(C) The total number of ways to draw $7$ balls from $11$ balls ($5$ white + $6$ green) is given by ${ }^{11}C_7$. The number of ways to choose $3$ whi

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$\frac{{ }^5C_3 \times { }^6C_4}{{ }^{11}C_7}$

(C) The total number of ways to draw $7$ balls from $11$ balls ($5$ white + $6$ green) is given by ${ }^{11}C_7$. The number of ways to choose $3$ white balls from $5$ white balls is ${ }^5C_3$. The number of ways to choose $4$ green balls from $6$ green balls is ${ }^6C_4$. The number of favorable outcomes is ${ }^5C_3 \times { }^6C_4$. Therefore, the required probability is $\frac{{ }^5C_3 \times { }^6C_4}{{ }^{11}C_7}$.

Class 11 Mathematics Probability Advanced Use of permutations and combinations in probability

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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:

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A
$\frac{7}{{ }^{11}C_7}$
B
$\frac{{ }^5C_3 + { }^6C_4}{{ }^{11}C_7}$
C
$\frac{{ }^5C_3 \times { }^6C_4}{{ }^{11}C_7}$
D
$\frac{{ }^6C_3 \times { }^5C_4}{{ }^{11}C_7}$

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Advanced Use of permutations and combinations in probability

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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:

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