Question

A bag of 29 tulip bulbs contains 12 red tulip​ bulbs, 8 yellow tulip​ bulbs, and 9 purple tulip bulbs.

​(a) What is the probability that two randomly selected tulip bulbs are both​ red? ​

(b) What is the probability that the first bulb selected is red and the second​ yellow? ​

(c) What is the probability that the first bulb selected is yellow and the second​ red? ​

(d) What is the probability that one bulb is red and the other​ yellow? ​

(Round to three decimal places as​ needed.)

Answer 1

a)

p( Red tulip bulb) = n(red tulip bulbs) / n ( total bulbs) = 12 / 29

Therefore,

p( Both red bulb) = p ( First red bulb) * p( second red bulb)

= 12 / 29 * 11 / 28

= 0.1626

p( Both red bulb) = 0.163

b)

p( Red tulip bulb) = n(red tulip bulbs) / n ( total bulbs) = 12 / 29

p( yellow tulip bulb) = n(Yellow tulip bulb) / n (total bulbs) = 8 / 29

p( first bulb is red ) = 12/29

Since first bulb is selected, 28 total bulbs left.

p( second bulb is yellow) = 8 / 28

p( first bulb red AND second bulb yellow) = p( first bulb red ) * p( second bulb yellow)

= 12 / 29 * 8 / 28

= 0.1182

p( first bulb red AND second bulb yellow) = 0.118

c)

p( first bulb is yellow) = 8 / 29

Since first bulb is selected, 28 total bulbs left.

p(second bulb is red) = 12 / 28

therefore,

p( first bulb is yellow AND the second​ red) = p( first bulb is yellow) * p( second bulb is red)

= 8 / 29 * 12 / 28

= 0.1182

p( first bulb is yellow AND the second​ red) = 0.118

d)

p( one bulb is red and the other​ yellow)

= p( first bulb red AND second bulb yellow) + p( first bulb is yellow AND the second​ red)

= 0.118+0.118

= 0.236

p( one bulb is red and the other​ yellow) = 0.236