P(B)= 9/20 P(A and B)= 9/100 P(A)=?

Apr 12, 2018

$P \left(A\right) = \frac{1}{5}$

Explanation:

Assuming events $A$ and $B$ are independent, we use

$\text{P"(A nn B) = "P"(A) * "P} \left(B\right)$

$\text{ "9/100 = "P} \left(A\right) \cdot \frac{9}{20}$

$\frac{20}{9} \cdot \frac{9}{100} = \text{P} \left(A\right)$

$\text{ "20/100="P} \left(A\right)$

$\text{ "1/5="P} \left(A\right)$

Apr 30, 2018

$P \left(A\right) = \frac{1}{5}$

Explanation:

multiplication rule for probabilities:

$P \left(A \mathmr{and} B\right) = P \left(A\right) \cdot P \left(B\right)$

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$P \left(B\right) = \frac{9}{20}$

$P \left(A \mathmr{and} B\right) = \frac{9}{100}$

$P \left(A\right) = \frac{P \left(A \mathmr{and} B\right)}{P \left(B\right)}$

$\frac{P \left(A \mathmr{and} B\right)}{P \left(B\right)} = \frac{9}{100} \div \frac{9}{20}$

$= \frac{9}{100} \cdot \frac{20}{9}$

$= \frac{\left(\cancel{9}\right) 1}{\left(\cancel{100}\right) 5} \cdot \frac{\left(\cancel{20}\right) 1}{\left(\cancel{9}\right) 1}$

$= \frac{1 \cdot 1}{5 \cdot 1}$

$= \frac{1}{5}$

hence, $P \left(A\right) = \frac{1}{5}$.

this can be checked by applying the multiplication rule again:

$P \left(A\right) \cdot P \left(B\right) = \frac{1}{5} \cdot \frac{9}{20} = \frac{1 \cdot 9}{5 \cdot 20} = \frac{9}{100}$

$P \left(A \mathmr{and} B\right) = \frac{9}{100}$