Question

A fair coin is tossed three times and the events A, B, and C are defined as follows: A: \{ At least one head is observed \} B: \{ At least two heads are observed \} C: \{ The number of heads observed is odd \} Find the following probabilities by summing the probabilities of the appropriate sample points (note that 0 is an even number):

(a) P(A) =
(b) P(A \textrm{ or } B) =
(c) P((\textrm{not }A) \textrm{ or } B \textrm{ or } (\textrm{not }C)) =

Answer 1

a)
`P(A)=0.875`

b)
`\text{P(A or B)}=0.875`

c)
`\text{P((not A)  or B  or (not C))}=0.625`

Explanation:

Given : A fair coin is tossed three times and the events A, B, and C are defined as follows: A: At least one head is observed, B: At least two heads are observed, C: The number of heads observed is odd.

To find : The following probabilities by summing the probabilities of the appropriate sample points ?

Solution :

The sample space is

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

n(S)=8

A: At least one head is observed

i.e. A={HHH,HHT,HTT,HTH,TTH,TTH,THH,THT}

n(A)=7

B: At least two heads are observed

i.e. B={HHH,HTT,TTH,THT}

n(B)=4

C: The number of heads observed is odd.

i.e. C={HHH,HTT,THT,TTH}

n(c)=4

a) Probability of A, P(A)

`P(A)=(n(A))/(n(S))`

`P(A)=(7)/(8)`

`P(A)=0.875`

b) P(A or B)

Using formula,

`\text{P(A or B)}=P(A)+P(B)-\text{P(A and B)}`

`\text{P(A or B)}=(n(A))/(n(S))+(n(B))/(n(S))-\frac{\text{n(A and B)}}{n(S)}`

`\text{P(A or B)}=(7)/(8)+(4)/(8)-(4)/(8)`

`\text{P(A or B)}=(7)/(8)`

`\text{P(A or B)}=0.875`

(c) P((not A) or B or (not C))

A={HHH,HHT,HTT,HTH,TTH,TTH,THH,THT}

not A = {TTT} = 1

B={HHH,HTT,TTH,THT}

C={HHH,HTT,THT,TTH}

not C = {HHT,HTH,THH,TTT} = 4

So, not A or B or not C = {HHH,HHT,HTH,THH,TTT}=5

`\text{P((not A)  or B  or (not C))}=(5)/(8)`

`\text{P((not A)  or B  or (not C))}=0.625`