Question

Suppose that you have 9 cards. 6 are green and 3 are yellow. The 6 green cards are numbered 1, 2, 3, 4, 5, and 6. The 3 yellow cards are numbered 1, 2, and 3. The cards are well shuffled. You randomly draw one card.

• G = card drawn is green
• Y = card drawn is yellow •
E = card drawn is even-numbered
a. List the sample space. (Type your answer using letter/number combinations separated by commas. Example: G1, Y1, ...)
b. Enter the probability as a fraction.
P(G) =
c. Enter the probability as a fraction.
P(G | E) =
d. Enter the probability as a fraction.
P(G AND E) =
e. Enter the probability as a fraction.
P(G OR E) =
f. Are G and E mutually exclusive?
Yes
No

Answer 1

In calculating probabilities of drawing green (G), yellow (Y), or even-numbered (E) cards, the sample space, individual probabilities, and joint probabilities are identified, showing that the events G and E are not mutually exclusive.

Step-by-step explanation:

The question involves the concept of probability and the calculation of probabilities for certain events when drawing from a set of distinctively marked and numbered cards. Given that there are 6 green cards numbered 1 through 6, and 3 yellow cards numbered 1 through 3, we can list the sample space and calculate the probabilities for different events.

• Sample Space: G1, G2, G3, G4, G5, G6, Y1, Y2, Y3.
• P(G): Probability of drawing a green card = number of green cards / total number of cards = 6/9 = 2/3.
• P(G | E): Probability of drawing a green card given that the card is even-numbered. There are three even-numbered green cards (G2, G4, G6) out of a total of three even-numbered cards, so P(G | E) = 3/3 = 1.
• P(G AND E): Probability of drawing a card that is both green and even-numbered = number of even-numbered green cards / total number of cards = 3/9 = 1/3.
• P(G OR E): Probability of drawing a card that is either green or even-numbered = P(G) + P(E) - P(G AND E). P(E) = 4/9 (even-numbered cards are G2, G4, G6, Y2). So, P(G OR E) = 2/3 + 4/9 - 1/3 = 8/9.
• G and E are not mutually exclusive because there are cards that are both green and even-numbered (G2, G4, G6).