# Problem Solving : Tossing/Throwing/Flipping coins

## Tossing multiple coins is the same as tossing a coin multiple times

In the experiment of tossing two or more coins, the act of each coin getting tossed is independent of the other coins getting tossed i.e. what appears on a coin is independent of what appears on the other coins.

The outcomes of the experiment of tossing two or more coins would be the same as the outcomes of the experiment of tossing a coin multiple times.

• Tossing the coin the first time would represent tossing the first coin.
• Tossing the coin a second time would represent, tossing the second coin.
• ...
• Tossing the coin for the nth time would represent, tossing the nth coin.

Therefore, in problem solving the experiment of tossing 'n' coins and the experiment of tossing a coing 'n' times should be viewed as the same.

## Total number of possible choices in tossing coins

The total number of possible choices in tossing a single coin is two, i.e. the elementary event of getting a head and the elementary event of getting a tail.

The experiment of towsing 'n' coins (or tossing a coin 'n' times), consists of 'n' independent trials of tossing a coin. The trials are independent since a coin showing up heads or tails is not dependent on what appears on the other coins.

To find the number of possible choices in tossing 'n' coins, we divide the event of tossing 'n' coins into 'n' independent events of tossing a coin each.

### Two Coins

 E : Tossing 2 Coins E1Tossing the 1st Coin E2Tossing the 2nd Coin

Total number of possible choices in tossing 2 coins

 ⇒ n(E) n(E1) × n(E2) = No. of possible events/choices in tossing the 1st coin × No. of possible events/choices in tossing the 2nd coin = = 2 × 2 = 22

### Three Coins

 E : Tossing 3 Coins E1Tossing the 1st Coin E2Tossing the 2nd Coin E3Tossing the 3rd Coin

Totall number of possible choices in tossing 3 coins

 ⇒ n(E) n(E1) × n(E2) × n(E3) = No. of possible events/choices in tossing the 1st coin × No. of possible events/choices in tossing the 2nd coin × No. of possible events/choices in tossing the 3rd coin = = 2 × 2 × 2 = 23

### n Coins

 E : Tossing n Coins E1Tossing the 1st Coin E2Tossing the 2nd Coin ... EnTossing the nth Coin

Totall number of possible choices in tossing n coins

 ⇒ n(E) n(E1) × n(E2) × ... × n(En) = No. of possible events/choices in tossing the 1st coin × No. of possible events/choices in tossing the 2nd coin × ... n times = = 2 × 2 × ... n times = 2n

Thus, the total number of possible choices in tossing

• 4 Coins = 16 (24)
• 5 Coins = 32 (25)
• 6 Coins = 64 (26)
• ...

## Throwing/Tossing/Rolling Two/Three coins - Favorable/Favourable choices for events

Listing out all the possible outcomes using symbols representing each possibility would be an easy way to find both the total number of possible choices as well as the number of favorable/favourable choices in relation to an experiment. If we are able to list the total possible choices, then the favourable choices would be none or more of those. This method can be adopted only where the total number of possible choices is small and there is no other simpler/convenient method available. Say for the experiment of tossing two, three or atmost four coins.

Where the number of coins involved is greater than three it would be convenient to follow the next method.

We use 'H' or 'T' to represent a head and tail respectively in relation to a coin.

The Word Form

• HT represents a head on the 1st and tail on the 2nd coins respectively.
• HTH represents a head on the 1st, a tail on the 2nd and a head on the 3rd coins respectively.

### Two Coins

Total number of possible choice = 4 {HH,HT,TH,TT}

Number of favorable choices for the event of getting

• both heads = 1 {HH}
• a head and a tail = 2 {HT, TH}
• atleast one head = 3 {HH, HT, TH}

### Three Coins

Total number of possible choice = 8 {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
Take the elements in the set representing the possible choices for two coins, prefix each element with 'H' to derive 4 choices and 'T' to derive another 4 choices.

Number of favorable choices for the event of getting

• two heads = 3 {HHT,HTH,THH}
• atleast a head and a tail = 6 {HHT, HTH, HTT, THH, THT, TTH}
• all tails = 1 {TTT}

### Four Coins

Total number of possible choice = 16
{THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT,
HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT}
Take the elements in the set representing the possible choices for three coins, prefix each element with 'H' to derive 8 choices and 'T' to derive another 8 choices.

Number of favorable choices for the event of getting

• two heads = 6 {THHT, THTH, TTHH, HHTT, HTHT, HTTH}
• atleast three heads = 6 {THHH, HHHH, HHHT, HTHH}
• all tails = 1 {HHHH}

## Tossing/Throwing/Rolling four or more coins - Favorable/Favourable choices for events

### Total number of possible choices

The total number of possible choices = (2)(Number of Coins)

This relation can be used for the experiment involving any number of coins.

