Q82 / 160 · TS EAMCET 2006 · Mathematics

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 :

  • A
    $\frac{7}{{ }^{11} C_7}$
  • B
    $\frac{{ }^5 C_3+{ }^6 C_4}{{ }^{11} C_7}$
  • Correct answer
    C
    $\frac{{ }^5 C_2{ }^6 C_2}{{ }^{11} C_7}$
  • D
    $\frac{{ }^6 C_3{ }^5 C_4}{{ }^{11} C_7}$

Solution

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 $=\frac{n(E)}{n(S)}$ $=\frac{{ }^5 C_2 \times{ }^6 C_2}{{ }^{11} C_7}$

Hence, the correct answer is option: 'C'.

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