Experiment:
An operation which can produce some well-defined outcome is
called an experiment.
Random Experiment:
An experiment in which all possible out comes are known and
the exact output cannot be predicted in advance is called a
random experiment.
EX:
1) Rolling an unbiased dice.
2) Tossing a fair coin.
3) Drawing a card from a pack of well-shuffled cards .
4)Picking up a ball of certain colour from a bag containing
balls of different colours.
Details:
1) When we thrown a coin ,then either a Head(H) or a Tail(T)
appears.
2)A dice is a solid cube ,having 6 faces,marked 1,2,3,4,5,6
respectively. When we throw a die ,the outcome is the number
that appears on its upper face.
3)A pack of cards has 52 cards.
It has 13 cards of each suit,namely spades,clubs,hearts and
diamonds. Cards of spades and clubs are black cards.
Cards of hearts and diamonds are red cards.
There are four hours of each suit.
These are Aces,Kings,queens and Jacks.
These are called Face cards.
Sample Space:
When we perform an experiment ,then the set of S of all
possible outcomes is called the Sample space .
EX:
1)In tossing a coin S= {H,T}.
2)If two coins are tossed then S= {HH,HT,TH,TT}.
3)In rolling a dice ,we have S={1,2,3,4,5,6}.
Event:Any subset of a sample space is called an Event.
Probability of occurrence of an Event:
Let S be the sample space.
Let E be the Event.
Then E cs i.e E is subset of S then
probability of E p(E) =n(E)/n(S).
Results on Probability:
1)P(S) =1.
2)0 < P(E) < 1
probability of an event lies between 0 and 1.
Max value of probability of an event is one.
3)P(Ф)=0.
4)For any events A and B we have .
P(AUB) =P(A) +P(B) -P(AnB).
5)If A denotes (not -A) then
P(A) =1-P(A)
P(A)+P(A) =1.
Problems:
1)An biased die is tossed.Find the probability of getting a
multiple of 3?
Sol: Here we have sample space S={1,2,3,4,5,6}.
Let E be the event of getting a multiple of 3.
Then E={3,6}.
P(E) =n(E)/n(S).
n(E) =2,
n(S) =6.
P(E) =2/6
P(E) =1/3.
2)In a simultaneous throw of a pair of dice,find the
probability of getting a total more than 7?
Sol: Here we have sample space n(S) =6*6 =36.
Let E be the event of getting a total more than 7.
={(1,6),(2,5),(3,4),(4,3)(5,2),(6,1)(2,6),(3,5),(4,4),
(5,3),(6,2),(4,5),(5,4),
(5,5),(4,6),(6,4)}
n(E) =15
P(E) = n(E)/n(S)
= 15/36.
P(E) = 5/12.
3)A bag contains 6 white and 4 black balls .Two balls are
drawn at random .Find the probability that they are of the
same color?
Sol: Let S be the sample space.
Number of ways for drawing two balls out of 6 white and
4 red balls = 10 C2
=10!/(8!*2!)
= 45.
n(S) =45.
Let E =event of getting both balls of the same color.
Then
n(E) =number of ways of drawing ( 2 balls out of 6) or
(2 balls out of 4).
= 6C2 +4C2
= 6!/(4!*2!) + 4!/(2! *2!)
= 6*5/2 +4 *3/2
=15+6 =21.
P(E) =n(E)/n(S) =21/45 =7/45.
4)Two dice are thrown together .What is the probability that
the sum of the number on
the two faces is divisible by 4 or 6?
Sol: n(S) = 6*6 =36.
E be the event for getting the sum of the number on the two
faces is divisible by 4 or 6.
E={(1,3)(1,5)(2,4?)(2,2)(3,5)(3,3)(2,6)(3,1)(4,2)(4,4)
(5,1)(5,3)(6,2)(6,6)}
n(E) =14.
Hence P(E) =n(E)/n(S)
= 14/36.
P(E) = 7/18
5)Two cards are drawn at random from a pack of 52 cards What
is the probability that either both are black or both are
queens?
Sol: total number of ways for choosing 2 cards from
52 cards is =52 C2 =52 !/(50!*2!)
= 1326.
Let A= event of getting both black cards.
Let B= event of getting both queens
AnB=Event of getting queens of black cards
n(A) =26 C2.
We have 26 black cards from that we have to choose 2 cards.
n(A) =26 C2=26!/(24!*2!)
= 26*25/2=325
from 52 cards we have 4 queens.
n(B) = 4C2
= 4!/(2!* 2!) =6
n(AnB) =2C2. =1
P(A) = n(A) /n(S) =325/1326
P(B) = n(B)/n(S) = 6/1326
P(A n B) = n(A n B)/n(S) = 1/1326
P(A u B) = P(A) +P(B) -P(AnB)
= 325/1326 + 6/1326 -1/1326
= 330/1326
P(AuB) = 55/221
