91Ó°ÊÓ

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.

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

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

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.

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

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

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

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.

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

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.

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

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.

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

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.

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$P\left(G∪E\right)=P\left(G\right)P\left(E\right)$

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

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