EduFarming
Your list · 0हिं

Type at least two letters. Press Esc to close.

↑↓ navigate↵ openTAB cycle filter

Conditional Probability & Bayes' Theorem · Question 110 of 125

Question 110

MediumOriginal question

Bag I has 3 red balls and 1 black ball; bag II has 2 red and 2 black balls. A bag is picked at random and a ball is drawn from it; RR is the event that the ball is red. Which of the following statements are true?

I. P(I∣R)=35P(\text{I} \mid R) = \frac{3}{5}.

II. P(I∣R)+P(II∣R)=1P(\text{I} \mid R) + P(\text{II} \mid R) = 1.

III. P(I∣R)+P(I∣R′)=1P(\text{I} \mid R) + P(\text{I} \mid R') = 1.

IV. P(II∣R′)=23P(\text{II} \mid R') = \frac{2}{3}.

Options for question 110

Show the answer and explanation
CorrectNot quiteAnswer A

Correct answer: A

Explanation

  1. P(R)=12×34+12×24=58P(R) = \frac{1}{2} \times \frac{3}{4} + \frac{1}{2} \times \frac{2}{4} = \frac{5}{8}, so P(I∣R)=3/85/8=35P(\text{I} \mid R) = \frac{3/8}{5/8} = \frac{3}{5}; P(R′)=38P(R') = \frac{3}{8}, so P(II∣R′)=2/83/8=23P(\text{II} \mid R') = \frac{2/8}{3/8} = \frac{2}{3}. I and IV are true.

  2. II is true: given RR, the ball came from exactly one bag. III is false: P(I∣R′)=13P(\text{I} \mid R') = \frac{1}{3}, and 35+13=1415\frac{3}{5} + \frac{1}{3} = \frac{14}{15}.

Worth noting

The chances of the two bags add to 1 under one condition; the chances of one bag under two different conditions need not.

Concept
Bayes' theorem
Typical time
120 s
Your time
—
The rule this testsBayes' theorem
Practise this chapter →

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.