Question

A point is selected at random inside the given figure.

Answer 1

P(B) = Area of B/(Area of A + B + C + D)
Area B = 12 in^2
Area A = 35 in^2
Area C = 16 in^2
Area D = 3 x (4+5) = 27 in^2
Hence, P(B) = 12/(12+35+16+27) = 2/15

Answer 2

Answer:  The required probability is $\dfrac{2}{15}.$

Step-by-step explanation:  Given that a point is selected at random from inside the given figure.

We are to find the probability that the point will be in the region labeled B.

From the figure, we note that the regions A, B and D are rectangles and the region C is a square.

The areas of all the regions are calculated as follows:

$\textup{Area of region A}=5\times (3+4)=5\times 7=35~\textup{sq. in.},\\\\\textup{Area of region B}=3\times 4=12~\textup{sq. in.},\\\\\textup{Area of region C}=4^2=16~\textup{sq. in.},\\\\\textup{Area of region D}=3\times(4+5)=3\times 9=27~\textup{sq. in.}$

Therefore, the probability that the randomly chosen point will lie in the region B is given by

$P\\\\\\=\dfrac{\textup{area of region B}}{\textup{area of region A+area of region B+area of region C+area of region D}}\\\\\\=\dfrac{12}{35+12+16+27}\\\\\\=\dfrac{12}{90}\\\\\\=\dfrac{2}{15}.$

Thus, the required probability is $\dfrac{2}{15}.$