A) \[\frac{5}{12}\]
B) \[\frac{7}{12}\]
C) \[\frac{1}{3}\]
D) \[\frac{1}{4}\]
Correct Answer: B
Solution :
[b] Total no. of ball \[=2+3+4=9\] Total no. of red balls = 3 E = An event of drawing two ball that containing at least one ball is red \[\therefore n(E){{=}^{3}}{{C}_{2}}{{+}^{3}}{{C}_{1}}{{\times }^{2}}{{C}_{2}}{{+}^{3}}{{C}_{1}}{{\times }^{4}}{{C}_{2}}=3+3\times 2+4\times 3=21\] \[\because \,\,Required\text{ }probability\,=P(E)=\frac{n(E)}{n(S)}=\frac{21}{^{9}{{C}_{2}}}=\frac{21}{\frac{9\times 8}{2}}=\frac{21\times 2}{9\times 8}=\frac{42}{72}=\frac{7}{12}\] Hence, option [b] is correct.
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