Basic Concepts of Probability, Business Mathematics and Statistics Video Lecture | SSC CGL Tier 2 - Study Material, Online Tests, Previous Year

if you roll a fair six-sided die there
are six possible outcomes and each of
these outcomes is equally likely a six
is as likely to come up as a three and
the same goes for the other four sides
of the die so what is the probability
that a one will come up since there are
six possible outcomes the probability is
1/6 what is the probability that either
a 1 or a 6 will come up
the two outcomes about which we are
concerned a one or a six coming up are
called favorable outcomes given that all
outcomes are equally likely we can
compute the probability of a 1 or 6 by
dividing the number of favorable
outcomes 2 in this case by the total
number of possible outcomes 6 in this
case so the probability of throwing
either a 1 or 6 is 1/3 don't be misled
by our use of the term favorable by the
way you should understand it to mean
favorable to the event in question
happening that event might not be
favorable to your well-being this
formula applies to many games of chance
for example what is the probability that
a card drawn at random from a deck of
playing cards will be an ace
since the deck has 4 aces there are 4
favorable outcomes since the deck has 52
cards there are 52 possible outcomes the
probability is therefore for over 52
which equals 113th what about the
probability that the card will be a club
since there are 13 clubs the probability
is 13 over 52 or 1/4
let's say you have a bag with 20
cherries 14 sweet and six sour
if you pick a cherry at random what is
the probability that it will be sweet
there are 20 possible charities that
could be picked so the number of
possible outcomes is 20 of these 20
possible outcomes 14 are favorable or
sweet so the probability that the cherry
will be sweet is 14 over 20 or 7/10
there is one potential complication
however it must be assumed that the
probability of picking any of the
cherries is the same as a probability of
picking any other this wouldn't be true
if let us imagine the sweet cherries are
smaller than the sour ones the sour
cherries would come to hand more readily
when you sampled from the bag let us
keep in mind therefore that when we
assess probabilities in terms of the
ratio of favorable to all potential
cases we rely heavily on the assumption
of equal probability for all outcomes
here is a more complex example you throw
two dice what is the probability that
the sum of the two dice will be six to
solve this problem list all the possible
outcomes there are 36 of them since each
die can come up one of six ways the 36
possibilities are shown here you can see
that five of the 36 possibilities total
six therefore the probability is five 36
if you know the probability of an event
occurring it is easy to compute the
probability that the event does not
occur if PA is the probability of event
a then 1 minus PA is the probability
that the event does not occur for the
last example the probability that the
total is 6 is 536 therefore the
probability that the total is not 6 is 1
minus 536 which equals 31 30 sixths
events a and B are independent events if
the probability of event B occurring is
the same whether or not event a occurs
let's take a simple example a fair coin
is tossed 2 times the probability that a
head comes up on the second toss is 1/2
regardless of whether or not a head came
up on the first toss the two events are
1 first toss is a head and 2 second toss
is a head so these events are
independent
consider the two events one it will rain
tomorrow in Houston and two it will rain
tomorrow in Galveston a city near
Houston these events are not independent
because it is more likely that it will
rain in Galveston on days that rains in
Houston than on days it does not
when two events are independent the
probability of both occurring is the
product of the probabilities of the
individual events more formally if
events a and B are independent then the
probability of both a and B occurring is
the probability of a time's the
probability of B if you flip a coin
twice what is the probability that it
will come up heads both times event a is
that the coin comes up heads on the
first flip and event B is that the coin
comes up heads on the second flip
since both PA and PB equal 1/2 the
probability that both events occur is
one-half times one-half which equals 1/4
let's take another example if you flip a
coin and roll a 6-sided die what is the
probability that the coin comes up heads
and the die comes up 6 since the two
events are independent the probability
is simply the probability of a head
which is one-half times the probability
of the die coming up 6 which is 1/6
therefore the probability of both events
occurring is 1/2 times 1/6 which equals
1/12
one final example you draw a card from a
deck of cards
put it back and then draw another card
what is the probability that the first
card is a heart and the second card is
black since there are 52 cards in a deck
and thirteen of them are hearts the
probability that the first card is heart
is 13 over 52 which equals one-fourth
since there are 26 black cards in the
deck the probability that the second
card is black is 26 over 52 or one-half
the probability of both events occurring
is therefore 1/4 times 1/2 which equals
1/8
if events a and B are independent the
probability that either event a or event
B occurs is the probability that event a
occurs plus the probability that event B
occurs minus the probability that both
events a and B occur in this discussion
when we say a or B occurs we include
three possibilities one a occurs and B
does not occur
- B occurs and a does not occur or three
both a and B occur
now for some examples if you flip a coin
two times what is the probability that
you will get a head on the first flip or
a head on the second flip or both
letting event a be a head on the first
flip and event B be head on the second
flip then PA equals one-half PB equals
one-half and PA and B equals one-fourth
therefore P a or B equals 1/2 plus 1/2
minus 1/4 equals 3/4
if you throw a six-sided die and then
flip a coin what is the probability that
you will get either a 1 on the die or a
head on the coin flip or both
the probability is 1/6 plus 1/2 minus
1/6 times 1/2
the result is 7 twelfths
an alternate approach to computing this
value is to start by computing the
probability of not getting either a six
or a head then subtract this value from
1 to compete the probability of getting
a six or a head although this is a
complicated method it has the advantage
of being applicable to problems with
more than two events here is a
calculation in the present case the
probability of not getting either a six
or a head can be recast as a probability
