PROBABILITY
Introduction:
Probability means the chances of
happening or occurring of an event. The probability of an event tells that how
likely the event will happen. Situations in which each outcome is equally
likely then we can find the probability using probability formula. Probability
is a chance of prediction. If the probability that an event will occur is “x”,
then the probability that the event will not occur is “1 – x”. if the
probability that one event will occur is “a” and the independent probability
that another event will occur is “b”, then the probability that both events
will occur is “ab”.
Definition:
a) An
experiment is a situation involving chance or probability that leads to results
called outcomes.
b) An outcome is the result of a single
trail of an experiment.
c) An event is one or more outcomes of
an experiment.
d) Probability is the measure of how
likely an event is.
Random Experiment:
An operation which produces an
outcome is known as experiment. When an experiment is conducted repeatedly
under the same conditions the results cannot be unique but may be one of the
various possible outcomes. Such an experiment is called a random experiment. In
a random experiment, we cannot predict the outcome. Tossing a coin is a random
experiment. When we toss a coin either head or tail may turn up. Some more
examples of random experiment:
a) Rolling a
die
b) Drawing a card from a pack of cards.
c) Taking out a ball from a bag
containing balls of different colours.
Trail: Performing
a random experiment is called a Trail.
Sample space: The set of all possibility outcomes of a random experiment is called a
sample space and is denoted by S.
When we roll a die, the possible
outcomes are 1, 2, 3, 4, 5, 6.
Sample space is S = {1, 2, 3, 4, 5,
6}
Event: Any
possible outcome or combination of outcome is called event. That is every
subset of the sample space S is called an event. Events are usually denoted by
A, B, C, D, E, F. When a coin is tossed, getting a head or trail is an event.
S = {H, T}, A = {H, B} = {T}
Here events A and B are subsets of
the sample space S.
Equally likely events: Two or more events are said to be equally likely if each one
of them has an equal chance of occurring. In tossing a coin, getting a head and
getting a tail are equally likely events.
Mutually exclusive events: Two or more events are said to be mutually exclusive when the
occurrence of anyone event excludes the occurrence of the other event. Mutually
exclusive events cannot occur simultaneously.
In throwing a die, let E be the event
of getting an odd number and F be the event of getting an even number.
E = {1, 3, 5} and F = {2, 4, 6}
The number that turns up cannot be
odd and even simultaneously. Therefore events E and F are mutually exclusive.
Exhaustive events: If two or more events together constitute the sample space S
then these events are said to be exhaustive events. In throwing a die, the
events of getting an odd number and the event of getting an even number
together from the sample space. So they are exhaustive events.
In an experiment of tossing three
coins, consider the following events.
A: exactly one head appears,
B: exactly two heads appear,
C: exactly three heads appear
D: at least two tails appear
S = {HHH, HHT, HTH, THH, HTT, THT,
TTH, TTT}
A = {HTT, THT, TTH}
B = {HHT, HTH, THH}
C = {HHH}
D = {TTH, THT, HTT, TTT}
The events A, B, C and D together
from the sample space S. That is S = A U B U C U D. Therefore A, B, C and D are
called exhaustive events.
Note: the events B, C and D are mutually
exclusive and Exhaustive.
Complementary events: Let E be an event of a random experiment and S the sample
space. All the other outcome which are not in E belong to the subset S–E. The
event S – E is called the complement of E. It is denoted by E.
In throwing a die, let E be an event
of getting a “multiple of 3”.
E = {3, 6} and E = {1, 2, 4,
5}
Sure Event: Since S is subset of S, S itself is an event and S is called sure or
certain event.
In tossing two coins simultaneously,
let E be an even of getting less than 3 tails. E = {HH, HT, TH, TT} = S. Therefore
E is the sure event.
Impossible event: Let F be an event of getting more than two heads in tossing
two coins simultaneously. F = {} = Ø. So F is an impossible event. Therefore, A
sample is a sure event. An empty set Ø is an impossible event.
Favourable outcomes: The outcomes corresponding to the desired event are called
the favourable outcomes. In rolling a die there are six outcomes. Let E be an
event of getting an even number. Then the outcomes 2, 4, 6 are favourable to
the event E.
Probability of an event: if a sample space contains n outcomes, m of which are
favourable to an event E, then the probability of an event E, denoted by P€, is
defined as the ratio of m to n.
Note:
a) The
probability of sure event is 1. That is P(S) = 1
b) The
probability of impossible event is 0. That is P(Ø) = 0
Important Formulae:
1) If a random
experiment is conducted in which there are n elementary events, all equally
likely, and A is an event, the probability of event A is:
MEMORY BASED SOLVED EXAMPLES BASED ON VARIOUS TYPES:
Type 1 – Based on Coins
1. When tossing two coins once, what is the probability of heads on
both the coins?
