Answer to Question #129413 in Statistics and Probability for javeria

Question

A box contains 10 red and 12 white rose flowers. Flowers are picked up at random one by one without replacement. What is probability that
a) The first 3 flowers are red.
b) Flowers are in alternate order starting from 1st one is red.

Answer 1

a) The probability that the first flower is red equals $\frac{10}{10+12}=\frac{10}{22}$. If the first flower is red, the probability for the second flower to be red equals $\frac{10-1}{9+12}=\frac{9}{21}$. For the third frower it equals $\frac{8}{20}$. Thus, overall probability to pick 3 red flower in a row equals $\frac{10}{22}\frac{9}{21}\frac{8}{20}=\frac{720}{9240}\approx0.078=7.8\%$ .

b) The probability of first flowe to be red is $\frac{10}{22}$. After that, the probability that we pick white flower is $\frac{12}{22-1}=\frac{12}{21}.$ The next flower is red with the probability of $\frac{9}{20}$. By this idea, the probability we will take red and white roses one by one is $\frac{10}{22}\frac{12}{21}\frac{9}{20}...\frac{4}{5}\frac{1}{4}\frac{3}{3}\frac{2}{2}\frac{1}{1}=\frac{10!12!}{22!}=1.5*10^{-6}=0.00015$ \%