PROBABILITY
Example 9.1
When a dice is rolled, find the probability to get the number which is greater than 4?
Solution:
$S=\{1,2,3,4,5,6\}$
$n(S)=6$
$A=\{5,6\}$
$n(A)=2$
$P(A)=\frac{n(A)}{n(S)}=\frac{2}{6}=\frac{1}{3}$
Example 9.2
In an office, where 42 staff members work, 7 staff members use cars, 20 staff members use two-wheelers and the remaining 15 staff members use cycles. Find the relative frequencies.
Solution:
Total staff = 42
The relative frequencies are :
Car users $=\frac{7}{42}=\frac{1}{6}$
Two wheeler users $=\frac{20}{42}=\frac{10}{21}$
Cycle users $=\frac{15}{42}=\frac{1}{6}$
Example 9.3
Team I and Team II play 10 cricket matches each of 20 overs. Their total scores in each match are tabulated in the table as follows:
| Match numbers | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| Team I | 200 | 122 | 111 | 88 | 156 | 184 | 99 | 199 | 121 | 156 |
| Team II | 143 | 123 | 156 | 92 | 164 | 72 | 100 | 201 | 98 | 157 |
Solution:
Team I wins 3 matches out of 10.
The relative frequency of Team I winning is $\frac{3}{10}=0.3$
Exercise 9.1
1. You are walking along a street. If you just choose a stranger crossing you, what is the probability that his next birthday will fall on a sunday?
Solution: S={Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}
n(S)=7
Let A denote the event of stranger's birthday falling on Sunday.
A={Sunday}
n(A)=1
$P(A)=\frac{n(A)}{n(S)}=\frac{1}{7}$
2. What is the probability of drawing a King or a Queen or a Jack from a deck of cards?
Solution: n(S)=52
Let A denote the event of drawing a King or a Queen or a Jack card.
n(A)=12
$P(A)=\frac{n(A)}{n(S)}=\frac{12}{52}=\frac{3}{13}$
3. What is the probability of throwing an even number with a single standard dice of six faces?
Solution: S={1,2,3,4,5,6}
n(S)=6
Let A denote the event of getting an even number.
A={2,4,6}
n(A)=3
$P(A)=\frac{n(A)}{n(S)}=\frac{3}{6}=\frac{1}{2}$
4. There are 24 balls in a pot. If 3 of them are Red, 5 of them are Blue and the remaining are Green then, what is the probability of picking out (i) a Blue ball, (ii) a Red ball and (iii) a Green ball?
Solution: n(S)=24
No. of red balls = 3
No. of blue balls = 5
No. of green balls = 16
(i) Let A denote the event of getting a blue ball.
n(A)=5
$P(A)=\frac{n(A)}{n(S)}=\frac{5}{24}$
(ii) Let B denote the event of getting a red ball.
n(B)=3
$P(B)=\frac{n(B)}{n(S)}=\frac{3}{24}=\frac{1}{8}$
(iii) Let C denote the event of getting a green ball.
n(C)=3
$P(C)=\frac{n(C)}{n(S)}=\frac{16}{24}=\frac{2}{3}$
5. When two coins are tossed, what is the probability that two heads are obtained?
Solution: S={HH, HT, TH, TT}
n(S)=4
Let A denote the event of getting two heads.
A={HH}
n(A)=1
$P(A)=\frac{n(A)}{n(S)}=\frac{1}{4}$
6. Two dice are rolled, find the probability that the sum is i) equal to 1 ii) equal to 4 iii) less than 13
Solution: S={(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}
n(S)=36
(i) Let A denote the event of getting the sum equal to 1.
A={}
n(A)=0
$P(A)=\frac{n(A)}{n(S)}=\frac{0}{36}=0$
(ii) Let B denote the event of getting the sum equal to 4.
B={(1,3), (2,2), (3,1)}
n(B)=3
$P(B)=\frac{n(B)}{n(S)}=\frac{3}{36}=\frac{1}{12}$
(iii) Let C denote the event of getting the sum less than 13.
n(C)=n(S)=36
$P(C)=\frac{n(C)}{n(S)}=\frac{36}{36}=1$
7. A manufacturer tested 7000 LED lights at random and found that 25 of them were defective. If a LED light is selected at random, what is the probability that the selected LED light is a defective one.
Solution: n(S)=7000
No. of defective bulbs = 25
Let A denote the event of getting defective bulb.
n(A)=25
$P(A)=\frac{n(A)}{n(S)}=\frac{25}{7000}=\frac{1}{280}$
8. In a football match, a goalkeeper of a team can stop the goal, 32 times out of 40 attempts tried by a team. Find the probability that the opponent team can convert the attempt into a goal.
Solution: n(S)=40
No. of times a team can stop the goal = 32
Let A denote the event of opponent team converting the attempt into a goal.
n(A)=8
$P(A)=\frac{n(A)}{n(S)}=\frac{8}{40}=\frac{1}{5}$
9. What is the probability that the spinner will not land on a multiple of 3?
Solution:
S={1,2,3,4,5,6,7,8}
n(S)=8
Let A denote the event of spinner not landing on a multiple of 3.
A={1,2,4,5,7,8}
n(A)=6
$P(A)=\frac{n(A)}{n(S)}=\frac{6}{8}=\frac{3}{4}$
10. Frame two problems in calculating probability, based on the spinner shown here.
Solution: What is the probability that the spinner will land on an odd number?
What is the probability that the spinner will land on an even number?
Example 9.4
The probability that it will rain tomorrow is $\frac{91}{100}$. What is the probability that it will not rain tomorrow?
Solution:
Let E denote the event that it will rain tomorrow.
