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Seven balls are drawn simultaneously from bag containing 5 white and 6 green balls. The probability of drawing 3 white and 4 green balls is :
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Verified Answer
The correct answer is:
$\frac{{ }^5 C_2{ }^6 C_2}{{ }^{11} C_7}$
Number of ways to get 3 white and 4 green balls from 5 white and 6 green balls
$={ }^5 C_3 \times{ }^6 C_4={ }^5 C_2 \times{ }^6 C_2$
and total number of ways $={ }^{11} C_7$
$\therefore$ Required probability $=\frac{n(E)}{n(S)}$
$=\frac{{ }^5 C_2 \times{ }^6 C_2}{{ }^{11} C_7}$
$={ }^5 C_3 \times{ }^6 C_4={ }^5 C_2 \times{ }^6 C_2$
and total number of ways $={ }^{11} C_7$
$\therefore$ Required probability $=\frac{n(E)}{n(S)}$
$=\frac{{ }^5 C_2 \times{ }^6 C_2}{{ }^{11} C_7}$
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