Tossing/Throwing/Flipping Two/Three Coins

Problem 1

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

Experiment :

Tossing two unbiased coins

Total Number of Possible Choices

= 2(Number of coins)

= 22

= 4

{(H, H), (H, T), (T, H), (T, T)}

⇒ n = 4

Let

A : the event of getting both tails

B : the event of getting a head and a tail

C : the event of getting at least one head

D : the event of not getting at least one tail

For Event A

Number of Favorable Choices

= 1

{(T, T)}

⇒ mA = 1

Probability of getting both tails

⇒ 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
=
1
4

Odds

Number of Unfavorable Choices

= Total Number of possible choices − Number of Favorable choices

⇒ mAc = n − mA
= 4 − 1
= 3

in favor

Odds in Favor of getting both tails

⇒ Odds in Favor of Event A

= Number of Favorable Choices : Number of Unfavorable Choices

= mA : mAc

= 1 : 3

against

Odds against getting both tails

⇒ Odds against Event A

= Number of Unfavorable Choices : Number of Favorable Choices

= mAc : mA

= 3 × 1

For Event B

Number of Favorable Choices

= 2

{(H, T), (T, H)}

⇒ mB = 2

Probability of getting a head and a tail

⇒ 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
=
2
4
=
1
2

For Event C

Number of Favorable Choices

= 3

{(H, H), (H, T), (T, H)}

⇒ mC = 3

Probability of getting at least one head

⇒ 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
=
3
4

Odds

Number of Unfavorable Choices

= Total Number of possible choices − Number of Favorable choices

⇒ mCc = n − mC
= 4 − 3
= 1

in favor

Odds in Favor of getting at least one head

⇒ Odds in Favor of Event C

= Number of Favorable Choices : Number of Unfavorable Choices

= mC : mCc

= 3 : 1

against

Odds against getting at least one head

⇒ Odds against Event C

= Number of Unfavorable Choices : Number of Favorable Choices

= mCc : mC

= 1 : 3

For Event D

Number of Favorable Choices

= 1

{(H, H)}

⇒ mD = 1

Probability of not getting at least one tail

⇒ Probability of occurrence of Event D

=
Number of Favorable Choices for the Event
Total Number of Possible Choices for the Experiment
⇒ P(D) =
mD
n
=
1
4

Problem 2

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

Experiment :

Tossing three coins

Total Number of Possible Choices

= 2(Number of coins)

= 23

= 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)}

⇒ n = 8

Let

P : the event of getting all heads

Q : the event of getting one head

R : the event of getting two heads

S : the event of getting all tails

T : the event of getting at least one head

U : the event of getting at least two heads

For Event P

Number of Favorable Choices

= 1

{(H, H, H)}

⇒ mP = 1

Probability of getting all heads

⇒ Probability of occurrence of Event P

=
Number of Favorable Choices for the Event
Total Number of Possible Choices for the Experiment
⇒ P(P) =
mP
n
=
1
8

Odds

Number of Unfavorable Choices

= Total Number of possible choices − Number of Favorable choices

⇒ mPc = n − mP
= 8 − 1
= 7

in favor

Odds in Favor of getting all heads

⇒ Odds in Favor of Event P

= Number of Favorable Choices : Number of Unfavorable Choices

= mP : mPc

= 1 : 7

against

Odds against getting all heads

⇒ Odds against Event P

= Number of Unfavorable Choices : Number of Favorable Choices

= mPc : mP

= 7 × 1

For Event Q

Number of Favorable Choices

= 3

{(H, T, T), (T, H, T), (T, T, H)}

⇒ mQ = 3

Probability of getting one head

⇒ Probability of occurrence of Event Q

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

For Event R

Number of Favorable Choices

= 3

{(H, H, T), (H, T, H), (T, H, H)}

⇒ mR = 3

Probability of getting two heads

⇒ Probability of occurrence of Event R

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

Odds

Number of Unfavorable Choices

= Total Number of possible choices − Number of Favorable choices

⇒ mRc = n − mR
= 8 − 3
= 5

in favor

Odds in Favor of getting two heads

⇒ Odds in Favor of Event R

= Number of Favorable Choices : Number of Unfavorable Choices

= mR : mRc

= 3 : 4

against

Odds against getting two heads

⇒ Odds against Event R

= Number of Unfavorable Choices : Number of Favorable Choices

= mRc : mR

= 4 × 3

For Event S

Number of Favorable Choices

= 1

{(T, T, T)}

⇒ mS = 1

Probability of getting all tails

⇒ Probability of occurrence of Event S

=
Number of Favorable Choices for the Event
Total Number of Possible Choices for the Experiment
⇒ P(S) =
mS
n
=
1
8

For Event T

Number of Favorable Choices

= 7

{(H, H, H), (H, H, T), (H, T, H), (H, T, T), (T, H, H), (T, H, T), (T, T, H)}

⇒ mT = 7

Probability of getting all tails

⇒ Probability of occurrence of Event T

=
Number of Favorable Choices for the Event
Total Number of Possible Choices for the Experiment
⇒ P(T) =
mT
n
=
7
8

Alternative

Number of Favorable Choices

= Total number of possible choices in the experiment − Number of Favorable Choices for event S

Since S and T are complimentary events

⇒ mT = n − mS
= 8 − 1
= 7

For Event T [Alternative]

Probability of getting at least 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
=
8 − 1
8
=
7
8

For Event U

Number of Favorable Choices

= 4

{(H, H, T), (H, T, H), (T, H, H), (H, H, H)}

⇒ mU = 4

Probability of getting at least two heads

⇒ Probability of occurrence of Event U

=
Number of Favorable Choices for the Event
Total Number of Possible Choices for the Experiment
⇒ P(U) =
mU
n
=
4
8
=
1
2

Practice Problems

  1. If two unbiased coins are tossed then the probability of getting two tails is
  2. The probability of getting heads in both trials, when a balanced coin is tossed twice will be ?

    (Or)

    In tossing a coin, the probability of getting head in two successive trials is?

    (Or)

    The probability of two half rupee coins falling heads up when tossed simultaneously is

  3. If a coin is tossed, two times, what is the probability of getting a tail at least once?
  4. If an unbiased coin is tossed twice then the probability that a head and a tail are obtained is
  5. When two coins are tossed, find the probability for a) getting one head b) not getting at least one head
  6. 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.
  7. 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

  8. 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

  9. 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 at least 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 : The Edifier