# Tossing/Throwing/Flipping Two/Three Coins

No. Problems & Solutions [Click Solution link to bring up the solutions page]
01.
What is the probability of getting a) both tails b) a head and a tail c) at least one head; d) not getting at least one tail, when two unbiased coins are tossed?

 Solution Hide/Show
In the experiment of tossing two unbiased coins (Or) tossing a coin twice,

Total No. of Possible Choices = 2 × 2

 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.

⇒ n = 4 {(H, H), (H, T), (T, H), (T, T)}
Let

• "A" be the events of getting both tails
• "B" be the event of getting a head and a tail
• "C" be the event of getting atleast one head
• "D" be the event of not getting atleast one tail

# For Event "A"

 ⇒ mA Number of Favourable/Favorable Choices = 1 {(T, T)} = 1

Probability of getting both tails
⇒ Probability of occurrence of Event "A" =
 Number of Favourable/Favorable Choices for the Event Total Number of Possible Choices for the Experiment
⇒ P(A) =
 mA n
=
 1 4

# • Odds

 ⇒ mAc n − mA No. of UnFavorable Choices = Total No. of possible choices − No. of Favorable choices = = 4 − 1 = 3

## » in favor/favour

Odds in Favour of getting both tails
 ⇒ Odds in Favor/Favour of Event "A" = No. of Favourable Choices : No. of UnFavorable Choices = mA : mAc = 1 : 3

## » against

Odds against getting both tails
 ⇒ Odds against Event "A" = No. of UnFavourable Choices : No. of Favorable Choices = mAc : mA = 3 : 1

# For Event "B"

 ⇒ mB Number of Favourable/Favorable Choices = 2 {(H, T), (T, H)} = 2

Probability of getting both tails
⇒ Probability of occurrence of Event "B" =
 Number of Favourable/Favorable Choices for the Event Total Number of Possible Choices for the Experiment
⇒ P(B) =
 mB n
=
 2 4
=
 1 2

# For Event "C"

 ⇒ mC Number of Favourable/Favorable Choices = 3 {(H, H), (H, T), (T, H)} = 3

Probability of getting atleast one head
⇒ Probability of occurrence of Event "C" =
 Number of Favourable/Favorable Choices for the Event Total Number of Possible Choices for the Experiment
⇒ P(C) =
 mC n
=
 3 4

# • Odds

Probability of non-occurrence of event C = 1 − Probability of Occurrence of Event C
⇒ P(Cc) = 1 − P(C)
=
1 −
 3 4
=
 1 4

## » in favor/favour

Odds in Favour of getting atleast one head
⇒ Odds in Favor/Favour of Event "C" = Probability of occurrence of the Event
: Probability of non-occurrence of the Event
= P(C) : P(Cc)
=
 3 4
:
 1 4
= 3 : 1

## » against

Odds against getting atleast one head
⇒ Odds against Event "C" = Probability of non-occurrence of the Event
: Probability of occurrence of the Event
= P(Cc) : P(C)
=
 1 4
:
 3 4
= 1 : 3

# For Event "D"

 ⇒ mD Number of Favourable/Favorable Choices = 1 {(H, H)} = 1

Probability of not getting atleast one tail
⇒ Probability of occurrence of Event "D" =
 Number of Favourable/Favorable Choices for the Event Total Number of Possible Choices for the Experiment
⇒ P(D) =
 mD n
=
 1 4

No. Problems & Solutions [Click Solution link to bring up the solutions page]
02.
Three unbiased coins are tossed, what is the probability of obtaining (a) all heads (b) one head (c) two heads (d) all tails (e) at least one head and (f) at least two heads

 Solution
In the experiment of tossing three unbiased coins (Or) tossing a coin thrice,

Total No. of Possible Choices = 2 × 2 × 2

 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.

⇒ n = 8 {(H, H, H), (H, H, T), (H, T, H), (H, T, T), (T, H, H), (T, H, T), (T, T, H), (T, T, T)}
Let

• "P" be the events of getting alll heads
• "Q" be the event of getting one head
• "R" be the event of getting two heads
• "S" be the event of getting all tails
• "T" be the event of getting atleast one head
• "U" be the event of getting atleast two heads

# For Event "P"

 ⇒ mP Number of Favourable/Favorable Choices = 1 {(H, H, H)} = 1

⇒ Probability of occurrence of Event "P" =
 Number of Favourable/Favorable Choices for the Event Total Number of Possible Choices for the Experiment
⇒ P(P) =
 mP n
=
 1 8

# • Odds

 ⇒ mPc n − mP No. of UnFavorable Choices = Total No. of possible choices − No. of Favorable choices = = 8 − 1 = 7

