Tossing/Throwing Single/One Coin Problems  Probability
Problem 1
Solution
 The event of getting a HEAD
 The event of getting a TAIL.
Problem 2
(Or) If a coin is tossed, what is the chance of a head?
Solution
Total Number of Possible Choices
= 2 {HEAD, TAIL}
⇒ n = 2
Let
A : the event of getting a head on throwing the coin.
For Event A
= 1 {HEAD}
⇒ m_{A} = 1
Probability of getting a head on throwing the coin
⇒ Probability of occurrence of Event A
= 

⇒ P(A)  = 
 
= 

Alternative
In an experiment with n elementary events all of which are equally likely, mutually exclusive and exhaustive, the probability of occurrence of each elementary event is1 
n 
In the experiment of tossing a coin,
There are two possible elementary events, the events of getting a HEAD and getting a TAIL.
⇒ n =2
These elementary events are
 Mutually exclusive
since only one of them can appear at a time
 Equally likely
since we can expect any one of them to appear and
 Exhaustive
since these are the only two possibilities in the experiment.
If A is the event of getting a HEAD,
For Event A
⇒ Probability of occurrence of the elementary event A
1 
2 
Problem 3
Solution
simultaneously
 at the same instant
In the experiment of tossing a coin,
Total Number of Possible Choices
= 2 {HEAD, TAIL}
⇒ n = 2
Let
A : the event of getting both head and tail simultaneously on throwing the coin.
For Event A
= 0 {Φ}
⇒ m_{A} = 0
Probability of getting both head and tail simultaneously on throwing the coin
⇒ Probability of occurrence of Event A
= 

⇒ P(A)  = 
 
= 
 
=  0 
Impossible Event
The event of getting a head and tail simultaneously is impossible since only one of these can appear at a time on throwing a coin.The probability of an impossible event is zero.
Problem 4
Solution
In the experiment of tossing a coin,
Total Number of Possible Choices
= 2 {HEAD, TAIL}
⇒ n = 2
Let
A : the event of getting a tail.
For Event A
= 1 {TAIL}
⇒ m_{A} = 1
Probability of getting both head and tail simultaneously on throwing the coin
⇒ Probability of occurrence of Event A
= 

⇒ P(A)  = 
 
= 
 
=  0 
Odds
Number of Unfavorable Choices= Total Number of possible choices − Number of Favorable choices
⇒ m_{A}^{c}  =  n − m_{A} 
=  2 − 1  
=  1 
in favor
Odds in Favor of getting a TAIL⇒ Odds in Favor of Event A
= Number of Favorable Choices : Number of Unfavorable Choices
= m_{A} : m_{Ac}
= 1 : 1
against
Odds against getting a TAIL⇒ Odds against Event A
= Number of Unfavorable Choices : Number of Favorable Choices
= m_{Ac} : m_{A}
= 1 : 1
Practice Problem 1
Ans [1/2]
Practice Problem 2
Ans [0]
Practice Problem 3
Ans [1]