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Conditional Probability & Bayes' Theorem · Question 94 of 125

Question 94

MediumOriginal question

Bag I has 3 red and 2 blue balls; bag II has 3 red and 1 blue ball. A bag is picked at random and two balls are drawn from it, one after the other, without replacement. What is the probability that both are red?

Options for question 94

Show the answer and explanation
CorrectNot quiteAnswer B

Correct answer: B

Explanation

  1. Both red from bag I: 35×24=310\frac{3}{5} \times \frac{2}{4} = \frac{3}{10}; from bag II: 34×23=12\frac{3}{4} \times \frac{2}{3} = \frac{1}{2}.

  2. 12×310+12×12=320+520=25\frac{1}{2} \times \frac{3}{10} + \frac{1}{2} \times \frac{1}{2} = \frac{3}{20} + \frac{5}{20} = \frac{2}{5}.

Worth noting

369800=12(925+916)\frac{369}{800} = \frac{1}{2}\left(\frac{9}{25} + \frac{9}{16}\right) puts the first ball back before the second draw.

Concept
The theorem of total probability
Typical time
90 s
Your time
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The rule this testsThe theorem of total probability
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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.