I dont get probability?!
Question: i need someone to check and help me in figuring out how to solve this; i have a test on it =(
a box contains 10 red, 6 green, and 3 blue marbles. Ron chooses 3 marbles from the box
assume he doesnt replace the marbles and fine each probability
a. chooses all 3 red
would this be: P(all 3 red) = (3/19)(2/18)(1/17) ??
b. chooses 3 red or 3 blue marbles
and this one: P(3 red or 3 blue) = (6/19)(5/18)(4/17)
c. chooses at least 1 green marble
no idea
d. chooses one of each color
:(
e. chooses exactly 2 green
helpp :)
Answers: i need someone to check and help me in figuring out how to solve this; i have a test on it =(
a box contains 10 red, 6 green, and 3 blue marbles. Ron chooses 3 marbles from the box
assume he doesnt replace the marbles and fine each probability
a. chooses all 3 red
would this be: P(all 3 red) = (3/19)(2/18)(1/17) ??
b. chooses 3 red or 3 blue marbles
and this one: P(3 red or 3 blue) = (6/19)(5/18)(4/17)
c. chooses at least 1 green marble
no idea
d. chooses one of each color
:(
e. chooses exactly 2 green
helpp :)
There are a total of 19 marbles
a) two ways to solve this:
P(all 3 red) = 10/19 * 9/18 * 8/17 = 0.123839
probability is the number of possible success divided by the possible number of failures.
another way to solve this problem is to use the Hypergeometric distribution.
Let X be the number of red marbles drawn. X has the hypergeometric distribution with the following parameters.
K = number of items to be drawn = 3
N = total objects = 19
M = number of objects of a given type = 10
The probability mass function for the hypergeometric distribution is defined as:
P(X = x | N, M, K) = ( M C x ) * ( (N - M) C (K - x) ) / ( N C K ) for x = {0, ..., K}
P(X = 0 | N, M, K) = 0 otherwise
Note that the constraints on x here are very generic and it is possible to have value of K, N and M such that for x in {0, ..., K} P(X = x) = 0.
If you have n objects and chose r of them, the number of combinations is:
n! / ( r! (n-r)! )
this can be written as nCr
the N C K is the total number of possible combinations of K objects drawn from N objects.
the M C x is the number of combinations of getting x objects of the given type
the (N - M) C ( K - x) is the number of combinations of non typed objects to be drawn.
Looking at the PMF you should be able to see that it is the ratio of the number of combination of selecting the X of the items of interest times the number of combinations of choosing K - X items from the remaining items and this is all divided by the total number of combination for choosing K items from N objects.
The Probability Mass Function, PDF,
f(X) = P(X = x) is:
P(X = 0 ) = 0.0866873
P(X = 1 ) = 0.371517
P(X = 2 ) = 0.4179567
P(X = 3 ) = 0.123839
=== ==== === ===== ==== === === === ==== == == ===
B)
P(3 red) = 0.123838
P(3 blue) = 3/19 * 2/18 * 1/17 = 0.001031992
For any two events A and B, the probability of P(A or B) = P(A U B) is:
P( A U B ) = P(A) + P(B) - P(A ∩ B)
P( 3 red U 3 blue ) = P( 3 red ) + P( 3 blue) - P(3 red ∩ 3 blue)
P( 3 red U 3 blue ) = 0.123838 + 0.001031992 - 0
P( 3 red U 3 blue ) = 0.12487
== == == == == == === == = = === === == === === = = ===
C)
Two ways to do this, the easy way is:
Let X be the number of green marbles:
P( X ≥ 1) = 1 - P(X = 0)
= 1 - (13/19 * 12/18 * 11/17)
= 1 - 0.2951496
= 0.7048504
another way is to use the hypergeometric distribution again. Let X be the number of green marbles.
K = number of items to be drawn = 3
N = total objects = 19
M = number of objects of a given type = 6
P(X = 0 ) = 0.2951496
P(X = 1 ) = 0.4829721
P(X = 2 ) = 0.2012384
P(X = 3 ) = 0.02063983
The Cumulative Distribution Function, CDF,
F(X) = P(X ≤ x) is:
x
∑ P(X = t) =
t = 0
P( X ≤ 0 ) = 0.2951496388028896
P( X ≤ 1 ) = 0.7781217750258
P( X ≤ 2 ) = 0.979360165118679
P( X ≤ 3 ) = 1
1 - F(X) is:
K
∑ P(X = t) =
t = x
P( X ≥ 0 ) = 1
P( X ≥ 1 ) = 0.70485036119711
P( X ≥ 2 ) = 0.2218782249742002
P( X ≥ 3 ) = 0.02063983488132093
== == == == ==== === ==== === == === == = ==== ===
D)
there are 3! = 6 ways to draw one of each color
RGB, RBG, BRG, BGR, GRB, GBR
each is equal probability.
P(RGB) = 10/19 * 6/18 * 3/17
= (10 * 6 * 3) / (19 * 18 * 17)
= 0.03095975
that is the probability for one permutation. the total probabilty for drawing one of marble of each color is:
6 * 0.03095975 = 0.1857585
== === == === === ===
e)
see second part to answer C
P(X = 2 ) = 0.2012384
HAHAHA...a good joke.
im laughing on the inside
yea i like it hahahaha.
yeah really wrong place its so hard we think its a joke, go to the math section thats where all the math nerds are they can help you lol!!!
P.S what grade (how old) r u cause that looks VERY hard unless it really is a joke...
I dont know if Merlyn is right or not, but if they are right they deserve best answer. And if they are wrong, they deserve best answer for that much BS in one answer.