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}
⇒ mA = 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}
⇒ mA = 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}
⇒ mA = 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
⇒ mAc | = | n − mA |
= | 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
= mA : mAc
= 1 : 1
against
Odds against getting a TAIL⇒ Odds against Event A
= Number of Unfavorable Choices : Number of Favorable Choices
= mAc : mA
= 1 : 1
Practice Problem 1
Ans [1/2]
Practice Problem 2
Ans [0]
Practice Problem 3
Ans [1]