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4. An urn contains 5 red and 6 white balls. Two balls are drawn in succession w...

Oct 29, 2024

Solution by Steps

step 1

Let A be the event that the first ball drawn is white, and B be the event that the second ball drawn is white. We need to find

P(A | B)

, which is given by the formula:

P(A | B) = \frac{P(A \cap B)}{P(B)}

step 2

To find

P(A \cap B)

, we consider that if the first ball is white, there are 5 red and 5 white balls left. Thus, the probability of drawing a white ball first and then a white ball second is:

P(A \cap B) = P(A) \cdot P(B | A) = \left(\frac{6}{11}\right) \cdot \left(\frac{5}{10}\right) = \frac{6}{11} \cdot \frac{1}{2} = \frac{3}{11}

step 3

Now, we calculate

P(B)

, the total probability of drawing a white ball second. This can happen in two ways: either the first ball is white or the first ball is red. Thus,

P(B) = P(A) \cdot P(B | A) + P(A^c) \cdot P(B | A^c)

where

P(A^c) = \frac{5}{11}

and

P(B | A^c) = \frac{6}{10}

. Therefore,

\[ P(B) = \left(\frac{6}{11} \cdot \frac{5}{10}\right) + \left(\frac{5}{11} \cdot \frac{6}{10}\right) = \frac{3}{11} + \frac{3}{11} = \frac{6}{11} \]

step 4 ⋮ Finally, substituting back into the formula for conditional probability, we have: ‖

P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{3}{11}}{\frac{6}{11}} = \frac{3}{6} = \frac{1}{2}

Answer

\frac{1}{2}

Key Concept

Conditional probability and the use of the law of total probability

Explanation

The answer

\frac{1}{2}

represents the probability that the first ball drawn was white given that the second ball drawn was white, calculated using the definitions of conditional probability and total probability.