a bag contains red and blue marbles. with 10% more blue marbles than red. a)what is the probability of selecting 5 blue marbels ? b) Of selecting 2 reds and 3 blues ?

"with 10% more blue marbles than red" could mean one of two things:

1. The percentages are: blue 55%, red 45%,
2. They are in the ratio blue:red = 11:10

I'll take the first interpretation. The procedure is the same for the second, except for the numbers.

I will further assume that the number of marbles is huge, so that picking 5 blue dos not affect the distribution remaining in the bag, or there was replacement.

1. probability of picking 1 blue: 0.55
probability of picking n blues: 0.55^n.

2. answer is represented by the sum of binomial expansion of r^2b^3 in (r+b)^5.
(from 1 5 10 10r²b³ 5 1) where r=0.45, b=0.55
P(2r3b)=10*0.45²0.55³=0.3369 (approx.)

Probability of

Your answer agrees with my homework.

BINOMIAL EXPANSION... this is what I know about it. (a+b)^n

Can you explain your steps a little more? I don't understand (r^2b^3):r to the power of 2b times 2b to the power of 3...

Thanks

Binomial expansion is typically finding the coefficients of the expansion of

(a+b)^n.
In this particular case, a+b=1, so 1^n is still n.
The individual terms therefore represent the respective probabilities of the combination of outcomes, 2R+3B, 0R+5B, etc. The beauty of this is that the probabilities of all the outcomes add up to 1, which is what it should be.

For the case of (r+b)^5 (where r+b=1), we have algebraically:
(r+b)^5 = r^5+5br^4+10b^2r^3+10b^3r^2+5b^4r+b^5
So by evaluating each term, we can get the probabilities of all possible outcomes. r^5 (0.45^4) represents the probability of drawing 5 reds. r^4b represents the probability of drawing 4 reds and 1 blue (0.45^4*0.55), etc.

Note that the answer for the case of 5 blues (b^5) is tucked at the end of the expression (b^5).

A continuation to above problem is this:

c)Selecting at least 3 reds?
d)Selecting Fewer than 4 reds?

c)Selecting at least 3 reds?

r^5+5br^4+10b^2r^3

d)Selecting Fewer than 4 reds?
10b^2r^3+10b^3r^2+5b^4r+b^5

To find the probability of selecting a certain number of marbles of a specific color from the bag, we need to know the total number of marbles in the bag. Unfortunately, the information provided does not specify the total number of marbles.

To solve these problems, we'll need to make some assumptions to proceed. Let's assume that there are a total of 100 marbles in the bag (50 red and 50 blue marbles). With this assumption, we can now calculate the probabilities.

a) Probability of selecting 5 blue marbles:
Given that there are 10% more blue marbles than red marbles, we can calculate the number of blue marbles in the bag:
Number of blue marbles = 1.1 * Number of red marbles = 1.1 * 50 = 55

Now, we need to find the probability of selecting 5 blue marbles out of the total blue marbles.

Probability of selecting 5 blue marbles = (number of ways to select 5 blue marbles) / (total number of ways to select any 5 marbles)

Using the concept of combinations, the number of ways to select 5 blue marbles out of 55 is:
55C5 = (55!)/(5!(55-5)!) = 341,055

The total number of ways to select any 5 marbles out of 100 is:
100C5 = (100!)/(5!(100-5)!) = 75,287,520

Therefore, the probability of selecting 5 blue marbles is:
341,055 / 75,287,520 ≈ 0.0045 or 0.45%

b) Probability of selecting 2 reds and 3 blues:
Using the same reasoning, we can find the probability of selecting 2 red marbles and 3 blue marbles.

Number of ways to select 2 red marbles out of 50:
50C2 = (50!)/(2!(50-2)!) = 1,225

Number of ways to select 3 blue marbles out of 55:
55C3 = (55!)/(3!(55-3)!) = 21,060

The total number of ways to select any 5 marbles out of 100 is still 75,287,520.

Therefore, the probability of selecting 2 red marbles and 3 blue marbles is:
(1,225 * 21,060) / 75,287,520 ≈ 0.3454 or 34.54%