Question

Two faces of a six-sided die are painted red, two are painted blue, and two are painted yellow. The die is rolled three times, and the colors that appear face up on the first, second, and third rolls are recorded.
(a) Find the probability of the event that exactly one of the colors that appears face up is red.
(b) Find the probability of the event that at least one of the colors that appears face up is red.

Answer 1

Answer:

(a) 4/9

(b) 19/27

Step-by-step explanation:

The probability of each die turning up red or not turning red are:

$P(R) =\frac{2}{6}= \frac{1}{3}\\P(O) =\frac{4}{6}= \frac{2}{3}$

(a). Exactly one face up is red.

The odds of only the first die being red are:

$P(A=R) =\frac{1}{3} *\frac{2}{3}* \frac{2}{3} \\P(1=R) =\frac{4}{27}$

The same odds are valid for only the second or only the third die being red. Therefore, the probability that exactly one die turns up red is:

$P(R=1) = 3*\frac{4}{27}\\P(R=1) = \frac{4}{9}$

(b). At least one face up is red.

The probability that at least one die turns up red is the sum of the probabilities of exactly one (found in the previous item), two or all dice turning up red.

Following the same logic as in the previous item, the probability that exactly two dice turn up red is:

$P(R=2) = 3*(\frac{1}{3}*\frac{1}{3}*\frac{2}{3})\\P(R=2) = \frac{2}{9}$

The probability that all die turn up red is:

$P(R=3) = \frac{1}{3}*\frac{1}{3}*\frac{1}{3}\\P(R=3)=\frac{1}{27}$

Thus, the probability that at least one die turns up red is:

$P= P(R=1)+P(R=2)+P(R=3) = \frac{4}{9}+\frac{2}{9}+\frac{1}{27}\\P=\frac{19}{27}$