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Conditional Probability & Bayes' Theorem · Question 101 of 125

Question 101

EasyOriginal question

Box I has 3 red and 7 white balls; box II has 6 red and 4 white balls. A box is picked at random and a ball drawn from it is red. What is the probability that it came from box I?

Options for question 101

Show the answer and explanation
CorrectNot quiteAnswer C

Correct answer: C

Explanation

  1. P(R)=12×310+12×610=920P(R) = \frac{1}{2} \times \frac{3}{10} + \frac{1}{2} \times \frac{6}{10} = \frac{9}{20}.

  2. P(I∣R)=3/209/20=13P(\text{I} \mid R) = \frac{3/20}{9/20} = \frac{1}{3}.

Worth noting

310\frac{3}{10} is P(R∣I)P(R \mid \text{I}), the chance of red from box I, the condition the other way round.

Concept
Bayes' theorem
Typical time
60 s
Your time
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The rule this testsBayes' theorem
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Ten questions from Conditional Probability & Bayes' Theorem, at your own pace.

Practise this chapter →

Ten questions from Conditional Probability & Bayes' Theorem, at your own pace.