Select Blue
Select Blue
Suppose there are 2 red balls and 4 blue balls in a bag. What is the probability of selecting 2 blue balls?
let’s first say we are sampling with replacement, meaning that we will draw one ball, replace it and then draw again.
Let X be the number of red balls drawn. X has the binomial distribution with n = 2 trials and success probability p = 1/3, assuming each ball is equally likely to be picked.
In general, if X has the binomial distribution with n trials and a success probability of p then
P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)
for values of x = 0, 1, 2, …, n
P[X = x] = 0 for any other value of x.
this is found by looking at the number of combination of x objects chosen from n objects and then a total of x success and n – x failures.
Or, to be more accurate, the binomial is the sum of n independent and identically distributed Bernoulli trials.
Here you have X ~ Binomial( n = 2, p = 1/3)
P(X = 0 ) = 0.4444444
P(X = 1 ) = 0.4444444
P(X = 2 ) = 0.1111111
———
if you draw without replacement then you have a different situation. Let X be the number of red balls drawn from the bag. X has the hyper geometric distribution.
you have
K = number of items to be drawn = 2
N = total objects = 6
M = number of objects of a given type = 2
You can use the hyper geometric distribution to find the solution
P(X = x | N, M, R) = ( M C x ) * ( (N – M) C (K – x) ) / ( N C K )
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.
P(X = 2) = ( M C x ) * ( (N – M) C (K – x) ) / ( N C K )
= (2 C 2 ) * ( 4 C 0) / (6 C 2)
= 1 * 1 / 15
= 1/15
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