Question

a bag of 29 tulip bulbs contains 11 red tulip bulbs, 11 yellow tulip bulbs, and 7 purple tulip bulbs. suppose two tulip bulbs are randomly selected without replacement from the bag.
(a) what is the probability that the 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?

(a) the probability that both bulbs are red is
(round to three decimal places as needed.)

Explanation:

Step1: Probability of first red

The probability that the first - selected tulip bulb is red is the number of red bulbs divided by the total number of bulbs. There are 11 red bulbs and 29 total bulbs, so $P(\text{first red})=\frac{11}{29}$.

Step2: Probability of second red given first red

Since the selection is without replacement, after one red bulb is selected, there are 10 red bulbs left and 28 total bulbs left. So $P(\text{second red}|\text{first red})=\frac{10}{28}$.

Step3: Probability of both red

By the multiplication rule for conditional probability $P(A\cap B)=P(A)\times P(B|A)$. Here, $A$ is the event that the first bulb is red and $B$ is the event that the second bulb is red. So $P(\text{both red})=\frac{11}{29}\times\frac{10}{28}=\frac{110}{812}\approx0.135$.

Answer:

0.135