Drawing/Selecting/Picking/Choosing Two/Three/More Coins from a Box/Bag/Urn
Problem 1
Solution
Total number of coins in the bag
=  12 Two Rupee Coins + 7 One Rupee Coins + 4 Half Rupee Coins  
=  23 Coins 
Number of coins selected = 3
Experiment : Selecting 3 coins from the bag
Total Number of Possible Choices
=  Number of ways in which three coins can be drawn from the total 23  
⇒ n  =  ^{23}C_{5}  

= 
 
=  23 × 11 × 7 × 19  
=  33,649 
Let
 A be the event of the sum of the three coins being maximum
 B be the event of the sum of the three coins being minimum
 C be the event of the three coins being of different values
For Event A
Event A ⇒ Event of the selecting three two rupee coins.
[Two Rupee]  [One Rupee]  [Half Rupee]  Total  

Available  12  7  4  23 
To Choose  3  0  0  3 
Choices  ^{12}C_{3}  ^{7}C_{0}  ^{4}C_{0}  ^{23}C_{3} 
Number of Favorable Choices
=  The number of ways in which the 3 two rupee coins can be selected from the total 12 two rupee coins  
⇒ m_{A}  =  ^{12}C_{3}  

= 
 
=  220 
Probability of the sum of the three coins being maximum
⇒ 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} 

=  33,649 − 220  
=  33,429 
» in favor
Odds in Favor of the sum of the three coins being maximum
⇒ Odds in Favor of Event A
=  Number of Favorable Choices : Number of Unfavorable Choices 
=  m_{A} : m_{A}^{c} 
=  6,160 : 9,344 
=  220 : 33,429 
=  20 : 3,039 
» against
Odds against the sum of the three coins being maximum
⇒ Odds against Event A
=  Number ofUnfavorable Choices : Number of Favorable Choices 
=  m_{Ac} : m_{A} 
=  33,429 : 220 
=  3,039 : 20 
For Event B
Event B ⇒ Event of the selecting three Half rupee coins.
[Two Rupee]  [One Rupee]  [Half Rupee]  Total  

Available  12  7  4  23 
To Choose  0  0  3  3 
Choices  ^{12}C_{0}  ^{7}C_{0}  ^{4}C_{3}  ^{23}C_{3} 
Number of Favorable Choices
=  The number of ways in which the 3 half rupee coins can be selected from the total 4 half rupee coins  
⇒ m_{A}  =  ^{4}C_{3} 

=  ^{4}C_{(4 − 3)}^{n}C_{r} = ^{n}C_{(n − r)}  
=  ^{4}C_{1}  
=  4 
Probability of the sum of the three coins being minimum
⇒ Probability of occurrence of Event B
= 
 
⇒ P(B)  = 
 

= 

For Event C
[Two Rupee]  [One Rupee]  [Half Rupee]  Total  

Available  12  7  4  23 
To Choose  1  1  1  3 
Choices  ^{12}C_{1}  ^{7}C_{1}  ^{4}C_{1}  ^{23}C_{3} 
Number of Favorable Choices
=  The number of ways in which the a two rupee coin, a one rupee coin and a half rupee coin can be selected from the total 23 coins  
⇒ m_{A}  =  (Number of ways in which one two rupee coin can be selected from the 12) × (Number of ways in which one rupee coin can be selected from the 7) × (Number of ways in which one half rupee coin can be selected from the 4)
 

=  ^{12}C_{1} × ^{7}C_{1} × ^{4}C_{1}  
=  12 × 7 × 4  
=  336 
Probability of the three coins being of different values
⇒ Probability of occurrence of Event C
= 
 
⇒ P(C)  = 
 

= 

Problem 2
Solution
Total number of coins in the box
=  2 Fifty Paise Coins + 5 Twenty Five Paise Coins + n (Ten paise + Five Paise) Coins 
=  (7 + n) Coins 
Number of coins selected = 5
Experiment : Picking 5 coins from the box
Total Number of Possible Choices
=  Number of ways in which five coins can be drawn from the total (7 + n) coins  
⇒ n  =  ^{(7 + n)}C_{5} 