Solution:
Total number of outcomes possible
when a coin is tossed = 2 (∵ Head or Tail)
Hence, total number of outcomes
possible when two coins are tossed, n(S) = 2 × 2 = 4
(∵ Here, S = {HH, HT, TH, TT})
E = event of getting heads on both
the coins = {HH}
Hence, n(E) = 1
P(E) = n(E)/n(S) = ¼
2. A fair coin is tossed 100 times, The probability of getting head
an odd number of times is?
Solution:
Solution:
∵ n(S) = (2)3 = 8
E = Event of getting 0, or 1 or 2
heads = {TTT, TTH, THT, HTT, HHT, HTH, THH}
⇒ n(E) = 7
∴ P(E) = n(E)/n(S) = 7/8
Type 2 - Based on Cards
1. One card is randomly drawn from a pack of 52 cards. What is the
probability that the card drawn is a face card (Jack, Queen or King)?
Solution:
Total number of cards, n(S) = 52
Total number of face cards, n(E) = 12
P(E) = 12/52 = 3/13
2. One card is drawn at random from a pack of 52 cards. What is the
probability that the card drawn is either a red card or a king?
Solution:
Clearly n(S) = 52. There are 26 red
cards (including 2 kings ) and there are 2 more kings.
Let (E) be the event of getting
either a red card or a king .
Then, n(E) = 28
∴ P(E) = n(E) / n(S) = 28/52 = 7/13
3. A card is drawn from a pack of 52 cards. A card is drawn at
random. What is the probability that it is neither a heart nor a king?
Solution:
There are 13 hearts and 3 more kings
∴ p (heart or a king ) = (13 + 3)/52 = 4/13
∴ P (neither a heart nor a king) = 1 - 4/13 = 9/13
4. From a set of 17 cards numbered 1, 2, 3 ....... 17 one is drawn
at random. The probability that the number is divisible by 3 or 7 is?
Solution:
Total number of cases is 17.
∵ Number divisible by 3 are 3, are 3, 6, 9, 12, 15 (These are
5 in number)
Number divisible by 7 are 7, 14.
(These are 2 in number)
There are two favourable numbers of
cases
Total no. of favourable number = 5 +
2
Required probability = 7/17.
5. In shuffling a pack of card 3 are accidentally dropped then the
chance that missing card should be of different suit is?
Solution:
Total ways 52C3
= 22100
There are 4 suit in a pack of cards
so three suit can be selected in 4C3 ways.
One card each from different unit can
selected as = 13C1 X 13C1 X 13C1
ways
So, favorable ways = 4C3
X 13C1 X 13C1 X 13C1
= 8788
∴ Required probability = 8788/22100 = 169/425
Type 3 – Based on Dice
1. A die is rolled twice. What is the probability of getting a sum
equal to 9?
Solution:
Total number of outcomes possible
when a die is rolled = 6 (∵ any one face out of the 6 faces)
Hence, total number of outcomes
possible when a die is rolled twice, n(S) = 6 × 6 = 36
E = Getting a sum of 9 when the two
dice fall = {(3, 6), {4, 5}, {5, 4}, (6, 3)}
Hence, n(E) = 4
P(E) = n(E)/n(S) = 4/36 = 1/9
2. A dice is rolled three times and sum of three numbers appearing
on the uppermost face is 15. The chance that the first roll was four is
Solution:
Total number of favourable outcomes
n(S) = 63 = 216
Combinations of outcomes for getting
sum of 15 on uppermost face = (4, 5, 6), (5, 4, 6), ( 6, 5, 4), (5, 6, 4),
(4, 6, 5), (6, 4, 5), (5, 5, 5),(6,
6, 3), (6, 3, 6), (3, 6, 6)
Now, outcomes on which first roll was
a four, n(E) = (4, 5, 6),(4, 6, 5)
∴ P(E) = n(E)/n(S) = 2/216 =1/108
3. The chance of throwing a total of 3 or 5 or 11 with two dice is?
Solution:
∵ Probability for 3 = (1, 2), (2, 1) = 2/36
⇒ Probability for 5 = (1, 4), (2, 3), (3, 2), (4, 1)= 4/ 36
⇒ Probability for 11 (5, 6), (6, 5) = 2/36
∴ Reqd. Probability = 2/36 + 4/36 + 2/36 = 8/36 = 2/9
4. A six - faced dice is so biased that is twice as likely to show
an even number as an odd number when throw. It is thrown twice. The probability
that the sum of two numbers thrown is even is?