### Number of Favorable choices

Instead of listing out all the possible events and locating the elements representing the favorable choices, we can use the following formulae to find the number of favorable choices

#### Formula Derivation with an example

Experiment : Tossing Four Coins

Let

A : The event of getting three heads

##### • For Event "A"
Word form representing the event - HHHT

In the word form

 L L(a) L(b) = Number of letters = 4 {H, H, H, T} = Number of H's [Letters of the first kind] = 3 = Number of T's [Letters of the second kind] = 1 = Number of Others [Letters which are all different] = 0 L = L(a) + L(b) + L(x)

Number of Favourable/Favorable Choices

= Number of arrangements possible with the word form representing the required event.
⇒ mA =
 L! L(a)! × L(b)!
i.e.
 Number of Coins (Number of Heads)! × (Number of Tails)!
=
 4! 3! × 1!
=
 4 × 3! 3! × 1
= 4

verification : {HHHT, HHTH, HTHH, THHH}

Since L = L(a) + L(b),

• L(a) = L − L(b)
(Or)
• L(b) = L − L(a)

Therefore,

 L! L(a)! × L(b)!
=
 L! L(a)! × (L − L(a))!
 n! r! × (n − r)!
= C(L, L(a)) C(n, r)
= C(Number of Coins, Number of Heads)
(Or) =
 L! (L − L(b)) × L(b)!
= C(L, L(b))
= C(Number of Coins, Number of Tails)

Taking the lesser of heads or tails would minimise the calculations.

### Examples

#### Tossing Five Coins

Experiment : Tossing Five Coins
1. A : The event of getting two heads
##### • For Event "A"
Number of Favourable/Favorable Choices
= C(Number of Coins, Number of Heads)
= C(5, 2)
=
 5 × 4 2 × 1
= 10
2. B : The event of getting atleast four tails
##### • For Event "B"
Event 'B' can be accomplished in two alternative ways
• BA1 : getting 4 tails
• BA2 : getting 5 tails
###### • For Event "BA1"
Number of Favourable/Favorable Choices
= C(Number of Coins, Number of Heads)
= C(5, 1)
=
 5 1
= 5

verification : {HTTTT, THTTT, TTHTT, TTTHT, TTTTH}

##### • For Event "BA2"
Number of Favourable/Favorable Choices
 = C(Number of Coins, Number of Heads) = C(5, 0) = 1

verification : {TTTT}

Number of Favourable/Favorable Choices

 ⇒ mB = Sum of Favourable/Favorable Choices for the alternatives = mB(A1) + mB(A2) = 5 + 1 = 6

### Tossing eight Coins

Experiment : Tossing Eight Coins
1. "F" : The event of getting five heads
##### • For Event "F"
Number of Favourable/Favorable Choices
= C(Number of Coins, Number of Tails)
= C(8, 3)
=
 8 × 7 × 6 3 × 2 × 1
= 56
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## Getting atleast/atmost the specified number of heads/tails in tossing multiple coins

If the event whose probability or odds is required to be found out may be an event with alternative possibilities, the total number of favorable choices for the event would be the sum of the favourable choices for its alternatives.

Such events can be identiifed by the presence of the phrases atleast or atmost in the event definitions. Using mathematical operators they can be interpreted as

### atleast (≥)

Atleast 'x' includes anything equal to or greater than x.
1. Getting atleast 4 heads on tossing 6 coins.

Experiment : Tossing 6 Coins

A : The event of getting atleast 4 heads

#### • For Event "A"

Event 'A' can be accomplished in three alternative ways.

• A1 : The event of getting 4 heads
• A1 : The event of getting 5 heads
• A3 : The event of getting 6 heads

Number of Favourable/Favorable Choices

 ⇒ mA = Sum of Number of Favourable/Favorable Choices for the alternatives = mA(1) + mA(2) + mA(3)

### atmost (≤)

Atmost 'x' includes anything equal to or less than x.
1. Getting atmost 3 tails on tossing 6 coins.

Experiment : Tossing 6 Coins

G : The event of getting atmost 3 tails

#### • For Event "G"

Atmost 3 tails ≡ 3 or less tails

Event 'G' can be accomplished in four alternative ways

• G1 : The event of getting 3 tails
• G2 : The event of getting 2 tails
• G3 : The event of getting 1 tail
• G4 : The event of getting 0 tails

Number of Favourable/Favorable Choices

 ⇒ mG = Sum of Number of Favourable/Favorable Choices for the alternatives = mG(1) + mG(2) + mG(3) + mG(4)

### Using Complimentary Events

In cases where we find the terms atleast and atmost and where the number of alternatives for the events are many, we may be able to minimise the calculations by using event complements.
• atleast (≥) and less than (<) are complements of each other.
Total number of possible choices  ⇒ n = Favorable/favourable choices for atleast + Favorable/favourable choices for less than = m(≥) + m(<)

Therefore, m(≥) = n − m(<)

• atmost (≤) and more than (>) are complements of each other.
Total number of possible choices  ⇒ n = Favorable/favourable choices for atleast + Favorable/favourable choices for less than = m(≤) + m(>)

Therefore, m(≤) = n − m(>)

#### atleast "x"

In cases involving atleast, where the value of 'x' is nearer to zero i.e. farther to number of trials in the experiment, it would be easier to find the number of favorable choices using complementary of the required event.
1. Getting atleast 2 heads on tossing 6 coins.

