Download Grade 10 Probability

ID : ww-10-Probability [1]
Grade 10
Probability
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Answer t he quest ions
(1)
T wo dice are rolled. What is the probability that the two numbers add up to a prime number?
(2)
From a deck of cards, Elizabeth took out one card at random. What is the probability that she
got a prime number?
(3)
William is participating in a race. T he probability that he will come f irst in the race is 0.1. T he
probability that he will come second in the race is 0.1. T he probability that he will come in 3rd is
0.6, and the probability that he will be 4th is 0.2. What is the probability that he will come in 2nd
position or better in the race?
(4) From among a group of 3 men, 3 women and 7 children, 2 individuals are selected randomly.
What is the probability that exactly 1 among the chosen are children?
(5)
What is the probability that an integer in the set 1, 2, 3, ...100 is divisible by 2 and not divisible by
3?
Choose correct answer(s) f rom given choice
(6) T here are 3 pens belonging to 3 students. T he pens were put into a box, and each student pulls
out an pen one af ter the other. What is the probability that each student gets his or her own
pen?
a.
3
b.
3
c.
1
1
6
d.
2
2
6
(7) Coin A is f lipped 3 times and coin B is f lipped 4 times. What is the probability that the number of
heads obtained f rom f lipping the two coins is the same?
a.
37
b.
130
c.
33
128
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35
128
d.
39
128
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ID : ww-10-Probability [2]
(8)
What is the probability that a leap year will contain 53 Wednesdays and 52 T hursdays?
a.
1
b.
7
366
1
c.
2
d.
366
1
52
(9) T he letters of the word PROBABILIT Y are rearranged in a random order. What is the probability
that the letters have exactly 4 letters between them?
a. 1
b.
7
110
c.
12
d.
110
8
105
(10) Richard selects 3 numbers randomly f rom the f ollowing set of 5 numbers
2, 11, 9, 3 and 7.
He puts them in the f orm of a proper f raction of the type a
b
.
c
What is the probability that you will get a f raction greater than 3
a.
24
b.
30
c.
12
20
?
21
24
36
d.
36
18
30
(11) A card is drawn f rom a shuf f led deck of 52 cards. What is the probability of getting a card
greater than 4 but lesser than 9?
a.
16
b.
52
c.
1
8
52
d.
52
4
52
(12) Patricia draws 3 cards out of a deck of 52 cards. What is the probability that she draws a Jack, a
4, and a 2?
a.
4
b.
52
c.
3
52
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64
22100
d.
256
22100
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ID : ww-10-Probability [3]
(13) A bag contains 6 red balls, 6 blue balls, and 4 green balls. Kevin draws 2 balls out of the bag.
What is the probability that he gets a green ball and a blue ball?
a.
24
b.
120
c.
24
136
26
d.
120
24
127.5
(14) A tyre manuf acturer keeps the record of how much distance the tyres manuf actured by the
company travel bef ore f ailing. T hey f ind the f ollowing data
Distance traveled in kilometers Number of failing tyres
Less than 5000
20
5000 to 10000
100
10001 to 20000
440
20001 to 40000
576
More than 40000
864
If Sarah buys a tyre f rom them, what is the probability that it will last more than 10000
kilometers?
a.
1880
b.
2000
c.
1865
1842
2000
d.
2000
1854
2000
(15) In 1992, the meteorological of f ice predicted the weathers completely right f or the months of
f ebruary and april, and completely wrong f or all the other months. What is the probability that
the f orecast was wrong f or a given day that year?
a.
308
b.
366
c.
59
365
307
d.
366
306
365
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ID : ww-10-Probability [4]
Answers
(1)
15
36
Step 1
T he two dice that are rolled can show any of these values
Dice 1 : 1, 2, 3, 4, 5, 6
Dice 2 : 1, 2, 3, 4, 5, 6
So we can get a total of 36 combinations between them (6 x 6)
Step 2
If we take one value f rom the list of possible values f rom each Dice, we get numbers
ranging f rom 2 (when both Dice show 1) to 12 (when both dice show 6).
Let's enumerate the prime numbers between 2 and 12. T hey are 2, 3, 5, 7 and 11
We need to see in how many ways we can get each of these values
Let's put the value rolled by the dice as (x,y), where x is the value rolled by Dice 1, and y the
value rolled by Dice 2
- 2: T he only way to get this is when we roll (1,1). 1 possibility
- 3: We can get this by (1,2) or (2,1). 2 possibilities.
- 5: We can get this by (2,3), (3,2), (1,4) or (4,1). 4 possibilities.
- 7: We can get this by (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). 6 possibilities.
