Question

Slips of paper are placed in a large hat and thoroughly mixed. Ten slips bear the number 1, 20 slips bear the number 2, 30 slips bear the number 3, and 5 slips bear the number 4. What is the probability of drawing

(a) A 1?

(b) A 2?

(c) A 3?

(d) A 4?

(e) A 1 or a 4?

(f) A 1 or a 2 or a 3 or a 4?

(g) A 5?

(h) A 2 and then a 3? (The number drawn first is replaced prior to the second draw.)

Answer 1

The total number of slips here: 10+20+30+5 = 65. We know the probability of drawing an event is equal to the number of favorable events divided by the total number of outcomes. Thus the probability of drawing

a) A 1 = 10/65 = 0.153846

b) A 2 = 20/65 = 0.307692

c) A 3 = 30/65 = 0.4615385

d) A 4 = 5/65 = 0.076923

e) A 1 or a 4 = (10+5)/65 = 0.230769

f) A 1 or 2 or 3 or 4 = 65/65 = 1

g) A 5 = 0/65 = 0

h) A 2 and the a 3 = P(A 2)*P(A 3 )= 10/65 * 30/65 = 0.071005