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Seven balls are drawn simultaneously from a bag containing 5 white and 6 green balls. The probability of drawing 3 white and 4 green balls is :

\[\begin {array} {1 1} (a)\;\frac{7}{^{11} C _7} & \quad (b)\;\frac{^{5} C_3+\;^6C_4}{^{11}C_7}\\ (c)\;\frac{^5C_2\;^6C_2}{^{11}C_7} & \quad (d)\;\frac{^6C_3\; ^5C_4}{^{11}C_7}\end {array}\]

1 Answer

$(c)\;\frac{^5C_2\;^6C_2}{^{11}C_7}$
answered Nov 7, 2013 by pady_1
 

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