Experiment : Tossing 6 Coins

D : The event of getting atleast 2 heads

#### • For Event "D"

Event 'D' can be accomplished in five alternative ways

• D1 : The event of getting 2 heads
• D2 : The event of getting 3 heads
• D3 : The event of getting 4 heads
• D4 : The event of getting 5 heads
• D5 : The event of getting 6 heads

Number of Favourable/Favorable Choices

 ⇒ mD = Sum of Number of Favourable/Favorable Choices for the alternatives = mD(1) + mD(2) + mD(3) + mD(4) + mD(5)

#### Using Complements [≥ and <]

• D : The event of getting atleast 2 heads
• Dc : The event of getting less than 2 heads
##### • For Event "Dc"
Event 'Dc' can be accomplished in two alternative ways
• Dc1 : The event of getting 0 heads
• Dc2 : The event of getting 1 head

Number of Favourable/Favorable Choices

 ⇒ mDc = Sum of Number of Favourable/Favorable Choices for the alternatives = mDc(1) + mDc(2)

##### • For Event "D"
Number of Favourable/Favorable Choices
 ⇒ mD = Total Number of possible Choices - Number of Favourable/Favorable Choices for the event complement = n - mDc
2. Getting atleast 1 tail on tossing 5 coins.

Experiment : Tossing 5 Coins

M : The event of getting atleast 1 tail

#### • For Event "M"

atleast 1 tail ≡ 1 or more tails

Event 'M' can be accomplished in five alternative ways

• M1 : The event of getting 1 tail
• M2 : The event of getting 2 tails
• M3 : The event of getting 3 tails
• M4 : The event of getting 4 tails
• M5 : The event of getting 5 tails

Number of Favourable/Favorable Choices

 ⇒ mD = Sum of Number of Favourable/Favorable Choices for the alternatives = mM(1) + mM(2) + mM(3) + mM(4) + mM(5)

#### Using Complements [≥ and <]

• M : The event of getting atleast 1 tail
• Mc : The event of getting less than 1 tails i.e. zero tails i.e. all heads
##### • For Event "Mc"
Number of Favourable/Favorable Choices
 ⇒ mMc = 1 {HHHHH} = 1
##### • For Event "M"
Number of Favourable/Favorable Choices
 ⇒ mM = Total Number of possible Choices - Number of Favourable/Favorable Choices for the event complement = n - mMc

#### atmost "x"

In cases involving atmost, where the value of 'x' is nearer to "n" (number of trials in the the experiment) i.e. farther from zero, it would be easier to find the number of favorable choices using complimentary of the required event.
1. Getting atmost 5 tails on tossing 7 coins.

Experiment : Tossing 7 Coins

R : The event of getting atmost 5 tails

#### • For Event "R"

atmost 5 tails ≡ 5 or less tails
Event 'R' can be accomplished in six alternative ways
• R1 : The event of getting 5 tails
• R2 : The event of getting 4 tails
• R3 : The event of getting 3 tails
• R4 : The event of getting 2 tails
• R5 : The event of getting 1 tail
• R6 : The event of getting 0 tails

Number of Favourable/Favorable Choices

 ⇒ mD = Sum of Number of Favourable/Favorable Choices for the alternatives = mR(1) + mR(2) + mR(3) + mR(4) + mR(5) + mR(6)

#### Using Compliments [≤ and >]

• R : The event of getting atmost 5 tails
• Rc : The event of getting more than 5 tails
##### • For Event "Rc"
Event 'Rc' can be accomplished in two alternative ways
• Rc1 : The event of getting 6 tails
• Rc2 : The event of getting 7 tails

Number of Favourable/Favorable Choices

 ⇒ mRc = Sum of Number of Favourable/Favorable Choices for the alternatives = mRc(1) + mRc(2)

##### • For Event "Rc"
Number of Favourable/Favorable Choices
 ⇒ mR = Total Number of possible Choices - Number of Favourable/Favorable Choices for the event compliment = n - mRc
2. Getting atmost 7 heads on tossing 8 coins.

Experiment : Tossing 7 Coins

V : The event of getting atmost 7 heads

#### • For Event "V"

Event 'V' can be accomplished in eight alternative ways
• V1 : The event of getting 7 heads
• V2 : The event of getting 6 heads
• V3 : The event of getting 5 heads
• V4 : The event of getting 4 heads
• V5 : The event of getting 3 tail
• V6 : The event of getting 2 heads
• V7 : The event of getting 7 heads
• V8 : The event of getting 8 heads

Number of Favourable/Favorable Choices

 ⇒ mD = Sum of Number of Favourable/Favorable Choices for the alternatives = mV(1) + mV(2) + mV(3) + mV(4) + mV(5) + mV(6) + mV(7) + mV(8)

#### Using Compliments [≤ and >]

• V : The event of getting atmost 7 heads
• Vc : The event of getting more than 7 heads i.e. 8 heads i.e. zero tails
##### • For Event "Vc"
Number of Favourable/Favorable Choices
 ⇒ mVc = 1 [HHHHHHHH] = 1
##### • For Event "Vc"
Number of Favourable/Favorable Choices
 ⇒ mV = Total Number of possible Choices - Number of Favourable/Favorable Choices for the event compliment = n - mVc
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