Let A be the event of the total value of the five coins being less than One rupee Fifty Paise.
For Event A
Event A can be accomplished in 13 alternative ways
 A_{1} : Picking 2 Fifty paise Coins + 1 Twenty Five paise Coin + 2 (Ten + Five) Paise Coins
 A_{2} : Picking 2 Fifty paise Coins + 0 Twenty Five paise Coin + 3 (Ten + Five) Paise Coins
 A_{3} : Picking 1 Fifty paise Coins + 4 Twenty Five paise Coin + 0 (Ten + Five) Paise Coins
 A_{4} : Picking 1 Fifty paise Coins + 3 Twenty Five paise Coin + 1 (Ten + Five) Paise Coins
 A_{5} : Picking 1 Fifty paise Coins + 2 Twenty Five paise Coin + 2 (Ten + Five) Paise Coins
 A_{6} : Picking 1 Fifty paise Coins + 1 Twenty Five paise Coin + 3 (Ten + Five) Paise Coins
 A_{7} : Picking 1 Fifty paise Coins + 0 Twenty Five paise Coin + 4 (Ten + Five) Paise Coins
 A_{8} : Picking 0 Fifty paise Coins + 5 Twenty Five paise Coin + 0 (Ten + Five) Paise Coins
 A_{9} : Picking 0 Fifty paise Coins + 4 Twenty Five paise Coin + 1 (Ten + Five) Paise Coins
 A_{10} : Picking 0 Fifty paise Coins + 3 Twenty Five paise Coin + 2 (Ten + Five) Paise Coins
 A_{11} : Picking 0 Fifty paise Coins + 2 Twenty Five paise Coin + 3 (Ten + Five) Paise Coins
 A_{12} : Picking 0 Fifty paise Coins + 1 Twenty Five paise Coin + 4 (Ten + Five) Paise Coins
 A_{13} : Picking 0 Fifty paise Coins + 0 Twenty Five paise Coin + 5 (Ten + Five) Paise Coins
[Fifty Paise]  [Twenty Five Paise]  [Ten Paise + Five Paise]  Total  

Available  2  5  n  (7 + n)  
To Choose  2  1  2  5  A_{1} 
Choices  ^{2}C_{2}  ^{5}C_{1}  ^{n}C_{2}  ^{(7 + n)}C_{5}  
To Choose  2  0  3  5  A_{2} 
Choices  ^{2}C_{2}  ^{5}C_{0}  ^{n}C_{3}  ^{(7 + n)}C_{5} 
Calculating the total number of favorable/favourable choices for the event is not possible for all these alternatives, specifically for these.
 A_{2} : Picking 2 Fifty paise Coins + 0 Twenty Five paise Coin + 3 (Ten + Five) Paise Coins
 A_{6} : Picking 1 Fifty paise Coins + 1 Twenty Five paise Coin + 3 (Ten + Five) Paise Coins
 A_{7} : Picking 1 Fifty paise Coins + 0 Twenty Five paise Coin + 4 (Ten + Five) Paise Coins
 A_{11} : Picking 0 Fifty paise Coins + 2 Twenty Five paise Coin + 3 (Ten + Five) Paise Coins
 A_{12} : Picking 0 Fifty paise Coins + 1 Twenty Five paise Coin + 4 (Ten + Five) Paise Coins
 A_{13} : Picking 0 Fifty paise Coins + 0 Twenty Five paise Coin + 5 (Ten + Five) Paise Coins
The number of ten and five paise coins together is given to be n (≥ 2). We do not know for sure whether there are 3 or more coins of this type.
Thus we take up an alternative route to find P(A).
Let B be the event of the total value of the five coins being greater than or equal to One rupee Fifty Paise.
For Event B
Event B can be accomplished in two alternative ways
 B_{1} : Picking 2 Fifty paise Coins + 3 Twenty Five paise Coin + 0 (Ten + Five) Paise Coins
 B_{2} : Picking 2 Fifty paise Coins + 2 Twenty Five paise Coin + 1 (Ten + Five) Paise Coins
[Fifty Paise]  [Twenty Five Paise]  [Ten Paise + Five Paise]  Total  

Available  2  5  n  (7 + n)  
To Choose  2  3  0  5  B_{1} 
Choices  ^{2}C_{2}  ^{5}C_{3}  ^{n}C_{0}  ^{(7 + n)}C_{5}  
To Choose  2  2  1  5  B_{2} 
Choices  ^{2}C_{2}  ^{5}C_{2}  ^{n}C_{1}  ^{(7 + n)}C_{5} 
» "B_{1}"
Number of Favorable Choices
=  The number of ways in which two fifty paise coins and three twenty five paise coins can be drawn from the total (7 + n) coins  
m_{B1}  =  (Number of ways in which 2 fifty paise coins can be drawn from the total 2) × (Number of ways in which 3 twenty five paise coins can be drawn from the total 5)  

=  ^{2}C_{2} × ^{5}C_{3}  
= 
 
=  10 
» "B_{2}"
Number of Favorable Choices
=  The number of ways in which two fifty paise coins, two twenty five paise coins and one (ten or five) paise coin can be drawn from the total (7 + n) coins  
m_{B2}  =  (Number of ways in which 2 fifty paise coins can be drawn from the total 2) × (Number of ways in which 2 twenty five paise coins can be drawn from the total 5) × (Number of ways in which 1 ten or five paise coin can be drawn from the total n)  

=  ^{2}C_{2} × ^{5}C_{2} × ^{n}C_{1}  
= 
 
=  10n 
Total number of Favorable/Favorable choices for Event B
⇒ m_{B}  =  m_{B1} + m_{B2} 

=  10 + 10n 
Probability of the total value of the five coins being greater than or equal to One rupee Fifty Paise.
⇒ Probability of occurrence of Event B
= 
 
⇒ P(B)  = 
 

= 

For Event A
⇒ Probability of occurrence of Event A
=  Probability of nonoccurrence of Event B  
⇒ P(A)  =  P(B_{c})  

=  1 − P(B)  
= 
 
= 
