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 |
|---|
| Available | 12 | 7 | 4 | 23 |
|---|
| To Choose | 3 | 0 | 0 | 3 |
|---|
| Choices | 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 |
|---|
| = | |
| = | 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) | = | |
|---|
| = | |
| = | |
• Odds
Number of Unfavorable Choices
| = | Total Number of possible choices − Number of Favorable choices |
| ⇒ mAc | = | n − mA |
|---|
| = | 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 |
|---|
| Available | 12 | 7 | 4 | 23 |
|---|
| To Choose | 0 | 0 | 3 | 3 |
|---|
| Choices | 12C0 | 7C0 | 4C3 | 23C3 |
|---|
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 |
| ⇒ mA | = | 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) | = | |
|---|
| = | |
For Event C | [Two Rupee] | [One Rupee] | [Half Rupee] | Total |
|---|
| Available | 12 | 7 | 4 | 23 |
|---|
| To Choose | 1 | 1 | 1 | 3 |
|---|
| Choices | 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) | = | |
|---|
| = | |
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
| = | Number of ways in which five coins can be drawn from the total (7 + n) coins |
| ⇒ n | = | (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 |
|---|
| Available | 2 | 5 | n | (7 + n) |
|---|
| To Choose | 2 | 1 | 2 | 5 | A1 |
|---|
| Choices | 2C2 | 5C1 | nC2 | (7 + n)C5 |
|---|
| To Choose | 2 | 0 | 3 | 5 | A2 |
|---|
| Choices | 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 |
|---|
| Available | 2 | 5 | n | (7 + n) |
|---|
| To Choose | 2 | 3 | 0 | 5 | B1 |
|---|
| Choices | 2C2 | 5C3 | nC0 | (7 + n)C5 |
|---|
| To Choose | 2 | 2 | 1 | 5 | B2 |
|---|
| Choices | 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 |
| = | |
| = | 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 |
| = | |
| = | 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) | = | |
|---|
| = | |
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) |
| = | |
| = | | ((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?
