Question

Suppose that a coin is tossed three times and the side showing face up on each toss is noted. Suppose also that on each toss heads and tails are equally likely. Let HHT indicate the outcome heads on the first two tosses and tails on the third, THT the outcome tails on the first and third tosses and heads on the second, and so forth.

A. List the eight elements in the sample space whose outcomes are all the possible head–tail sequences obtained in the three tosses.
B. Write each of the following events as a set and find its probability:
i) The event that exactly one toss results in a head.
ii) The event that at least two tosses result in a head.
iii) The event that no head is obtained.

Answer 1

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

B

• P(
  `E_{h1` ) =
  `(3)/(8)`
• P(
  `E_{h2` ) =
  `(1)/(2)`
• P(
  `E_(no \ head)` ) =
  `(1)/(8)`

Explanation:

From the given information:

Suppose that a coin is tossed three times and on each toss heads and tails are equally likely.

Hint:

For the first two tosses ; Let HHT indicate the outcome heads and tails on the third toss

For the first and the third tosses: Let THT indicate the outcome tails and heads on the second toss. (and so on and so forth)

The objectives are to :

A. List the eight elements in the sample space whose outcomes are all the possible head–tail sequences obtained in the three tosses.

So; if a coin is tossed thrice, and sample S is the list of the 8 possible outcomes, Then :

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

B. Write each of the following events as a set and find its probability:

i) The event that exactly one toss results in a head.

For a toss to result into one head:

Let consider the above sample S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

n(S) = 8

So; Let represent
`E_{h1` to be the event for one head to appear, we have ;

`E_{h1` = { HTT, THT, TTH}

n(
`E_{h1`) = 3

Probability P =
`(number \ of favourable \ outcome)/(Total \ number \ of \ possible \ outcomes)`

P(
`E_{h1` ) =
`(3)/(8)`

ii) The event that at least two tosses result in a head.

Let represent
`E_{h2` to be the event that at least two tosses result in a head., we have ;

`E_{h2` = { HHH, HHT, HTH, THH }

n(
`E_{h2` ) = 4

P(
`E_{h2` ) =
`(4)/(8)`

P(
`E_{h2` ) =
`(1)/(2)`

iii) The event that no head is obtained.

Let
`E_(no \ head)` be the event that no head is obtained.

Then
`E_(no \ head)` = {TTT}

n(
`E_(no \ head)` ) = 1

P(
`E_(no \ head)` ) =
`(1)/(8)`