Solution:
∵ Probability for odd = p
∴ Probability for even = 2p
∵ p + 2p = 1
⇒ 3p = 1
⇒ p = 1/3
∴ Probability for odd = 1/3, Probability for even = 2/3.
Sum of two nos. is even means either
both are odd or both are even
∴ Reqd. probability = 1/3 x 1/3 + 2/3 x 2/3 = 1/9 + 4/9 = 5/9
[∵ die is thrown twice ]
Type 4 - Miscellaneous
1. The probability that a man lives after 10 years is 1/4 and that
his wife is alive after 10 years is 1/3. The probability that neither of them
is alive after 10 years is ?
Solution:
P(M) = ¼; P(W) = 1/3
P(M) = 1 – ¼ = 3/4 ; P(W)
= 1 – 1/3 = 2/3
Required probability = P(M) P(W)
= ¾ × 2/3 = ½
2. There are 100 students in a college class of which 36 are boys
studying statistics and 13 girls not studying statistics. If there are 55 girls
in all, then the probability that a boy picked up at random is not studying
statistics, is
Solution:
There are 55 girls and 45 boys in the
college
Out of 45 boys, 36 are studying
Statistics and 9 are not studying statistics.
The probability that a boy picked up
at random is not studying Statistics = 9/45 = 1/5
3. A bag contains 2 yellow, 3 green and 2 blue balls. Two balls are
drawn at random. What is the probability that none of the balls drawn is blue?
Solution:
Total number of balls = 2 + 3 + 2 = 7
Let S be the sample space.
n(S) = Total number of ways of
drawing 2 balls out of 7 = 7C2 = 21
Let E = Event of drawing 2 balls,
none of them is blue.
n(E) = Number of ways of drawing 2
balls, none of them is blue
= Number of ways of drawing 2 balls
from the total 5 (=7-2) balls = 5C2 = 10
(∵There are two blue balls in the total
7 balls. Total number of non-blue balls = 7 - 2 = 5)
P(E) = n(E)/n(S) = 10/21
4. Out of 15 students studying in a class, 7 are from Maharastra, 5
are from Karnataka and 3 are from Goa, Four students are to be selected at
random. What are the chances that at least one is from Karnataka?
Solution:
Total possible ways of selecting 4
students out of 15 students = 15C4
= (15 x 14 x 13 x 12) / (1 x 2 x 3 x
4) = 1365
The no. of ways of selecting 4
students in which no students belongs to karnataka = 10C4
∴ Number of ways of selecting atleast one student from
karnataka = 15C4 - 10C4 = 1155.
∴ Required probability = 1155 / 1365 = 77 / 91 = 11 / 13
5. If the probability for A fail in an examination is 0.2 and that
for B is 0.3, then the probability that either A or B fails is?
Solution:
Solution:
The probability that head is show in
one coin is 1/2.
The probability that the sum of the
number on the dice is a prime = the probability that the following pair of
number on the dice is a getting on
the dice, namely (1, 1), (1, 2), (2, 1), (1, 4 ), (4, 1), (2, 3), (3, 2), (1,
6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3), (6, 5), (5, 6) = 15/36.
∴ The required probability = 1/2 x 1/2 x 15/36 = 5/48.
7. A speaks truth in 60% of the cases and B in 80% of the cases. In
what percentage of cases are they likely to contradict each other, narrating
the same incident?
Solution:
Solution:
Total number of cases = 10C4
Favourable number of cases = 4C2.
6C2
{Since, we are to select 2 children
out of 4 and remaining 2 persons are to be selected from remaining 6 persons (
2W + 4M)}
∴ required Probability = 4C2 . 6C2
/ 10C4
= [{(4 x 3) / (2 x 1)} x {(6 x 5) /
(2 x 1)}] / [(10 x 9 x 8 x 7) / (4 x 3 x 2 x 1)]
= [(12/2) x (30/2)] / 210 = 90 / 210
= 9 /12
9. The probability that a man can hit a target is 3/4. He tries 5
times. The probability that he will hit the target at least three times, is
Solution:
Required Probability
Given, n = 5 and r = 3
Then, Success P = ¾
Failure, q = 1 - 3/4 = 1/4
Man hit the target thrice = 5C3
(3/4)3 (1/4)2 + 5C4 (3/4)3
(1/4) + 5C3(3/4)5
= (270/1024) + (405/1024) +
(243/1024) = 918/1024 = 459/512
10. The letters B, G, I, N, R are rearranged to form the word
'BRING'. Find its probability?
Solution:
The five letters could be arrange in
5! ways.
One of them is 'BRING'.
∴ Required probability = 1/5! = 1/(5 x 4 x 3 x 2 x 1) = 1/120
