A bowl contains 9 red balls and 9 blue balls. A woman selects balls at random without looking at them. (a) How many balls must she select (minimum) to be sure of having at least three blue balls?
(b) How many balls must she select (minimum) to be sure of having at least three balls of the same color? (1 point)
A family has 3 children. Assume that each child is as likely to be a boy as it is to be a girl.
Find the probability that the family has 3 girls if it is known the family has at least one girl.

I'll do the first problem to get you started

Part A

There are 9 red and 9 blue.

The worst case scenario is that she selects 3 red in a row. This means that to be guaranteed to have 3 blue, she must select 3+3 = 6 balls.

It is possible that the first three could be blue, but it likely won't be the case. It's best to plan for the worst case scenario.

Part B

The worst case scenario is that she pulls out say a red ball, then a blue, then red, and so on. The alternating color selections mean that we don't have three balls of the same color until we get to selection 5.

We could write the sequence like this

RB RB B

or like this

RB RB R

and we see that regardless of what the fifth selection is, we are guaranteed to have three of the same color. The first four items don't really matter. We could have RB changed to BR and it would be the same idea.

1

step-by-step explanation:

it's the only multiple of both and the least

y=1/4x-4

step-by-step explanation: