Question

A box holds 8 red and 4 blue beads. Three beads are taken from the box and not replaced. Determine: The probability that all three beads taken are red The probability that there are at least two blue beads from the three that were taken

Answer 1

Answer: $\dfrac{14}{55}, \dfrac{13}{55}$

Step-by-step explanation:

Given

There are 8 red and 4 blue beads

Three beads were taken from the box

(i)No of ways of choosing three red beads out of 12 beads is $^8C_3$

Total no of ways $^{12}C_3$

Probability is

$P=\dfrac{^8C_3}{^{12}C_3}=\dfrac{8\times 7\times 6}{12\times 11\times 10}=\dfrac{2\times 7}{11\times 5}\\P=\dfrac{14}{55}$

(ii)atleast two beads refers to minimum two beads

There can be two possibilities (2B,1 R), (3B, 0R)

$\Rightarrow P=\dfrac{^4C_2\times ^8C_1}{^{12}C_3}+\dfrac{^4C_3}{^{12}C_3}\\\\\Rightarrow P=\dfrac{6\times 8}{220}+\dfrac{4}{220}=\dfrac{52}{220}\\\\\Rightarrow P=\dfrac{13}{55}$