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Question Number 113605 by bobhans last updated on 14/Sep/20

$$\mathrm{A}\mathrm{bag}\mathrm{contains}6\mathrm{red},5\mathrm{white}\mathrm{and}4\mathrm{black}\mathrm{balls}$$$$\mathrm{If}\mathrm{twl}\mathrm{balls}\mathrm{are}\mathrm{drawn},\mathrm{what}\mathrm{is}\mathrm{the}\mathrm{probability}$$$$\mathrm{that}\mathrm{none}\mathrm{of}\mathrm{them}\mathrm{are}\mathrm{red}?$$$$$$

Commented by bemath last updated on 14/Sep/20

$$method\mathrm{\_}2$$$$p\left(A\right)=\frac{9}{15}\times \frac{8}{14}=\frac{3}{5}\times \frac{4}{7}=\frac{12}{35}$$

Answered by bemath last updated on 14/Sep/20

$$method\mathrm{\_}1$$$$n\left(S\right)=C_2^{15}=\frac{15\times 14}{2\times 1}=105$$$$noneofthemarered\to \left(2W\right),\left(2B\right),(1W,1B)$$$$n\left(A\right)=C_2^5+C_2^4+C_1^5\times C_1^4$$$$n\left(A\right)=10+6+20=36$$$$sop\left(A\right)=\frac{36}{105}=\frac{12}{35}$$

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