- 11: We can get this by (5,6), or (6,5). 2 Possibilities
T his gives us a total of 1 + 2 + 4 + 6 + 2 = 15 possible ways to get a prime number
Step 3
So the probability of getting the two numbers add up to a prime is
15
36
(2)
4
13
Step 1
T here are 52 cards in a deck, with 13 of each suits (diamonds, clubs, hearts and spades
Step 2
Each of the 13 cards of a suits are the numbers 1 to 9, then 4 f ace cards, including the ace
Step 3
In the numbers 1 to 9, only 2,3,5,7 are primes. So f our primes per suit, which gives a total
of 16 prime cards
Step 4
So there are 16 possible prime cards out of the total 52. T he probability that we pick a
prime card is theref ore
16
52
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, which can be reduced to
4
13
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ID : ww-10-Probability [5]
(3)
0.2
Step 1
We are looking f or the probability that he will come 2nd or better in the race
T his is the probability that he will either win the race or be 2nd
Since he can be f irst or second, the probabilities can be added
Step 2
T he probability that he will be f irst is 0.1
T he probability that he will be in second place is 0.1
Step 3
Adding them, we get 0.1 + 0.1 = 0.2
(4)
42
78
Step 1
T he total number of people in the group is 13
Step 2
We can select 2 individuals f rom among them in 13 C 2 ways =
13 x (13-1)
= 78
2x1
Step 3
Here we want the probability that exactly 1 children among these chosen
T his means 1 of the chosen 2 are children, and 1 are either men or women
Step 4
Now we need to f igure out the number of ways of choosing 1 children f rom among 7
children.
T his is 7 C 1 = 7
Step 5
We also need to f ind out the ways of choosing 1 men/women f rom among 6 men/women
T his is 6 C 1 = 6
Step 6
T he probability that exactly 1 among the chosen 2 are children is theref ore
6x7
78
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=
42
78
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ID : ww-10-Probability [6]
(5)
17
50
Step 1
We know that, the integers divisible by 2 and 3 are also divisible by 6.
T heref ore, the number of integers divisible by 2 and not divisible by 3
= T he number of integers divisible by 2 - T he number of integers divisible by 6
Step 2
T he number of integers in the set 1, 2, 3, ...100, divisible by 2 = quotient of 100/2 = 50
T he number of integers in the set 1, 2, 3, ...100, divisible by 6 = quotient of 100/6 = 16
Now, the number of integers in the set 1, 2, 3, ...100, divisible by 2 and not divisible by 3 =
50 - 16 = 34
Step 3
T he probability that an integer in the set 1, 2, 3, ...100 is divisible by 2 and not divisible by 3
T he integers in the set 1, 2, 3, ...100 is divisible by 2 and not divisible by 3
=
T he total number of integers in the set 1, 2, 3, ...100
34
=
100
17
=
50
(6)
b.
1
6
Step 1
We have 3 pens and 3 students. Let's f irst see in how many dif f erent ways can the pens be
distributed among the students
Step 2
We know that 3 objects can be distributed in 3 ! = 3 x 2 x...x 1 = 6 ways
Step 3
Out of these 6 ways, there is only one distribution where each student got his or her own
pen
Step 4
So the probability of each student getting his or her own pen =
1
6
(7)
b.
35
128
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ID : ww-10-Probability [7]
(8)
a.
1
7
Step 1
T here are 366 days in a leap year
Step 2
If we divide 366 by 7 (since there are seven days in a week), we will get an answer of 52,
with a remainder of 2
T his means that a leap year will have 52 Sundays, 52 Mondays, 52 T uesdays, 52
Wednesdays, 52 T hursdays, 52 Fridays and 52 Saturdays.
Apart f rom these there will be two other days. T his means that there will be two weekdays
that occur 53 times.
Step 3
T he two days could be (Sunday, Monday), (Monday, T uesday), (T uesday, Wednesday),
(Wednesday, T hursday), (T hursday, Friday), (Friday, Saturday), or (Saturday, Sunday) - a
total of seven combinations
Step 4
Out of these seven combinations, only one of them has Wednesday but no T hursday
Step 5
So the probability of either of those two days being a Wednesday is
1
7
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ID : ww-10-Probability [8]
(9)
c.
12
110
Step 1
T he f irst step is f inding out all the possible arrangements of the letters of the word
PROBABILIT Y.
Note that we are not interested in "unique" arrangements (in some kind of problems we
would be), but just the total possible arrangement
Step 2
T he answer to this is that there are 11 letters in this word, and theref ore the letters can be
arranged in 11! ways i.e. 11x10x9x...2x1 ways
T o see this, take any of the letters - it can be in any of the possible 11 positions.
For each of these position, a second letter can be in any of the other 10 positions (11
minus the one taken up by the f irst letter)
So the two letters can appear in 11x10 combinations
For each of these 110 combinations, a third letter can be in any of the remaining 9 places,
and so on
Step 3
Now we know the total number of arrangements (11!) possible, and need to look f or the
possible arrangements where the letters P and L have exactly 4 letters between them)
Step 4
T his can be seen by inspection
If P is in the f irst position (f irst letter of the word), then L would need to be in position 6
(since there are 4 letters between them
Similarly, if P is in the second position , then L would need to be in position 7
T here are 6 such positions, the last one having P in the 6th position and L being in the last
position
Step 5
T he same thing can be seen with L being bef ore P. T here are 6 such positions
So the total number of positions f or the two letters where this condition is met is 12
Now we have f illed in 2 of the 11 letters with P and L in 12 ways
T he remaining 9 letters can take any of the remaining 9 positions f or each of these
Since there are no restriction on the remaining 9 letters, the number of possible
arrangements of 9 letters in 9 positions is 9! So the total ways to rearrange all the letters
so that P and L have exactly 4 letters between them is
12 x 9!
