TS EAMCET · Maths · Probability
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 \(\frac{{ }^5 C_3+{ }^6 C_4}{{ }^{11} C_7}\)
- C \(\frac{{ }^5 C_2{ }^6 C_2}{{ }^{11} C_7}\)
- D \(\frac{{ }^6 C_3{ }^5 C_4}{{ }^{11} C_7}\)
Answer & Solution
Correct Answer
(C) \(\frac{{ }^5 C_2{ }^6 C_2}{{ }^{11} C_7}\)
Step-by-step Solution
Detailed explanation
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 \(=\frac{n(E)}{n(S)}\)…
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