Answer to Question #147384 in Statistics and Probability for rehan tahir

Question

A box contains 10 red and 12 white rose flowers. Four flowers are picked up at random one by one without replacement. What is the probability that;
a. The first 3 flowers are red?
b. three are 2 red and 2 white flowers
c. at least 2 are red.

Answer 1

a. P(the first flower is red) = $\frac{10}{10 + 12} = \frac{10}{22}$

P(the probability for the second flower to be red, if the first flower is red) = $\frac{10-1}{9 + 12} = \frac{9}{21}$

P(the third flower is red, if the first and the second flowers are red) = $\frac{8}{20}$

P(the first 3 flowers are red) $= \frac{10}{22} \times \frac{9}{21} \times \frac{8}{20} = \frac{720}{9240} = 0.078$ or 7.8 %

b. P(2 red and 2 white flowers) $= 6 \times \frac{10}{22} \times \frac{9}{21} \times \frac{12}{20} \times \frac{11}{19} = \frac{71280}{175560} = 0.4060$ or 40 %

c. At least 2 are red out of 4. The number of red roses can be 2, 3 or 4.

Total number of red roses are 10.

If 2 roses are red:

¹⁰C₂ $\times$ ¹²C₂ = $\frac{10!}{8!2!} \times \frac{12!}{10!2!} = 2970$

If 3 roses are red:

¹⁰C₃ $\times$ ¹²C₁ = $\frac{10!}{7!3!} \times \frac{12!}{11!} = 1440$

If 4 roses are red:

¹⁰C₄ $\times$ ¹²C₀ = $\frac{10!}{6!4!} \times 1 = 210$

P(at least 2 are red) $= \frac{2970 + 1440 +210}{C^{22}_4} = \frac{4620}{7315} = 0.631$ or 63 %