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