A) \[\frac{3}{10}\]
B) \[\frac{4}{5}\]
C) \[\frac{1}{2}\]
D) \[\frac{1}{6}\]
Correct Answer: B
Solution :
Since the item are choosen without replacement. \ \[P(X=x)=\frac{^{3}{{C}_{x}}{{+}^{7}}{{C}_{4-x}}}{^{10}{{C}_{4}}}\] Putting \[x=1,\,2\] we have \[P(0<x<3)\]\[=\frac{^{3}{{C}_{1}}{{\times }^{7}}{{C}_{3}}}{210}+\frac{^{3}{{C}_{2}}{{\times }^{7}}{{C}_{2}}}{210}\] \[=\frac{3\times 35+3\times 21}{210}=\frac{105+63}{210}=\frac{168}{210}=\frac{4}{5}\].
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