$P(E)=\frac{91}{100}=0.91$
Let E' denote the event that it will not rain tomorrow.
$P(E')=1-P(E)=1-0.91=0.09$
Example 9.5
In a recent year, of the 1184 centum scorers in various subjects in tenth standard public exams, 233 were in mathematics. 125 in social science and 106 in science. If one of the student is selected at random, find the probability of that selected student, (i) is a centum scorer in Mathematics (ii) is not a centum scorer in Science
Solution:
n(S)=1184
(i) Let A denote the event of selecting a centum scorer in Mathermatics.
n(A)=233
$P(A)=\frac{n(A)}{n(S)}=\frac{233}{1184}$
(ii) Let B denote the event of selecting a centum scorer in Science.
n(B)=106
$P(B)=\frac{n(B)}{n(S)}=\frac{106}{1184}$
Let B' denote the event of selecting a non centum scorer in Science.
$P(B')=1-P(E)=1-\frac{106}{1184}=\frac{1184-106}{1184}=\frac{1078}{1184}$
Exercise 9.2
1. A company manufactures 10000 Laptops in 6 months. Out of which 25 of them are found to be defective. When you choose one Laptop from the manufactured, what is the probability that selected Laptop is a good one.Solution:
n(S)=10000
No. of defective laptops = 25
Let A denote the event of selecting a defective laptop.
n(A)=25
$P(A)=\frac{n(A)}{n(S)}=\frac{25}{10000}=0.0025$
Let A' denote the event of selecting a good laptop.
$P(A')=1-P(A)=1-0.0025=0.9975$
Another Method:
n(S)=10000
No. of defective laptops = 25
No. of good laptops = 10000-25 = 9975
Let A denote the event of selecting a good laptop.
n(A)=9975
$P(A)=\frac{n(A)}{n(S)}=\frac{9975}{10000}=0.9975$
2. In a survey of 400 youngsters aged 16-20 years, it was found that 191 have their voter ID card. If a youngster is selected at random, find the probability that the youngster does not have their voter ID card.
Solution:
n(S)=400
No. of youngsters having voter ID = 191
Let A denote the event of selecting an youngster having their voter ID.
n(A)=191
$P(A)=\frac{n(A)}{n(S)}=\frac{191}{400}$
Let A' denote the event of selecting an youngster not having their voter ID.
$P(A')=1-P(A)=1-\frac{191}{400}=\frac{400-191}{400}=\frac{209}{400}$
Another Method :
n(S)=400
No. of youngsters having voter ID = 191
No. of youngsters not having voter ID = 400 - 191 = 209
Let A denote the event of selecting an youngster not having their voter ID.
n(A)=209
$P(A)=\frac{n(A)}{n(S)}=\frac{209}{400}$
3. The probability of guessing the correct answer to a certain question is $\frac{x}{3}$. If the probability of not guessing the correct answer is $\frac{x}{5}$, then find the value of x.
Solution:
$P(A)=\frac{x}{3}$
$P(A')=\frac{x}{5}$
$P(A')=1-P(A)$
$\frac{x}{5}=1-\frac{x}{3}$
$\frac{x}{5}=\frac{3-x}{3}$
$3x=5(3-x)$
$3x=15-5x$
$3x+5x=15$
$8x=15$
$x=\frac{15}{8}$
4. If a probability of a player winning a particular tennis match is 0.72. What is the probability of the player loosing the match?
Solution:
P(A)=0.72
P(A')=1-P(A)
P(A')=1-0.72=0.28
5. 1500 families were surveyed and following data was recorded about their maids at homes
| Type of maids | Only part time | Only full time | Both |
| Number of families | 860 | 370 | 250 |
Solution:
n(S)=1500
(i) Let A denote the event of selecting both type of maids
n(A)=250
$P(A)=\frac{n(A)}{n(S)}=\frac{250}{1500}=\frac{1}{6}$
(ii) Let B denote the event of selecting part time maids
n(B)=860
$P(B)=\frac{n(B)}{n(S)}=\frac{860}{1500}=\frac{43}{75}$
(iii) Part time maid + Full time maid + Both = 860 + 370 + 250 = 1480
No. of families with no maids = 20
Let C denote the event of selecting No maids
n(C)=20
$P(B)=\frac{n(B)}{n(S)}=\frac{20}{1500}=\frac{1}{75}$
Exercise 9.3
Multiple choice questions1. A number between 0 and 1 that is used to measure uncertainty is called
(1) Random variable (2) Trial (3) Simple event (4) Probability
2. Probability lies between
(1) −1 and +1 (2) 0 and 1 (3) 0 and n (4) 0 and ∞
3. The probability based on the concept of relative frequency theory is called
(1) Empirical probability (2) Classical probability
(3) Both (1) and (2) (4) Neither (1) nor (2)
4. The probability of an event cannot be
(1) Equal to zero (2) Greater than zero (3) Equal to one (4) Less than zero
5. The probability of all possible outcomes of a random experiment is always equal to
(1) One (2) Zero (3) Infinity (4) Less than one
6. If A is any event in S and its complement is A’ then, P(A′) is equal to
(1) 1 (2) 0 (3) 1−A (4) 1−P(A)
7. Which of the following cannot be taken as probability of an event?
(1) 0 (2) 0.5 (3) 1 (4) −1
8. A particular result of an experiment is called
(1) Trial (2) Simple event (3) Compound event (4) Outcome
9. A collection of one or more outcomes of an experiment is called
(1) Event (2) Outcome (3) Sample point (4) None of the above
10. The six faces of the dice are called equally likely if the dice is
(1) Small (2) Fair (3) Six-faced (4) Round