## » in favor/favour

Odds in Favour of getting all heads
 ⇒ Odds in Favor/Favour of Event "P" = No. of Favourable Choices : No. of UnFavorable Choices = mP : mPc = 1 : 7

## » against

 ⇒ Odds against Event "P" = No. of UnFavourable Choices : No. of Favorable Choices = mPc : mP = 7 : 1

# For Event "Q"

 ⇒ mQ Number of Favourable/Favorable Choices = 3 {(H, T, T), (T, H, T), (T, T, H)} = 3

⇒ Probability of occurrence of Event "Q" =
 Number of Favourable/Favorable Choices for the Event Total Number of Possible Choices for the Experiment
⇒ P(Q) =
 mQ n
=
 3 8

# For Event "R"

 ⇒ mR Number of Favourable/Favorable Choices = 3 {(H, H, T), (H, T, H), (T, H, H)} = 3

⇒ Probability of occurrence of Event "R" =
 Number of Favourable/Favorable Choices for the Event Total Number of Possible Choices for the Experiment
⇒ P(R) =
 mR n
=
 3 8

# • Odds

Probability of non-occurrence of event R = 1 − Probability of Occurrence of Event R
⇒ P(Rc) = 1 − P(R)
=
1 −
 3 8
=
 5 8

## » in favor/favour

Odds in Favour of getting two heads
⇒ Odds in Favor/Favour of Event "R" = Probability of occurrence of the Event
: Probability of non-occurrence of the Event
= P(R) : P(Rc)
=
 3 8
:
 5 8
= 3 : 5

## » against

⇒ Odds against Event "R" = Probability of non-occurrence of the Event
: Probability of occurrence of the Event
= P(Rc) : P(R)
=
 5 8
:
 3 8
= 5 : 3

# For Event "S"

 ⇒ mS Number of Favourable/Favorable Choices = 1 {(T, T, T)} = 1

Probability of getting all tails
⇒ Probability of occurrence of Event "S" =
 Number of Favourable/Favorable Choices for the Event Total Number of Possible Choices for the Experiment
⇒ P(S) =
 mS n
=
 1 8

# For Event "T"

 ⇒ mT Number of Favourable/Favorable Choices = 7 {(H, H, H), (H, H, T), (H, T, H), (H, T, T), (T, H, H), (T, H, T), (T, T, H)} = 7

## • Alternative

 ⇒ mT Number of Favourable/Favorable Choices = Total number of possible choices in the experiment − Number of Favourable/Favorable Choices for event "S" [Since "S" and "T" are complimentary events] = n − mS = 8 − 1 = 7

Probability of getting atleast one head
⇒ Probability of occurrence of Event "T" =
 Number of Favourable/Favorable Choices for the Event Total Number of Possible Choices for the Experiment
⇒ P(T) =
 mT n
=
 7 8

# For Event "T" [Alternative]

Probability of getting atleast one head = Probability of not getting all tails
⇒ Probability of occurrence of Event "T" = 1 − Probability of getting all tails
⇒ P(T) = 1 − P(S)
=
1 −
 1 8
=
 7 8

# For Event "U"

 ⇒ mU Number of Favourable/Favorable Choices = 4 {(H, H, T), (H, T, H), (T, H, H), (H, H, H)} = 4

Probability of getting atleast two heads
⇒ Probability of occurrence of Event "U" =
 Number of Favourable/Favorable Choices for the Event Total Number of Possible Choices for the Experiment
⇒ P(U) =
 mU n
=
 4 8
=
 1 2

No. Problems for Practice
01.
If two unbiased coins are tossed then the probability of getting two tails is
02.
The probability of getting heads in both trails, when a balanced coin is tossed twice will be ?
(Or) In tossing a coin, the probability of getting head in two successive trails is
(Or) The probability of two half rupee coins falling heads up when tossed simultaneously is
03.
If a coin is tossed, two times. What is the probability of getting a tail atleast once?
04.
If an unbiased coin is tossed twice then the probability that a head and a tail are obtained is
05.
When two coins are tossed, find the probability for a) getting one head b) not getting at least one head
06.
If a coin is tossed, what is the chance of a ‘Tail’? if three coins are tossed, find the chance that they are all Tails.
07.
A coin tossed 3 times. The probability of getting head once and tail two times is.
(Or) If three coins are tossed, the probability of the event showing exactly one head on them is
08.
When an unbiased coin is three times, the probability of falling all heads is
(Or) The probability of three half – rupee coins falling all heads up when tossed simultaneously is
09.
The probability of getting at least two heads when tossing a coin three times is ...
10.
When an unbiased coin is tossed three times, the probability of getting head atleast once is
11.
When a coin is tossed three times, the probability of getting exactly one tail or two tails is
12.
When 3 different coins are tossed, write the failures for the event of getting at least two heads.
 Author Credit : The Edifier