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Sagot :
Answer: [tex]\dfrac{14}{55}, \dfrac{13}{55}[/tex]
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 [tex]^8C_3[/tex]
Total no of ways [tex]^{12}C_3[/tex]
Probability is
[tex]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}[/tex]
(ii)atleast two beads refers to minimum two beads
There can be two possibilities (2B,1 R), (3B, 0R)
[tex]\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}[/tex]
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