of not getting a six and not getting a
head this follows because if you do not
get a six and you do not get a head then
you do not get a six or a head the
probability of not getting a six is one
minus 1/6 or 5/6 the probability of not
getting a head is 1 minus 1/2 which
equals 1/2 the probability of not
getting a 6 and not getting a head is
5/6 times 1/2 which equals 5 twelfths
the probability of getting a 6 or a head
is therefore once again 1 minus 512
which equals 7 twelfths
if you throw a die three times what is
the probability that one or more of your
throws will come up a six that is what
is the probability of getting a six on
the first throw or a six on the second
throw or a six on the third throw the
easiest way to approach this problem is
to compute the probability of not
getting a six on the first throw and not
getting a six on the second throw and
not getting a six on the third throw the
answer will be 1 minus this probability
the probability of not getting a 6 on
any of the three throws is 5/6 times 5/6
times 5/6 which equals 125 over 216
therefore the probability of getting a 6
on at least one of the throws is 1 minus
125 over 216 which equals 91 over 216
often it is required to compute the
probability of an event given that
another event has occurred for example
what is the probability that two cards
drawn at random from a deck of playing
cards we'll both be aces it might seem
that you could use the formula for the
probability of two independent events
and simply multiply for over 52 times 4
over 52 which equals 1 over 169
this would be incorrect however because
the two events are not independent if
the first card drawn is an ace then the
probability that the second card is also
an ace would be lower because there
would be only three aces left in the
deck
once the first card chosen is an ace the
probability that the second card chosen
is also an ace is called the conditional
probability of drawing an ace in this
case the condition is that the first
card is an ace symbolically we write
this as shown the vertical bar is read
as given so the above expression is
short for the probability that an ace is
drawn on the second draw given that an
ace was drawn on the first draw what is
this probability after an ace is drawn
on the first draw there are three aces
at a fifty-one total cards left this
means the probability that one of these
aces will be drawn is three over 51
which equals 117
if events a and B are not independent
then the probability of a and B is the
probability of a times the conditional
probability of B given a applying this
to the problem of two aces the
probability of drawing two aces from a
deck is four over fifty two times three
over fifty one which equals one over 221
one more example if you draw two cards
from a deck what is the probability that
you will get the ace of diamonds and a
black card there are two ways you can
satisfy this condition a you can get the
ace of diamonds first and then a black
card or B you can get a black card first
and then the ace of diamonds
let's calculate case a the probability
that the first card is the ace of
diamonds is 1 over 52 the probability
that the second card is black given that
the first card is the ace of diamonds is
26 over 51 because 26 of the remaining
51 cards are black the probability is
therefore 1 over 52 times 26 over 51
which equals 1 over 102 now for case B
the probability that the first card is
black is 26 over 52 which equals 1/2 the
probability that the second card is the
ace of diamonds given that the first
card is black is 1 over 51 the
probability of case 2 is therefore 1/2
times 1 over 51 which equals 1 over 102
the same as the probability of case 1
recall that the probability of A or B is
the probability of a plus the
probability of B minus the probability
of a and B in this problem the
probability of a and B equals zero since
a card cannot be the ace of diamonds and
be a black card therefore the
probability of case a or case B is 1
over 102 plus 1 over 102 which equals 2
over 102 therefore 2 over 102 which
equals 1 over 51 is the probability that
you will get the ace of diamonds and a
black card
when drawing two cards from a deck
if there are 25 people in a room what is
the probability that at least two of
them share the same birthday if your
first thought is that it is 25 over 365
which equals 0.068 you will be surprised
to learn that it is much higher than
that this problem requires the
application of the sections on
probability of a and B and conditional
probability
this problem is best approached by
asking what is the probability that no
two people have the same birthday once
we know this probability we can simply
subtract it from one to find the
probability that two people share a
birthday if we choose two people at
random what is the probability that they
do not share a birthday of the 365 days
on which the second person could have a
birthday 364 of them are different from
the first person's birthday therefore
the probability is 364 over 365 let's
define p2 as the probability that the
second person drawn does not share a
birthday with the person drawn
previously p2 is therefore 364 over 365
now define p3 is a probability that the
third person drawn does not share a
birthday with anyone drawn previously
given that there are no previous
birthday matches p3 is therefore a
conditional probability
if there are no previous birthday
matches then two of the 365 days have
been used up leaving 363 non-matching
days therefore P 3 equals 363 over 365
in like manner P 4 equals 362 over 365 P
5 equals 361 over 365 and so on up to P
25 equals 341 over 365
in order for there to be no matches the
second person must not match any
previous person and the third person
must not match any previous person and
the fourth person must not match any
previous person etc to find the
probability of no matches all we have to
do is multiply P 2 P 3 P 4 up through p
25 together the result is 0.4 3 1
therefore the probability of at least
one match is 1 minus 0.4 3 1 which
equals 0.5 6 1
a fair coin is flipped five times and
comes up heads each time what is the
probability that it will come up heads
on the sixth flip the correct answer of
course is 1/2
but many people believe that a tail is
more likely to occur after throwing five
heads their faulty reasoning may go
something like this in the long run
the number of heads and tails will be
the same so the tails have some catching
up to do the flaws in this logic are
exposed in our simulation on the
gamblers fallacy in short the proportion
of heads approaches 0.5 as a number of
flips increases however the difference
between the number of heads and the
number of tails does not
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