Question

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)?

Answer 1

Answer:

a) S = {1, 2, 3}

b) P(odd number) = $\frac{2}{3}$

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) = $\frac{number of odd digits}{number of faces}=\frac{4}{6}=\frac{2}{3}$

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 $x$. Then that of 3 is $2x$(because it is twice the others). The sum of probabilities must be 1.

$x+x+x+x+x+2x=1$

$7x=1$

$x=\frac{1}{7}$

P(odd number) = $3\times$Prob(1) + Prob(3)

= $3\times\frac{1}{7}+\frac{2}{7}$ = $\frac{5}{7}$