A bag contains 8 white counters, 3 black counters, and 6 green counters. What is the probability of drawing

Question

(a) a white counter or a green counter?

(b) a black counter or a green counter?

(c) not a green counter?

Answer 1

There are a total of 17 counters (8 white, 3 black, and 6 green) in the bag.

W = picking a white counter (8/17)

B = picking a black counter (3/17)

G = picking a green counter (6/17)

a.) P(W ∪ G) = P(W) + P(G) - P(W ∩ G) = (8/17) + (6/17) - 0 = 14/17 = 0.8235

b.) P(B ∪ G) = P(B) + P(G) - P(B ∩ G) = (3/17) + (6/17) - 0 = 9/17 = 0.5294

c.) P(G^c) = 1 - P(G) = 1 - (6/17) = 11/17 = 0.6471

For parts a and b, use the addition rule and two events are mutually exclusive because you can't have a counter that have both colors at the same time. For part c, use the complement rule.

Answer 2

The probability is the number of favorable outcomes divided by the number of total possible outcomes. For part a, how many counters are white OR green, and how many counters are there total?