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a bag of 29 tulip bulbs contains 11 red tulip bulbs, 11 yellow tulip bu…

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 0.128
(round to three decimal places as needed.)
(b) the probability that the first bulb is red and the second is yellow is 0.149
(round to three decimal places as needed.)
(c) the probability that the first bulb is yellow and the second is red is 0.149
(round to three decimal places as needed.)
(d) the probability that one bulb is red and one is yellow is
(round to three decimal places as needed.)

Explanation:

Step1: Calculate the probability of first - red then - yellow

The probability that the first bulb is red is $\frac{11}{11 + 11+7}=\frac{11}{29}$. After one red bulb is selected, there are 28 bulbs left. The probability that the second bulb is yellow is $\frac{11}{28}$. So the probability of first - red then - yellow is $\frac{11}{29}\times\frac{11}{28}$.

Step2: Calculate the probability of first - yellow then - red

The probability that the first bulb is yellow is $\frac{11}{29}$. After one yellow bulb is selected, there are 28 bulbs left. The probability that the second bulb is red is $\frac{11}{28}$. So the probability of first - yellow then - red is $\frac{11}{29}\times\frac{11}{28}$.

Step3: Calculate the probability that one is red and one is yellow

The probability that one is red and one is yellow is the sum of the probability of first - red then - yellow and the probability of first - yellow then - red.
$P=\frac{11}{29}\times\frac{11}{28}+\frac{11}{29}\times\frac{11}{28}=2\times\frac{11\times11}{29\times28}=\frac{2\times121}{812}\approx0.298$

Answer:

$0.298$