we have to leave the hearts and the kings out. There are 13 hearts and 3 kings o t h e r t h a n t h a t o f h e a r t s
. Hence, the probability of drawing neither a heart nor a king is: 52 -13 -3
/52 = 9/13
v i
There are 13 spades and 3 aces o t h e r t h a n t h a t o f s p a d e s
. Hence the probability is: 13 +3
/52 = 4/13
v i i
This means that we have to leave the aces and the kings out. There are 4 aces and 4 kings. Hence, the probability of drawing neither an ace nor a king is: (52 -
4 -
4)/52 = 11/13.
v i i i
This means that we have to leave the red cards and the queens out. There are 26 red cards and 2 queens o n l y b l a c k q u e e n s a r e c o u n t e d s i n c e t h e r e d s a r e a l r e a d y c o u n t e d a m o n g t h e r e d c a r d s
. Hence, the probability of drawing neither a red card nor a queen is: 52 -26 -2
/52 = 6/13
i x
It means that we have to leave out the aces. Since there are 4 aces, then the probability is (52 -
4)/52 = 12/13
x
Since there are four 10s, the probability is: 4/52 = 1/13
x i
Since there are 13 spades, the probability is: 13/52 = 1/4
x i i
Since there are 26 black cards, the probability is: 26/52 = 1/2
x i i i
There is only one card named seven of clubs. Hence, the probability is 1/52.
x i v
Since there are 4 jacks, the probability is: 4/52 = 1/13
x v
There is only 1 card named ace of spade. Hence, the probability is 1/52.
x v i
Since there are 4 queens, the probability is: 4/52 = 1/13
x v i i
Since there are 13 hearts, the probability is: 13/52 = 1/4
x v i i i
Since there are 26 red cards, the probability is 26/52 = 1/2
Q u e s t i o n : 6
An urn contains 10 red and 8 white balls. One ball is drawn at random. Find the probability that the ball drawn is white.
S o l u t i o n :
Number of red balls = 10Number of white balls = 8Total number of balls in the urn = 10 + 8 = 18Therefore, the total number of cases is 18 and the number of favourable cases is 8. ?
Q u e s t i o n : 7
A bag contains 3 red balls, 5 black balls and 4 white balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is:
i
white?
i i
red?
i i i
black?
i v
not red?
S o l u t i o n :
Number of red balls = 3Number of black balls = 5 Number of white balls = 4Total number of balls = 3 + 5 + 4 = 12Therefore, the total number of cases is 12. i Since there are 4 white
Q u e s t i o n : 8
What is the probability that a number selected from the numbers 1, 2, 3, ..., 15 is a multiple of 4?
S o l u t i o n :
There are 15 numbers from 1, 2, ?, 15. Hence, the total number of cases is 15. Again, the multiples of 4 are 4, 8 and 12. Therefore, the total number of favourable cases is 3. ? P(the number
( )
Q u e s t i o n : 9
A bag contains 6 red, 8 black and 4 white balls. A ball is drawn at random. What is the probability that ball drawn is not black?
S o l u t i o n :
Number of red balls = 6Number of black balls = 8 Number of white balls = 4Total number of balls = 6 + 8 + 4 = 18 ? Total number of cases = 18Again, number of balls that are not
Q u e s t i o n : 1 0
A bag contains 5 white and 7 red balls. One ball is drawn at random. What is the probability that ball drawn is white?
S o l u t i o n :
Number of white balls = 5Number of red balls = 7Total number of balls = 5 + 7 = 12 ? The total number of cases = 12Again, there are 5 white balls. Therefore, the number of favourable
Q u e s t i o n : 1 1
A bag contains 4 red, 5 black and 6 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is:
i
white
i i
red
i i i
not black
i v
red or white
S o l u t i o n :
Number of red balls = 4Number of black balls = 5Number of white balls = 6Total number of balls in the bag = 4 + 5 + 6 = 15Therefore, the total number of cases is 15. i Let A denote
Q u e s t i o n : 1 2
A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is:
i
red
i i
black
S o l u t i o n :
Number of red balls = 3Number of black balls = 5Total number of balls = 3 + 5 = 8 i Let A be the event of drawing a red ball. ? P(A) =
Number of favourable outcomes
Total number of outcomes
=
3
8
ii Let B be the event
Q u e s t i o n : 1 3
A bag contains 5 red marbles, 8 white marbles, 4 green marbles. What is the probability that if one marble is taken out of the bag at random, it will be
i
red
i i
white
i i i
not green
S o l u t i o n :
Number of red marbles = 5Number of white marbles = 8Number of green marbles = 4Total number of marbles in the bag = 5 + 8 + 4 = 17 ? Total number outcomes = 17 i Let A be
Q u e s t i o n : 1 4
If you put 21 consonants and 5 vowels in a bag. What would carry greater probability? Getting a consonant or a vowel? Find each probability.
S o l u t i o n :
Number of consonants = 21Number of vowels = 5Total number of possible outcomes = 21 + 5 = 26Let C be the event of getting a consonant and V be the event of getting a vowel. ?
Q u e s t i o n : 1 5
If we have 15 boys and 5 girls in a class which carries a higher probability? Getting a copy belonging to a boy or a girl. Can you give it a value?
S o l u t i o n :
Number of boys in the class = 15Number of girls in the class = 5Total number of students in the class = 15 + 5 = 20 ? Number of possible outcomes = 20Since the number of boys
Q u e s t i o n : 1 6
If you have a collection of 6 pairs of white socks and 3 pairs of black socks. What is the probability that a pair you pick without looking is
i
white?
i i
black?
S o l u t i o n :
Number of pairs of white socks = 6 Number of pairs of black socks = 3 Total number of pairs of socks = 6 + 3 = 9 ? Number of possible outcomes = 9 i Let A be the event of getting
Q u e s t i o n : 1 7
If you have a spinning wheel with 3-green sectors, 1-blue sector and 1-red sector. What is the probability of getting a green sector? Is it the maximum?
S o l u t i o n :
Number of green sectors in the wheel = 3Number of blue sectors in the wheel = 1 Number of red sectors in the wheel = 1Total number of sectors in the wheel = 3 + 1 + 1 = 5 ? Number
Q u e s t i o n : 1 8
When two dice are rolled:
i
List the outcomes for the event that the total is odd.
i i
Find probability of getting an odd total.
i i i
( )
( ) ( )
( )
( )
List the outcomes for the event that total is less than 5.
i v
Find the probability of getting a total less than 5?
S o l u t i o n :
Possible outcomes when two dice are rolled: S = {(1, 1), (1, 2), (1, 3), (1, 4), ?, (6, 5), (6, 6)}Therefore, the number of possible outcomes in the sample space is 36. i The outcomes for the
( )
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