TS EAMCET · Maths · Probability
A bag \(X\) contains 2 white and 3 black balls and another bag \(Y\) contains 4 white and 2 black balls. One bag is selected at random and a ball is drawn from it. Then, the probability for the ball chosen be white, is :
- A \(\frac{2}{15}\)
- B \(\frac{7}{15}\)
- C \(\frac{8}{15}\)
- D \(\frac{14}{15}\)
Answer & Solution
Correct Answer
(C) \(\frac{8}{15}\)
Step-by-step Solution
Detailed explanation
Probability of selecting a white ball from \(X\) bag \(=\frac{2}{5}\) Probability of selecting a white ball from \(Y\) bag \(=\frac{4}{6}=\frac{2}{3}\) Probability of selecting a white ball from \(X\) or \(Y\) bags \(=\frac{2}{5}+\frac{2}{3}=\frac{6+10}{15}=\frac{16}{15}\)…
See the Complete Solution
Get step-by-step explanations for this and 2.5 Lakh+ more JEE, NEET & CET questions.
- Unlock all solutions
- Practice the full chapter
- Track accuracy across PYQs
4.8 rated on Google Play · 14,000+ reviews
More questions from Maths
- The combined equation of the direct common tangents of the circles \(x^2+y^2+2 x=0\) and \(x^2+y^2-2 y-3=0\) isTS EAMCET 2022 Medium
- If \(a=\frac{1-i \sqrt{3}}{2}\), then the correct matching of 'List-I from List-II is List-I List-II (i) \(a \bar{a}\) (A) \(-\frac{\pi}{3}\) (ii) \(\arg \left(\frac{1}{\bar{a}}\right)\) (B) \(-i \sqrt{3}\) (iii) \(a-\bar{a}\) (C) \(2 i / \sqrt{3}\) (iv) \(\operatorname{Im}\left(\frac{4}{3 a}\right)\) (D) 1 (E) \(\pi / 3\) (F) \(\frac{2}{\sqrt{3}}\) correct match is (i) (ii) (iii) (iv)TS EAMCET 2007 Hard
- If \(\omega \neq 1\) is a cube root of unity, then one root among the \(7^{\text {th }}\) roots of \((1+\omega)\) isTS EAMCET 2025 Medium
- \(E_1: a+b+c=0, \quad\) if 1 is a root of \(a x^2+b x+c=0, \quad E_2: b^2-a^2=2 a c\), if \(\sin \theta\), \(\cos \theta\) are the roots of \(a x^2+b x+c=0\) Which of the following is true?TS EAMCET 2005 Medium
- If thenTS EAMCET 2021 Medium
- Let \(f(x)=x^2 e^{-2 x}, x>0\). The maximum value of \(f(x)\) isTS EAMCET 2018 Easy
More PYQs from TS EAMCET
- Iron crystalizes in with an edge length of . If it contains Schottky defects, calculate its approximate density [ of ]TS EAMCET 2022 Hard
- \(f:[1,3] \rightarrow R\) is a function defined as \(f(x)=x^3+a x^2+b x\). If \(f(\mathrm{l})-f(3)=0\) and \(f^{\prime}\left(\frac{2 \sqrt{3}+1}{\sqrt{3}}\right)=0\). Then, \(a-b\) is equal toTS EAMCET 2021 Medium
- If \(\frac{2 x^3+3 x^2+3 x+5}{\left(x^2+1\right)\left(x^2+2\right)}\) is expanded in terms of the powers of \(x\), then the coefficient of \(x^5\) isTS EAMCET 2023 Medium
- The radio horizon of a transmitting antenna of height 39.2 m is (Radius of the earth \(=6400 \mathrm{~km}\) )TS EAMCET 2025 Medium
- Two concentric circular coils, one of small radius \(r\) and the other of large radius \(R\) are placed co-axially with centres coinciding. If the radius \(r\) is changed by \(2 \%\), then the change in mutual inductance of the arrangement is (assume, \(r< < R\) )TS EAMCET 2020 Medium
- If \(2 \hat{i}-\hat{j}+\hat{k}, \hat{i}-3 \hat{j}-5 \hat{k}\) are the position vectors of the points \(\mathrm{A}\) and \(\mathrm{B}\) respectively, \(\mathrm{C}\) divides \(\mathrm{AB}\) in the ratio \(2: 3\) and \(\mathrm{M}\) is the mid-point of \(\mathrm{AB}\), then 5 (position vector of \(C)-2(\) position vector of \(M)=\)TS EAMCET 2022 Medium