Question

An experiment consists of first rolling a die and then tossing a coin. (a) List the sample space. {(3, H), (4, H)} {(1, T), (2, T), (3, T), (4, T), (5, T), (6, T)} {(1, H), (1, T), (2, H), (2, T), (3, H), (3, T), (4, H), (4, T), (5, H), (5, T), (6, H), (6, T)} {(3, H), (3, T), (4, H), (4, T)} {(1, H), (2, H), (3, H), (4, H), (5, H), (6, H)} (b) Let A be the event that either a three or a four is rolled first, followed by landing a head on the coin toss. Find P(A). (Enter your probability as a fraction.)

Answer 1

(a)

`S = \{(1,H),(2,H),(3,H),(4,H),(5,H),(6,H),(1,T),(2,T),(3,T),(4,T),(5,T),(6,T)\}`

(b)

`P(A) = (1)/(6)`

Explanation:

Given

The above sample space

A = Event of an occurrence of 3 or 4 then a head

(a): List out the sample space:

The rolls of the dice come first, followed by the toss of the coin.

So, the sample space are:

`S = \{(1,H),(2,H),(3,H),(4,H),(5,H),(6,H),(1,T),(2,T),(3,T),(4,T),(5,T),(6,T)\}`

The number of the sample space is:

`n(S) = 12`

(b): Find P(A)

First, we list out the outcomes of event A

`A = \{(3,H),(4,H)\}`

The number of outcome is:

`n(A) = 2`

The probability of A is then calculated as:

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

`P(A) = (2)/(12)`

Simplify fraction

`P(A) = (1)/(6)`