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Question

An experiment consists of tossing a coin and rolling a die.
(a) Find the sample space.
(b) Find the probability of getting heads and an even number.
(c) Find the probability of getting heads and a number greater than.4.
(d) Find the probability of getting tails and an odd number.


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Solution

Step- 1: Find the sample space:

Given an experiment consists of tossing a coin and rolling a die.

If an experiment consists of tossing a coin and rolling a die and recording the results in order, the sample space of an experiment is the set of all possible outcomes.

If we let H stands for heads and T for tails, then the sample space of the tossing a coin and rolling a dying experiment is,

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

Hence, the sample space is S={(H,1),(H,2),(H,3),(H,4),(H,5),(H,6),(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)}.

Step-2 : Find the probability of getting heads and an even number:

Given an experiment consists of tossing a coin and rolling a die.

Let S be the sample space of an experiment in which all outcomes are equally likely and let E be an event.

The probability of E, written P(E), is
P(E)=n(E)n(S)P(E)=number of elemnets inEnumber of elemnets inS

The sample space S of this experiment is S=(H,1),(H,2),(H,3),(H,4),(H,5),(H,6),(T,1),(T,2),(T,3),(T,4),(T,5),(T,6) contains twelve outcomes, and the event E of getting "heads and even number" contains three outcomes, {(H,2),(H,4),(H,6)}, so by the definition of probability,
P(E)=n(E)n(S)=312=14
Hence, the probability of getting heads and an even number is 14.

Step-3: Find the probability of getting heads and a number greater than 4:

Let S be the sample space of an experiment in which all outcomes are equally likely and let F be an event. The probability of F, written P(F), is
P(F)=n(F)n(S)P(F)=number of elemnets inFnumber of elemnets inS
The sample space of this experiment is S=(H,1),(H,2),(H,3),(H,4),(H,5),(H,6),(T,1),(T,2),(T,3),(T,4),(T,5),(T,6).
It contains twelve outcomes, and the event F of getting "heads and a number greater than 4" contains two outcome, {(H,5),(H,6)}, so by the definition of probability,
P(F)=n(F)n(S)=212=16
Hence, the probability of getting heads and a number greater than 4 is 16.

Step- 4: Find the probability of getting tails and an odd number:

Let S be the sample space of an experiment in which all outcomes are equally likely and let G be an event.

The probability of G, written P(G), is
P(G)=n(G)n(S)P(G)=number of elemnets inGnumber of elemnets inS

The sample space S of this experiment is S=(H,1),(H,2),(H,3),(H,4),(H,5),(H,6),(T,1),(T,2),(T,3),(T,4),(T,5),(T,6) contains twelve outcomes, and the event G of getting "tails and odd number" contains three outcome, {(T,1),(T,3),(T,5)}, so by the definition of probability,
P(G)=n(G)n(S)=312=14
Hence, the probability of getting tails and an odd number is 14.

Hence,

a) The sample space is S={(H,1),(H,2),(H,3),(H,4),(H,5),(H,6),(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)}

b) The probability of getting heads and an even number is 14.

c) The probability of getting heads and a number greater than 4 is 16.

d) The probability of getting tails and an odd number is 14.


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