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Suppose that $$P(A | B)=0.4 \quad$$ and $$\quad P(B)=0.5$$
Determine the following:

(a) $$P(A \cap B)$$
(b) $$P\left(A^{\prime} \cap B\right)$$

$$ A, B$$
$$P(A | B)=0.4$$
$$P(B)=0.5$$

(a) $$P(A \cap B)=P(A | B) P(B)=0.4 \times 0.5=0.2$$

(b) $$P\left(A^{\prime} \cap B\right)=P(A^\prime | B) P(B)$$

$$P(A^\prime | B)=1-P(A | B)=1-0.4=0.6$$

$$P\left(A^{\prime} \cap B\right)=0.6 \times 0.5=0.3$$

Suppose that $$P(A | B)=0.2, P\left(A | B^{\prime}\right)=0.3,$$ and
$$P(B)=0.8 .$$ What is $$P(A) ?$$

$$P(A | B)=0.2$$
$$P\left(A | B^{\prime}\right)=0.3$$

$$P(B)=0.8$$
$$P(A) ? ?$$

$$P(A)=P(A \cap B)+P\left(A \cap B^{\prime}\right)$$

$$P(A \cap B)=P(A | B) \times P(B)$$

$$=0.2 \times 0.8=0.16$$

$$P(A \cap B^\prime)=P\left(A | B^{\prime}\right) \cdot P(B^{\prime})$$

$$P(B^{\prime})=1-P(B)=1-0.8=0.2$$

$$P(A \cap B^\prime)=0.3 \times 0.2=0.06$$

$$P(A)=0,16+0.06=0.22$$

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