Bag 1: 2R, 1B
Bag 2: 1R, 1B, 2Y
Draw 1 from each bag
a) P(same color)
We have 2/3 chance of drawing R from bag 1 and a 1/4 chance of drawing R from bag 2
P(both R) = (2/3)(1/4) = 1/6
We have a 1/3 chance of drawing B from bag 1 and 1/4 chance of B from bag 2
P(both B) = (1/3)(1/4) = 1/12
P(same color) =1/6 + 1/12 = 3/12 = 1/4
Answer: a) 1/4
b) P(exactly one R)
Same story except we look at the complement in bag 2
We have 2/3 chance of drawing R from bag 1 and a 1/4 chance of drawing R from bag 2, so a 3/4 chance of NOT drawing R.
P(1st red, second not red) = (2/3)(3/4) = 1/2
We have 1/3 chance of drawing B from bag 1, so 1/3 chance of drawing NOT red from bag 1. We have a 1/4 chance of drawing R from bag 2, so
P(1st blue, second not blue) = (1/3)(1/4) = 1/12
P = 1/2 + 1/12 = 6/12 + 1/12 = 7/12
Answer: 7/12
Answer:
0, 1
Step-by-step explanation:
The discriminant is b² - 4ac where we have ax² + bx + c, in this case a = -4, b = 12, c = -9 so D = 12² - 4(-4)(-9) = 0.
Because the discriminant is 0 that means that f(x) only has one solution for x meaning that it has only 1 x-intercept.
Answer:
it shows commutative property
Domain of f = R .............
Step-by-step explanation:
