To solve for the missing probability [tex]\( P(B) \)[/tex], given [tex]\( P(A) = \frac{7}{20} \)[/tex] and [tex]\( P(A \cap B) = \frac{49}{400} \)[/tex], we'll use the concept of conditional probability and the formula for the intersection of two events.
The formula for the intersection of two events [tex]\( A \)[/tex] and [tex]\( B \)[/tex] is:
[tex]\[
P(A \cap B) = P(A) \cdot P(B|A)
\][/tex]
However, if we isolate [tex]\( P(B) \)[/tex] on the right-hand side, we get:
[tex]\[
P(A \cap B) = P(A) \cdot P(B)
\][/tex]
since [tex]\( P(B|A) \cdot P(A) \)[/tex] is [tex]\( P(A \cap B) \)[/tex].
Rearranging this formula to solve for [tex]\( P(B) \)[/tex] gives:
[tex]\[
P(B) = \frac{P(A \cap B)}{P(A)}
\][/tex]
Substitute the given values into this formula:
[tex]\[
P(B) = \frac{\frac{49}{400}}{\frac{7}{20}}
\][/tex]
To simplify this, we need to divide [tex]\( \frac{49}{400} \)[/tex] by [tex]\( \frac{7}{20} \)[/tex]. Dividing fractions is the same as multiplying by the reciprocal:
[tex]\[
P(B) = \frac{49}{400} \times \frac{20}{7}
\][/tex]
Simplify the multiplication:
[tex]\[
P(B) = \frac{49 \cdot 20}{400 \cdot 7} = \frac{980}{2800}
\][/tex]
Now, simplify [tex]\( \frac{980}{2800} \)[/tex]:
[tex]\[
P(B) = \frac{49}{140} = \frac{7}{20}
\][/tex]
Checking for mistakes, realizing actually miscalculated the simple product, re-evaluating:
[tex]\[
P(B) = \frac{49/400}{7/20} = \frac{49}{400} \times \frac{20}{7} = 7 \frac{10}{400} = \frac{49}{1400}
\][/tex]
Reducible gives [tex]\(
P(B) = \frac{49}{1400} =\frac{7}{20}
\)[/tex]
Thus, the probability [tex]\( P(B) \)[/tex] is correctly revaluating showing:
Finally,
Thus the correct option is:
[tex]\[
\boxed{\frac{7}{10}}
\][/tex]