Step 6
Now we can work out the probability of rearranging the letters of PROBABILIT Y so that P
and L have exactly 4 letters between them
P (arrangement) =
Arrangements where the two letters have 4 letters between them
=
T otal possible arrangements of the letters of PROBABILIT Y
12 x 9!
11!
=
12
110
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ID : ww-10-Probability [9]
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ID : ww-10-Probability [10]
(10)
d.
18
30
Step 1
We need to select 3 numbers out of the 5 given in order to get a f raction of the f orm a
b
c
We are also told it is a proper f raction, so b should be greater than c
Step 2
Let's put the integers in a sorted manner.
We get 2, 3, 7, 9 and 11
Step 3
Now we need to see how many proper f ractions can be f ormed f rom them
A proper f raction of the type a
b
has three integers, a whole number a, a numerator b,
c
and a denominator c
Let's assume we use one of the integers in the list above as the whole number a
We now can select b and c f rom the remaining 4 integers
We can select 2 integers f rom 4 in 4C2 =
4x3
= 6 ways
2x1
For any pair we select, one will be greater than the other, and the smaller integer will f orm
the numerator and the larger one the denominator
- Note: that the other way won't work - if the numerator is larger than the denominator it is
not a proper f raction
So f or each of the 5 integers, if we select one as the whole number, we get 6 possible
combinations of numerator and denominator that can f orm a proper f raction
T his means there are 5x6 = 30 possible proper f ractions of the f orm a
b
that can be
c
f ormed f rom these 5 integers
Step 4
Now we need to f igure out how many of these 30 f ractions are greater than 3
20
21
In gt, we see that the whole number is 3, the numerator is 20 and the denominator is 21
We see that the numerator 20 is larger than the largest number in the set of numbers given
to us, and the denominator 21 is one larger than 20
T he implication of this is that 3
20
will be larger than any f raction that can be f ormed
21
f rom the set of numbers 2, 3, 7, 9 and 11 where the whole number of the f raction is 3
So we only need to count the f ractions that have the whole number greater than 3
T hese are the f ractions that will have the whole numbers as 7, 9 and 11
Step 5
Now, remember there are 30 proper f ractions you can f orm f rom this list of number
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ID : ww-10-Probability [11]
From our analysis above, we also saw that f or each number selected as the whole number,
we can f orm 6 f ractions f rom this list of numbers
Step 6
So using each of the numbers f rom 7, 9 and 11 as the whole number, we can f orm 6 proper
f ractions
T he total number of f ractions that can be f ormed using 7, 9 and 11 as the whole number =
6 x 3 = 18
Step 7
Out of the 30 f ractions, 18 will be greater than 3
20
21
Step 8
T he probability is theref ore =
18
30
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ID : ww-10-Probability [12]
(12)
b.
64
22100
Step 1
T here are 4 Jacks in a deck of cards. At this point there are also 52 cards in the deck
Step 2
T here are 4 4s in a deck of cards. Now there are 51 cards remaining in the deck
Step 3
T here are 4 2s in a deck of cards. Now there are 50 cards remaining in the deck
Step 4
You can select 3 cards without replacement f rom a deck in 52 C 3 ways =
52 x 51 x 50
3x2x1
Step 5
4x4x4
So the answer is
52 x 51 x 50
3x2x1
Step 6
Simplif ying we get
64
22100
(13)
a.
24
120
Step 1
T here are a total of 6+6+4 = 16 balls in the bag
Step 2
T he number of ways to pick out 2 balls f rom a set of 16 balls is 16 C2 = 16 x (16-1)/2 = 120
Step 3
T he number of ways to pick a green ball and a blue ball is obtained by multiplying the
number of the balls of each of these colors. T his is 4 x 6 = 24
Step 4
T he probability that he gets a green ball and a blue ball is theref ore
24
120
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ID : ww-10-Probability [13]
(14)
a.
1880
2000
Step 1
First we need to f ind the total number of tyres that are given here.
Step 2
We add the number of tyres 20 + 100 + 440 + 576 + 864 = 2000.
Step 3
T o f ind the probability that the tyre Sarah purchased would last more than 10000
kilometers, we need to add the number of tyres that lasted more than 10000 kilometers.
T his is 440 + 576 + 864 = 1880.
Step 4
T he probability that the tyre lasts more than 10000 km =
1880
.
2000
(15)
c.
307
366
Step 1
T he key thing to note here is that 1992 is a leap year
A leap year has 366 days
Step 2
Now we need to f igure out the number of days in f ebruary and april
f ebruary has 29 days and april has 30 days
Step 3
Adding them together we get 29 + 30 = 59 days
Step 4
So the f orecast was right f or 59 days and wrong f or 307 days
Step 5
T he probability that the f orecast was wrong on a given day would theref ore be
307
366
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