3/4 I think, but it also depends on the order of the birth, so it could be 7/8 as well.
Answer:
6 AND 12
Step-by-step explanation:
3 x 2 = 6
3 x 4 = 12
Answer/Step-by-step explanation:
1. The figure is composed of a triangle and a rectangle.
Area of the triangle = ½*base*height
base = 4 ft
height = 12 - 8 = 4ft
Area of triangle = ½*4*4 = 8 ft²
Area of rectangle = length * width
Length = 8 ft
Width = 4 ft
Area of rectangle = 8*4 = 32 ft²
✔️Area of the figure = 8 + 32 = 40 ft²
2. The figure is composed of a semicircle and a triangle
Area of the semicircle = ½(πr²)
radius (r) = 3 cm
π = 3
Area = ½(3*3²) = 13.5 cm²
Area of triangle = ½*base*height
base = 3*2 = 6 cm
height = 6 cm
Area = ½*6*6 = 6 cm²
✔️Area of the figure = 13.5 + 6 = 19.5 cm²
The probability of an event is expressed as

Given:

The probability of drwing two blue balls one after the other is expressed as

For the first draw:

For the second draw, we have only 1 blue ball left out of a total of 6 balls (since a blue ball with drawn earlier).
Thus,

The probability of drawing two blue balls one after the other is evaluted as

The probablity that none of the balls drawn is blue is evaluted as

Hence, the probablity that none of the balls drawn is blue is evaluted as