Question

Three coins are tossed. Find the chance of tossing: i) one heads and two tails ii) exactly two heads iii) at least two tails iv) no heads

Answer 1

“If three coins are tossed, what is the probability of getting two heads and one tail?”

You can calculate it, but for such a small number of possible combinations of independent events (8), let’s look at them all.

H = heads

T = tails

Possible events with equal probability (order matters):

HHH

HHT

HTH

HTT

THH

THT

TTH

TTT

Number with 2 heads: 3

Total number: 8

From the definition of probability, the number you are looking for is 3/8 = 0.375 = 37.5%

Explanation:

Answer 2

i)

The probability that getting one heads and two tails = 0.375 = 37.5%

ii)

The probability that getting Exactly two heads = 0.375 = 37.5%

iii)

The probability that getting at -least two tails = 0.5 =50%

iv)

The probability that getting no heads = 0.125 = 12.5%

Explanation:

Explanation:-

Given Three coins are tossed

The number of exhaustive cases

= { HHH,TTT,HTT,THT,TTH,HTH,THH,HHT} =8

n(S) = 8

i)

Let 'E₁' be the event of getting one heads and two tails

n ( E₁) = {HTT ,THT , TTH} = 3

The probability that getting one heads and two tails

`P(E_(1) ) = (n(E_(1) ))/(n(S)) = (3)/(8) = 0.375`

The chance of tossing one heads and two tails = 37.5%

ii)

Let 'E₂' be the event of getting Exactly two heads

n(E₂) = {HTH,THH,HHT} =3

The probability that getting Exactly two heads

`P(E_(2) ) = (n(E_(2) ))/(n(S)) = (3)/(8) = 0.375`

The chance of tossing Exactly two heads = 37.5%

iii)

Let 'E₃' be the event of getting at -least two tails

n(E₃) = {TTT,HTT,THT,TTH} = 4

The probability that getting at -least two tails

`P(E_(3) ) = (n(E_(3) ))/(n(S)) = (4)/(8) = 0.5`

The chance of tossing at -least two tails = 50%

iv)

Let 'E₄' be the event of getting no heads

n(E₄) = { TTTT} =1

The probability that getting no heads

`P(E_(4) ) = (n(E_(4) ))/(n(S)) = (1)/(8) = 0.125`

The chance of tossing no heads = 12.5%