A bit flower vase has a 5 begonias, 6 carnations and 7 tulips. Two flowers are randomly selected without replacement. What is the probability of selecting two carnations? Express in simplest form of fractions.
There are a total of 5 + 6 + 7 = 18 flowers in the vase.
The probability of selecting a carnation on the first draw is 6/18, or 1/3.
Since we are drawing without replacement, there will be 5 + 7 = 12 flowers left in the vase for the second draw.
The probability of selecting another carnation on the second draw, given that the first draw was a carnation, is 5/12.
To find the probability of both events occurring, we multiply the probabilities:
P(selecting 2 carnations) = (1/3) x (5/12) = 5/36
Therefore, the probability of selecting two carnations is 5/36.
To find the probability of selecting two carnations, we need to find the total number of possible outcomes and the number of favorable outcomes.
Total number of flowers in the vase: 5 begonias + 6 carnations + 7 tulips = 18 flowers
Number of possible outcomes for the first flower: 18 (since any of the 18 flowers can be selected)
Number of possible outcomes for the second flower (after one carnation has already been selected): 17 (since only 17 flowers remain in the vase)
Number of favorable outcomes (2 carnations): There are 6 carnations in the vase, so the first carnation can be selected in 6 ways and the second carnation in 5 ways (without replacement).
Now, let's calculate the probability:
Probability = (Number of favorable outcomes) / (Number of possible outcomes)
= (6 * 5) / (18 * 17)
= 30 / 306
= 5 / 51
Therefore, the probability of selecting two carnations is 5/51.