Mathematics: 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.)

btw B is not the anwserStep-by-step explanation:...

An experiment consists of first rolling a die and

TKwan

13.03.2021

(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)

$P(A) = \frac{1}{6}$

Step-by-step 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:

$P(A) = \frac{n(A)}{n(S)}$

$P(A) = \frac{2}{12}$

Simplify fraction

$P(A) = \frac{1}{6}$

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