Question

Suppose that you have eight cards. Five are green and three are yellow. The five green cards are numbered $1,2,3,4\text{and}5$. The three yellow cards are numbered $1,2,\text{and}3$. The cards are well shuffled. You randomly draw one card.

• G = card drawn is green

• E = card drawn is even-numbered

a. List the sample space.

b. $P\left(G\right)=$_____

c. $P\left(G\right|E)=$_____

d. $P(GANDE)=$_____

e. $P(GORE)=$_____

f. Are G and E mutually exclusive? Justify your answer numerically

Short answer

(a) Sample:

$S=\left\{\right(G1),(G2),(G3),(G4),(G5),(Y1),(Y2),(Y3\left)\right\}$

localid="1648906675528" $$\begin{gathered} \text{(b)}\hspace{0.17em}P\left(G\right)=\frac{5}{8} \\ \text{(c)}P\left(G|E\right)=\frac{2}{3} \\ \text{(d)}P\left(GANDE\right)=\frac{1}{4} \\ \text{(e)}P\left(GORE\right)=\frac{3}{4} \\ \text{(f)}G\text{and}E\text{arenotmutuallyexclusive.} \end{gathered}$$

Step-by-step solution

Given information (part a)

Suppose that one eight cards. Five are green and three are yellow. The five green cards are numbered $1,2,3,4,5$. The three yellow cards are numbered $1,2,3$. The cards are well shuffled. One card is drawn randomly. Let G= card drawn is green and E= card drawn is even-numbered.

Explanation (part a)

Let G= card drawn is green.

E= card drawn is even-numbered.

Sample space consists of all the possible outcomes.

Thus, sample space is defined as

$S=\left\{\right(G1),(G2),(G3),(G4),(G5),(Y1),(Y2),(Y3\left)\right\}$

Given information (part b)

Suppose that one eight cards. Five are green and three are yellow. The five green cards are numbered $1,2,3,4,5$. The three yellow cards are numbered $1,2,3$. The cards are well shuffled. One card is drawn randomly. Let G= card drawn is green and E= card drawn is even-numbered.

Explanation (part b)

Sample space consists of all the possible outcomes.

Thus, sample space is defined as

$S=\left\{\right(G1),(G2),(G3),(G4),(G5),(Y1),(Y2),(Y3\left)\right\}$

Total number of possible outcomes$=8$

Let G = card drawn is green

Possible outcomes of the event$=\left\{\right(G1),(G2),(G3),(G4),(G5\left)\right\}$

Total number of outcomes$=5$

Thus, the probability that the card drawn is green is calculated as

$P\left(E\right)=\frac{\text{Number of Favourable Outcomes}}{\text{Total Number of Possible Outcomes}}$

$P\left(G\right)=\frac{5}{8}$

Given information (part c)

Suppose that one eight cards. Five are green and three are yellow. The five green cards are numbered $1,2,3,4,5$. The three yellow cards are numbered $1,2,3$. The cards are well shuffled. One card is drawn randomly. Let G= card drawn is green and E= card drawn is even-numbered.

Explanation (part c)

Sample space consists of all the possible outcomes.

Thus, sample space is defined as

$S=\left\{\right(G1),(G2),(G3),(G4),(G5),(Y1),(Y2),(Y3\left)\right\}$

Total number of possible outcomes$=8$

Let G= card drawn is green

E= card drawn is even-numbered.

Possible outcomes that card is even numbered$=\left\{\right(G2),(G4),(Y2\left)\right\}$

$P\left(G|E\right)$denotes the probability that card drawn is green given that card is even-numbered.

Possible outcomes of the event that $S=\left\{\left(G2\right),\left(G4\right)\right\}$

Total number of outcomes$=2$

Thus, $P\left(G|E\right)$is calculated as

localid="1648195838169" $$\begin{gathered} P\left(G\right|E)=\frac{\text{Number of Favourable Outcomes}}{\text{Total Number of Possible Outcomes}} \\ P\left(G|E\right)=\frac{2}{3} \end{gathered}$$

Given information (part d)

Suppose that one eight cards. Five are green and three are yellow. The five green cards are numbered $1,2,3,4,5$. The three yellow cards are numbered $1,2,3$. The cards are well shuffled. One card is drawn randomly. Let G= card drawn is green and E= card drawn is even-numbered.

