TS EAMCET · Maths · Probability
A bag contains 6 red, 2 white and 8 blue balls. Three are drawn at random from the bag. Match the items of List-I with those items of List-II. 
The correct match is \(\begin{array}{lllllllll}A & B & C & D & A & B & C & D\end{array}\)
- A III \(\quad 1 \quad I N \quad\) II
- B III IV \(\mathrm{V}\) II
- C \(\begin{array}{llll}\text { IV } & \text { III } & \text { I } & \text { V }\end{array}\)
- D \(\begin{array}{llll}\| & \text { I } & \mathrm{V} & \mathrm{IV}\end{array}\)
Answer & Solution
Correct Answer
(A) III \(\quad 1 \quad I N \quad\) II
Step-by-step Solution
Detailed explanation
Since, the bag contains 6 red, 2 white and 8 blue balls. So, probability of drawn three balls, such that none of the balls is white \(=\frac{{ }^{14} C_3}{{ }^{16} C_3}=\frac{14 \times 13 \times 12}{16 \times 15 \times 14}=\frac{13}{20}\) Probability of getting 2 white and 1…
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