Question 1166857: Suppose that you have 7 cards. 3 are green and 4 are yellow. The 3 green cards are numbered 1, 2, and 3. The 4 yellow cards are numbered 1, 2, 3, and 4. 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
List the sample space. (Type your answer using letter/number combinations separated by commas. Example: G1, Y1, ...)
G1, G2, G3, Y1, Y2, Y3, Y4
Enter the probability as a fraction.
P(G) = 3/7
Enter the probability as a fraction.
P(G | E) = 1/3
Enter the probability as a fraction.
P(G AND E) = 1/7
Enter the probability as a fraction.
P(G OR E) = ??? (I need help solving this one)
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
P(G OR E)....
(1) an elementary solution
Count the number that are either green (3), or are not green (i.e., are yellow) and are even (2):
P(G OR E) = 3/7 + 2/7 = 5/7
(2) by a more sophisticated method
Count the number that are green (3), add the number that are even (3), then subtract the number that are both green and even (1) (since you have counted them twice):
P(G OR E) = P(G) + P(E) - P(G AND E) = 3/7 + 3/7 - 1/7 = 5/7
(3) by yet a different method
Count the number that are NOT green AND NOT even (i.e., that are NEITHER green NOR even = that are yellow and odd (2)), and subtract that from the total (7):
P(G OR E) = 1 - P(NOT(G) AND NOT(E)) = 7/7 - 2/7 = 5/7
|
|
|
