Skip to main contentSkip to solution

A bag contains 4 balls. Let E be an event in which two balls are drawn at random and found to be white, then

Match List-I with List-II

List-IList-II
(A) Probability that all balls are white(I) 16\dfrac{1}{6}
(B) Probability of drawing 2 white ball at random given that there are 3 white ball(II) 35\dfrac{3}{5}
(C) P(E)(III) 12\dfrac{1}{2}
(D) Probability of drawing 2 white ball at random given that there is only 2 white ball(IV) 59\dfrac{5}{9}

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 2

Take the number of white balls as 2, 3 or 4, each with probability 13\dfrac13. Then P(E∣2)=2C24C2=16P(E\mid2)=\dfrac{^2C_2}{^4C_2}=\dfrac16, P(E∣3)=36=12P(E\mid3)=\dfrac36=\dfrac12, P(E∣4)=1P(E\mid4)=1. So P(E)=13(16+12+1)=59P(E)=\dfrac13\left(\dfrac16+\dfrac12+1\right)=\dfrac59, and by Bayes' theorem P(all white∣E)=1/35/9=35P(\text{all white}\mid E)=\dfrac{1/3}{5/9}=\dfrac35.

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question