AP EAMCET · Maths · Probability
Box \(A\) contains 2 black and 3 red balls, while Box \(B\) contains 3 black and 4 red balls. Out of these two boxes one is selected at random; and the probability of choosing Box \(A\) is double that of Box \(B\). If a red ball is drawn from the selected box, then the probability that it has come from Box \(B\), is
- A \(\frac{21}{41}\)
- B \(\frac{10}{31}\)
- C \(\frac{12}{31}\)
- D \(\frac{13}{41}\)
Answer & Solution
Correct Answer
(B) \(\frac{10}{31}\)
Step-by-step Solution
Detailed explanation
Let \(P(B)=p\) according to given condition \(\begin{aligned} P(A) & =2 P(B)=2 p \\ P\left(\frac{R}{A}\right) & =\frac{{ }^3 C_1}{{ }^5 C_1}=\frac{3}{5} \end{aligned}\) and \(P\left(\frac{R}{B}\right)=\frac{{ }^4 C_1}{{ }^7 C_1}=\frac{4}{7}\) Using Baye's theorem…
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