The word PROBABILITY is used to indicate an unclear possibility that something might happen. It is also used simultaneously with chance.
Definition of Probability
If an event ‘E’ can happen in ‘m’ ways & fail in ‘k’ ways out of a total of ‘n’ ways, & each of them is equally likely, then the probability of the happening of ‘E’ is .
P(E) = m/(m+k) = m/n, where n = (m+k)
In other words, if a random, experiment is conducted ‘n’ times & ‘m’ of them are favorable to event ‘E’, thenP(E) = m/(m+k) = m/n, where n = (m+k)
Probability always lies between 0 to 1. If your answer exceeds 1 then your answer is incorrect.
Random Experiment
A random experiment is an experiment or a process for which the outcome cannot be predicted with certainty.
Example: - Drawing 2 cards from a well shuffled pack is a random experiment while getting an Ace & a King are events.
Mutually Exclusive Events
In probability theory, two events) are mutually exclusive if they cannot both be true or occur at the same time.Example: The events of getting a head or a tail when a coin is tossed are mutually exclusive.
Equally Likely Events
Example: - When a die is thrown, any number from 1 to 6 may turn up. In this trial, the six events are equally likely.
Independent Events
Two events E1 and E2 are said to be independent, if the occurrence of the event E 2 is not affected by the occurrence or the non-occurrence of the event E 1 .To find the probability of two independent events that occur in sequence, find the probability of each event occurring separately, and then multiply the probabilities. This multiplication rule is defined symbolically below. Note that multiplication is represented by AND.
Compound Events
Example: - When a die is thrown and a coin is flipped the occurring events are called compound events.
Exercise with Explanation
Solution:-
When a coin is tossed we either get head or tail upwards.
So, total number of cases= 2 = n,
number of favorable cases to get H = 1 = m
P (H) = No. of favorable cases/ Total no. of outcomes = (m/n)
=1/2
Solution:-
When a die is rolled we can get any one of the numbers from 1 to 6.
Total number of cases = n = 6
a) Let event A= getting a multiple of 3
Then A= {3.6}.
Therefore m=2
P (A) = m/n = 2/6 = 1/3
b) Let event B = getting a prime number
Then B= {2, 3, 5}.
so, m = 3
P (A) =m/n=3/6=1/2
Before solving this problem, let us recall the game of cards. One deck of cards contains totally 52 cards. Among them we have 13 spades & 13 clubs which are black in color, 13 diamonds & 13 hearts all of which are red. In the 13 spades, 9 are numbered cards, numbered from 2 to 10, one ace card & 3 face cards namely J, Q & K. Similarly for the clubs, diamonds & hearts also.
Totally there are 52 cards & any one can be drawn
So, total number of cases = n = 52
a) There are 13 diamonds & 13 hearts which are red
Number of red cards = m = 26
P (getting red) =26/52=1/2
b) There are 4 queens = 4
P (getting queen) = m/n=4/52
=1/13
c) The king of spade & clubs are black
No. of cards which is king & black=m=2
P (king & black) =m/n =2/56
=1/26
d) The 9 number cards of hearts & 9 number cards of diamonds are red.
No. of cards which are red & number cards = m = 18
P (red & number)=m/n=18/52
=9/26
e) There are 4 queens & kings
No. of favorable case = m = 8
P (queen or king) = m/n = 8/52
= 2/13
Solution: Total no. of beads in the bag = 6+4 =10. n= 10
Any one of the 6 white beads can be selected, m=6
Therefore, P (getting white bead) = m/n=6/10
=3/5
Solution:-
Total number of marbles= 8+6+10 =24, n=24
There are 6 green & 8 red marbles
Therefore, number of favorable cases=6+8=14
P (red or green) = 14/24=7/12
Solution:-
Sample space S= {TT, TH, HT, HH}
No. of total cases=n=4
a) Getting two heads is only one case i.e. HH
No. of favorable case=m=1
P (getting 2 heads)=m/n= 1/4
b) At least one head means one or more heads. In this case it is one or 2 heads, in 3 cases i.e. TH,HT,HH
Therefore, P (atleast one head) =3/4
c) Exactly one head is two cases TH, HT
P (exactly one head) =2/4=1/2
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