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Worked Examples · Example 7

Q.Bag I contains 4 white and 6 black balls; Bag II contains 6 white and 4 black balls. A bag is selected at random and a ball is drawn from it; the ball drawn is found to be white. Find the probability that it was drawn from Bag I.

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Step 1 — Prior probabilities of selecting each bag. A bag is chosen at random from two bags, so:

P(Bag I)=12,P(Bag II)=12P(\text{Bag I})=\dfrac{1}{2}, \qquad P(\text{Bag II})=\dfrac{1}{2}

Step 2 — Conditional probability of drawing white from each bag.

P(white∣Bag I)=410=25,P(white∣Bag II)=610=35P(\text{white}\mid\text{Bag I})=\dfrac{4}{10}=\dfrac{2}{5}, \qquad P(\text{white}\mid\text{Bag II})=\dfrac{6}{10}=\dfrac{3}{5}

Step 3 — Total probability of drawing a white ball (Law of Total Probability).

P(white)=P(Bag I)P(white∣Bag I)+P(Bag II)P(white∣Bag II)=12×25+12×35=15+310=210+310=510=12P(\text{white})=P(\text{Bag I})P(\text{white}\mid\text{Bag I})+P(\text{Bag II})P(\text{white}\mid\text{Bag II})=\dfrac{1}{2}\times\dfrac{2}{5}+\dfrac{1}{2}\times\dfrac{3}{5}=\dfrac{1}{5}+\dfrac{3}{10}=\dfrac{2}{10}+\dfrac{3}{10}=\dfrac{5}{10}=\dfrac{1}{2}

Step 4 — Apply Bayes' Theorem.

P(Bag I∣white)=P(Bag I)P(white∣Bag I)P(white)=12×2512=1/51/2=25P(\text{Bag I}\mid\text{white})=\dfrac{P(\text{Bag I})P(\text{white}\mid\text{Bag I})}{P(\text{white})}=\dfrac{\frac{1}{2}\times\frac{2}{5}}{\frac{1}{2}}=\dfrac{1/5}{1/2}=\dfrac{2}{5} …

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