Explanation (part d)

Sample space consists of all the possible outcomes.

Thus, sample space is defined as

$S=\left\{\right(G1),(G2),(G3),(G4),(G5),(Y1),(Y2),(Y3\left)\right\}$

Total number of possible outcomes$=8$

Let G = card drawn is green

Possible outcomes that card green $=\left\{\right(G1),(G2),(G3),(G4),(G5\left)\right\}$

E= card drawn is even-numbered.

Possible outcomes that card is even-numbered$=\left\{\right(G2),(G4),(Y2\left)\right\}$

localid="1648906722942" $P\left(GANDE\right)$denotes the probability that the card drawn is green and the card is even-numbered.

Possible outcomes of the event that$=\left\{\left(G2\right),\left(G4\right)\right\}$

Total number of outcomes $=2$

Thus, localid="1648906732755" $P\left(GANDE\right)$is calculated as

localid="1648906750982" $$\begin{gathered} P(GANDE)=\frac{\text{card drawn is green and card is even-numbered}}{\text{total possible outcomes}} \\ P\left(GANDE\right)=\frac{2}{8} \\ P\left(GANDE\right)=\frac{1}{4} \end{gathered}$$

Given information (part e)

Suppose that one eight cards. Five are green and three are yellow. The five green cards are numbered $1,2,3,4,5$. The three yellow cards are numbered $1,2,3$. The cards are well shuffled. One card is drawn randomly. Let G= card drawn is green and E= card drawn is even-numbered.

Explanation (part e)

Sample space consists of all the possible outcomes.

Thus, sample space is defined as

$S=\left\{\right(G1),(G2),(G3),(G4),(G5),(Y1),(Y2),(Y3\left)\right\}$

Total number of possible outcomes$=8$

Let G= card drawn is green and E = card drawn is even number.

Possible outcomes that card is even-numbered$=3$

localid="1648906854050" $$\begin{gathered} P(GORE)=P\left(G\right)+P\left(E\right)-P(G\cup E) \\ P(GORE)=\frac{5}{8}+\frac{3}{8}-\frac{1}{4} \\ P(GORE)=\frac{6}{8} \\ P(GORE)=\frac{3}{4} \end{gathered}$$

Given information (part f)

Suppose that one eight cards. Five are green and three are yellow. The five green cards are numbered $1,2,3,4,5$. The three yellow cards are numbered $1,2,3$. The cards are well shuffled. One card is drawn randomly. Let G= card drawn is green and E= card drawn is even-numbered.

Explanation (part f)

Sample space:

$S=\left\{\right(G1),(G2),(G3),(G4),(G5),(Y1),(Y2),(Y3\left)\right\}$

Total number of possible outcomes$=8$

Let $G=\text{carddrawnisgreen}$

Possible outcomes that card is even number$=\left\{\right(G2),(G4),(Y2\left)\right\}$

E = card drawn is an even number

Possible outcomes that card is greenlocalid="1648197211136" $=\{\left(G1\right),(G2),\left(G3\right),(G4),(G5\left)\right\}$

If two events are mutually exclusive then$P\left(G\cup E\right)=P\left(G\right)P\left(E\right)$

$$\begin{gathered} P\left(G\cup E\right)=\frac{1}{4} \\ P\left(G\right)=\frac{5}{8} \\ P\left(E\right)=\frac{3}{8} \\ P\left(G\right)P\left(E\right)=\frac{15}{64} \\ P\left(G\cup E\right)\bcancel{=}P\left(G\right)P\left(E\right) \end{gathered}$$

Thus, the events G and E are not mutually exclusive.