Drawing/Picking/Choosing Single/One black/white ball from a bag/urn/box
Problem 2
4 
9 
5 
9 
Solution
Total number of balls in the bag
= 12 Black + 15 White
= 27
Experiment :
Drawing a ball at random from the bag containing 12 black and 15 white balls
Total Number of Possible Choices
= Number of ways in which one ball can be drawn from the total 27
⇒ n  =  ^{27}C_{1}  
= 
 
=  27 
Let
A : the event of the ball drawn being black
B : the event of the ball drawn being white
For Event A
Favorable (Black)  Unfavorable (Others)  Total  

Available  12  15  27 
To Choose  1  0  1 
Choices  ^{12}C_{1}  ^{15}C_{0}  ^{27}C_{1} 
Number of Favorable Choices
= Number of ways in which one black ball can be drawn from the total 12
⇒ m_{A}  =  ^{12}C_{1}  
= 
 
=  12 
Probability of the ball drawn being black
⇒ Probability of occurrence of Event A
= 

⇒ P(A)  = 
 
= 
 
= 

Odds
Number of Unfavorable Choices= Total Number of possible choices − Number of Favorable choices
⇒ m_{A}^{c}  =  n − m_{A} 
=  27 − 12  
=  15 
in favor
Odds in Favor of the ball drawn being black⇒ Odds in Favor of Event A
= Number of Favorable Choices : Number of Unfavorable Choices
= m_{A} : m_{Ac}
= 12 : 15
= 4 : 5
against
Odds against the ball drawn being black⇒ Odds against Event A
= Number of Unfavorable Choices : Number of Favorable Choices
= m_{Ac} : m_{A}
= 15 : 12
= 5 : 4
For Event B
Favorable (White)  Unfavorable (Others)  Total  

Available  15  12  27 
To Choose  1  0  1 
Choices  ^{15}C_{1}  ^{12}C_{0}  ^{27}C_{1} 
Number of Favorable Choices
= Number of ways in which one white ball can be drawn from the total 15
⇒ m_{B}  =  ^{15}C_{1}  
= 
 
=  15 
Probability of the ball drawn being white
⇒ Probability of occurrence of Event B
= 

⇒ P(B)  = 
 
= 
 
= 

Odds
Number of Unfavorable Choices= Total Number of possible choices − Number of Favorable choices
⇒ m_{B}^{c}  =  n − m_{B} 
=  27 − 15  
=  12 
in favor
Odds in Favor of the ball drawn being white⇒ Odds in Favor of Event B
= Number of Favorable Choices : Number of Unfavorable Choices
= m_{B} : m_{Bc}
= 15 : 12
= 5 : 4
against
Odds against the ball drawn being white⇒ Odds against Event B
= Number of Unfavorable Choices : Number of Favorable Choices
= m_{Bc} : m_{B}
= 12 : 15
= 4 : 5
For Event B [Alternative I]
Probability of the ball drawn being white
= Probability of the ball drawn not being black.
⇒ P(B)  =  P (A^{c})  
=  1 − P (A)  
=  1 −
 
= 
 
= 

For Event B [Alternative II]
The number of favourable choices for drawing a black ball would form the number of unfavorable choices for drawing a white ball and viceversa
Number of Favorable Choices
= Number of Unfavorable Choices for Event A
⇒ m_{B}  =  m_{Ac} 
=  1 − m_{A}  
=  1 − 12  
=  15 
Probability of the ball drawn being white
⇒ Probability of occurrence of Event B
= 

⇒ P(B)  = 
 
= 
 
= 
