9: 8. Conditional Probability and Independence

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Nov 10, 2019
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The following topics are included in this series of seven videos.

Conditional Probability, Example 1
Conditional Probability, Example 2
Conditional Probability, Example 3
Conditional Probability, Example 4
Independent Events, Example 1
Independent Events, Example 2
Independent Events, Example 3

Prework:

Suppose $P r (A) = 0.4$ , $P r (B) = 0.6$ , and $P r (A \cap B) = 0.15$ . Compute $P r (A | B)$ and $P r (A | B^{c})$ .
Suppose that $P r (A) = 0.3$ , $P r (B) = 0.5$ , $P r (A \cap B^{c}) = 0.16$ . Are $A$ and $B$ independent? Explain.
Suppose $A$ and $B$ are independent events, with $P r (A) = .4$ and $P r (B) = .7$ . Compute $P r (A^{c} | B)$ and $P r (A | B^{c})$ .
There 5 first-years and 3 sophomores in a classroom. Two are selected at random. What is the probability that both are first-years given that at least one is a first-year?

Solutions:

Using a Venn diagram, we can determine that $P r (A \cap B) = 0.15$ , $P r (A \cap B^{c}) = 0.25$ , $P r (A^{c} \cap B) = 0.45$ , and $P r (A^{c} \cap B^{c}) = 0.15$ . Then $P r (A | B) = \frac{P r (A \cap B)}{P r (B)} = \frac{.15}{.6}$ and $P r (A | B^{c}) = \frac{P r (A \cap B^{c})}{P r (B^{c})} = \frac{.25}{.4}$ .
We need to check if $P r (A) \cdot P r (B) = P r (A \cap B) .$ The left hand side of this equation just becomes $0.3 \cdot 0.5 = 0.15$ . If we create and fill in a Venn diagram, we determine that $P r (A \cap B) = 0.14$ . Since $0.15 \neq 0.14$ , $A$ and $B$ are NOT independent.
Since $A$ and $B$ are independent events, $P r (A \cap B) = P r (A) \cdot P r (B) = 0.4 \cdot 0.7 = 0.28$ . We can use this information to fill out a Venn diagram and determine that $P r (A^{c} | B) = \frac{P r (A^{c} \cap B)}{P r (B)} = \frac{.42}{.7} = 0.6$ and $P r (A | B^{c}) = \frac{P r (A \cap B^{c})}{P r (B^{c})} = \frac{.12}{.3} = 0.4$ . Another way to do this problem is to realize that since $A$ and $B$ are independent, so are $A^{c}$ and $B$ , and $A$ and $B^{c}$ (and, though it isn't relevant for this problem, $A^{c}$ and $B^{c}$ ). Therefore, for the first part of the problem, we can rewrite the numerator as $P r (A^{c} | B) = \frac{P r (A^{c} \cap B)}{P r (B)} = \frac{P r (A^{c}) \cdot P r (B)}{P r (B)} = P r (A^{c}) = 1 - P r (A) = .6$ . We can do the same thing for the second part of the problem: $P r (A | B^{c}) = \frac{P r (A \cap B^{c})}{P r (B^{c})} = \frac{P r (A) \cdot P r (B^{c})}{P r (B^{c})} = P r (A) = 0.4$ .
We use the conditional probability formula: $P r (F F | at least one F) = \frac{P r (F F \cap at least one F)}{P r (at least one F)} = \frac{P r (F F)}{P r (F F or F S)}$ . The numerator is then $P r (F F) = \frac{C (5, 2)}{C (8, 2)}$ and the denominator is $P r (F F o r F S) = \frac{C (5, 2) + C (5, 1) \cdot C (3, 1)}{C (8, 2)}$ . Combining these, we get that $P r (F F | at least one F) = \frac{C (5, 2)}{C (5, 2) + C (5, 1) \cdot C (3, 1)}$ .

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