# Problem 1

A bag contains 12 two rupee coins, 7 one rupee coins and 4 half rupee coins. If 3 coins are selected at random, then the probability that a) the sum of the 3 coins is maximum b) the sum of the 3 is maximum c) each coins is of different value.

# 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 = 23C5
=
 23 × 22 × 21 × 20 × 19 5 × 4 × 3 × 2 × 1
= 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

The sum of the three coins is maximum when the three coins chosen are Two Rupee Coins

Event A ⇒ Event of the selecting three two rupee coins.

 [Two Rupee] [One Rupee] [Half Rupee] Total 12 7 4 23 3 0 0 3 12C3 7C0 4C0 23C3

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
⇒ mA = 12C3
=
 12 × 11 × 10 3 × 2 × 1
= 220

Probability of the sum of the three coins being maximum

⇒ Probability of occurrence of Event A

=
 Number of Favorable Choices for the Event Total Number of Possible Choices for the Experiment
⇒ P(A) =
 mA n
=
 220 33,649
=
 20 3,059

## • Odds

Number of Unfavorable Choices

 ⇒ mAc n − mA = Total Number of possible choices − Number of Favorable choices = = 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 = mA : mAc = 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 = mAc : mA = 33,429 : 220 = 3,039 : 20

## For Event B

The sum of the three coins is minimum when the three coins chosen are Half Rupee Coins

Event B ⇒ Event of the selecting three Half rupee coins.

 [Two Rupee] [One Rupee] [Half Rupee] Total 12 7 4 23 0 0 3 3 12C0 7C0 4C3 23C3

Number of Favorable Choices

 ⇒ mA = The number of ways in which the 3 half rupee coins can be selected from the total 4 half rupee coins = 4C3 = 4C(4 − 3)nCr = nC(n − r) = 4C1 = 4

Probability of the sum of the three coins being minimum

⇒ Probability of occurrence of Event B

=
 Number of Favorable Choices for the Event Total Number of Possible Choices for the Experiment
⇒ P(B) =
 mB n
=
 4 33,649

## For Event C

 [Two Rupee] [One Rupee] [Half Rupee] Total 12 7 4 23 1 1 1 3 12C1 7C1 4C1 23C3

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
⇒ mA = (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)
 Fundamental Counting Theorem (of Multiplication): Where an event can be sub divided into two or more independent sub-events, the total number of ways in which the total event can be accomplished is equal to the product of the number of ways in which the sub-events can be accomplished.
= 12C1 × 7C1 × 4C1
= 12 × 7 × 4
= 336

Probability of the three coins being of different values

⇒ Probability of occurrence of Event C

=
 Number of Favorable Choices for the Event Total Number of Possible Choices for the Experiment
⇒ P(C) =
 mC n
=
 336 33,649

# Problem 2

A box contains 2 fifty paise coins, 5 twenty five paise coins and a certain number n ( ≥ 2) of ten and five paise coins. Five coins are taken out of the box at random. The probability that the total value of these 5 coins is less than one rupee and fifty paise is!

# 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

 ⇒ n = Number of ways in which five coins can be drawn from the total (7 + n) coins = (7 + n)C5

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

• A1 : Picking 2 Fifty paise Coins + 1 Twenty Five paise Coin + 2 (Ten + Five) Paise Coins
• A2 : Picking 2 Fifty paise Coins + 0 Twenty Five paise Coin + 3 (Ten + Five) Paise Coins
• A3 : Picking 1 Fifty paise Coins + 4 Twenty Five paise Coin + 0 (Ten + Five) Paise Coins
• A4 : Picking 1 Fifty paise Coins + 3 Twenty Five paise Coin + 1 (Ten + Five) Paise Coins
• A5 : Picking 1 Fifty paise Coins + 2 Twenty Five paise Coin + 2 (Ten + Five) Paise Coins
• A6 : Picking 1 Fifty paise Coins + 1 Twenty Five paise Coin + 3 (Ten + Five) Paise Coins
• A7 : Picking 1 Fifty paise Coins + 0 Twenty Five paise Coin + 4 (Ten + Five) Paise Coins
• A8 : Picking 0 Fifty paise Coins + 5 Twenty Five paise Coin + 0 (Ten + Five) Paise Coins
• A9 : Picking 0 Fifty paise Coins + 4 Twenty Five paise Coin + 1 (Ten + Five) Paise Coins
• A10 : Picking 0 Fifty paise Coins + 3 Twenty Five paise Coin + 2 (Ten + Five) Paise Coins
• A11 : Picking 0 Fifty paise Coins + 2 Twenty Five paise Coin + 3 (Ten + Five) Paise Coins
• A12 : Picking 0 Fifty paise Coins + 1 Twenty Five paise Coin + 4 (Ten + Five) Paise Coins
• A13 : 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 2 5 n (7 + n) 2 1 2 5 2C2 5C1 nC2 (7 + n)C5 2 0 3 5 2C2 5C0 nC3 (7 + n)C5

Calculating the total number of favorable/favourable choices for the event is not possible for all these alternatives, specifically for these.

• A2 : Picking 2 Fifty paise Coins + 0 Twenty Five paise Coin + 3 (Ten + Five) Paise Coins
• A6 : Picking 1 Fifty paise Coins + 1 Twenty Five paise Coin + 3 (Ten + Five) Paise Coins
• A7 : Picking 1 Fifty paise Coins + 0 Twenty Five paise Coin + 4 (Ten + Five) Paise Coins
• A11 : Picking 0 Fifty paise Coins + 2 Twenty Five paise Coin + 3 (Ten + Five) Paise Coins
• A12 : Picking 0 Fifty paise Coins + 1 Twenty Five paise Coin + 4 (Ten + Five) Paise Coins
• A13 : 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

• B1 : Picking 2 Fifty paise Coins + 3 Twenty Five paise Coin + 0 (Ten + Five) Paise Coins
• B2 : 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 2 5 n (7 + n) 2 3 0 5 2C2 5C3 nC0 (7 + n)C5 2 2 1 5 2C2 5C2 nC1 (7 + n)C5

## » "B1"

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
mB1 = (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)
= 2C2 × 5C3
=
1 ×
 5 × 4 × 3 3 × 2 × 1
= 10

## » "B2"

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
mB2 = (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)
= 2C2 × 5C2 × nC1
=
1 ×
 5 × 4 2 × 1
× n
= 10n

Total number of Favorable/Favorable choices for Event B

 ⇒ mB mB1 + mB2 = = 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

=
 Number of Favorable Choices for the Event Total Number of Possible Choices for the Experiment
⇒ P(B) =
 mB n
=
 10 + 10n (7 + n)C5

## For Event A

Probability of the total value of the five coins being less than One rupee Fifty Paise.

⇒ Probability of occurrence of Event A

= Probability of non-occurrence of Event B
⇒ P(A) = P(Bc)
= 1 − P(B)
=
1 −
 10 + 10n (7 + n)C5
=
 ((7 + n)C5) − (10 + 10n) (7 + n)C5

# Practice Problem 1

A man’s pocket contains 5 fifty paise coins, 4 twenty five paise coins and 4 ten paise coins. A boy is asked to draw two coins at random. What is the probability of the boy drawing the maximum possible amount?

# Practice Problem 2

A man's pocket contains 10 one Rupee coins 6 half rupee coins and 4 twenty five paise coins. If a child is asked to select 3 coins at random, then the probability that he selects the (i) maximum amount, (ii) minimum amount is?

# Practice Problem 3

In a bag there are 5 half rupee coins, 4 twenty five paise coins and 4 ten paise coins. If two coins are drawn from a bag at random, then the probability for the amount drawn to be minimum is?