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A die with six faces has the number 1 painted on three of its faces, the number 2 painted on 2 of its faces, and number 3 on one face. Assume each face is equally likely to come up.
a) Find a sample space.
b) Find P(odd number).
c) If die were loaded so that the face with 3 on it were twice as likely to come up as the other 5 faces, would this change the sample space?
d) If die were loaded so that the face with 3 on it were twice as likely to come up as the other 5 faces, would this change the value of P(odd number)?

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adelekeademolu

Answer:

a) S = {1, 2, 3}

b) P(odd number) = [tex]\frac{2}{3}[/tex]

c) No

d) Yes

Step-by-step explanation:

a) The sample space is the set of all possible outcomes. By definition, the elements of a set should not be repeated. Hence, the sample space S = {1, 2, 3}

However, the sample is not equiprobable because each element has different probabilities.

b) P(odd number) = [tex]\frac{number of odd digits}{number of faces}=\frac{4}{6}=\frac{2}{3}[/tex]

Note that the odd numbers are 1 (on three faces) and 3 (on one face).

c) The fact the die has been biased does not change the possible outcomes. It only changes the probability of getting any given number.

d) Because the 3-face has been loaded, this probability changes. In fact, it is calculated thus:

Let's assume the probability for 1 or 2 is [tex]x[/tex]. Then that of 3 is [tex]2x[/tex](because it is twice the others). The sum of probabilities must be 1.

[tex]x+x+x+x+x+2x=1[/tex]

[tex]7x=1[/tex]

[tex]x=\frac{1}{7}[/tex]

P(odd number) = [tex]3\times[/tex]Prob(1) + Prob(3)

= [tex]3\times\frac{1}{7}+\frac{2}{7}[/tex] = [tex]\frac{5}{7}[/tex]

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