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 0.135 (round to three decimal places as needed.) (b) the probability that the first bulb is red and the second is yellow is (round to three decimal places as needed.)

Explanation:

Step1: Calculate probability of first - red

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

Step2: Calculate probability of second - yellow

After one red bulb is selected, there are 28 bulbs left and 11 yellow bulbs. So the probability of selecting a yellow bulb second given that the first was red is $P(\text{second yellow}|\text{first red})=\frac{11}{28}$.

Step3: Use multiplication rule

By the multiplication rule for conditional probability $P(A\cap B)=P(A)P(B|A)$. Here, $A$ is the event of selecting a red bulb first and $B$ is the event of selecting a yellow bulb second. So $P(\text{first red and second yellow})=\frac{11}{29}\times\frac{11}{28}=\frac{121}{812}\approx0.149$.

Answer:

0.149