INTRO
Total number of red and white balls = 100
Number of red balls = 92
Number of white balls = 8
Question A - Express a probability for white balls
A) My answer:
Pr (of white balls) = 0.08
My calculations:
Pr (red balls) = 0.92 = 92 % , 23/25 = 92/100
Pr (white balls) = 0.08 = 8 % , 2/25 = 8/100
Question B - Express a distribution of an random variable X X = number of balls (Red & White) n=1,2,3,4,5 Use the probability you from question 1
B) My thinking :
Red balls = x1 White balls = x2
X = (x1+ x2)
Question C
Calculate the probability from the following events:
C1 - Will find precisely 2 red balls within the 5 selected? Yes/No, explain your reasoning!
C2 - You will find at least 2 red balls within the 5 selected? Yes/No, explain your reasoning!
My thinking B:
So say I use this equation:
Pr(X=x) :nCr (n,x) p^x (1-p)^(n-x)
But if I am to take 1 ball from the first try, there will be 99 balls let in the second try and 98 balls in the third try and so on… in the 10:th try there will be 90 balls left which means that the entire probability of balls would change after each try, no matter the colour... Is there a formula that take THIS in account? if not, do you know which formula works in this case?
Help on this one would mean ALOT...
Part A) Your answer is correct.
Part B) I'm not entirely sure what's being asked, so please keep that in mind with my response.
n = 1 ball drawn:
P(r) = 92%
P(w) = 8%
n = 2 balls drawn:
P(r,r) = ${\frac{92}{100}}$ $\cdot$ ${\frac{91}{99}}$
P(w,w) = ${\frac{8}{100}}$ $\cdot$ ${\frac{7}{99}}$
P (r,w) = ${\frac{92}{100}}$ $\cdot$ ${\frac{8}{99}}$
Note that red then white is no different than white followed by red. This is explained better in the explanation for Part C1.
n = 3 balls drawn:
P(r,r,r) = ${\frac{92}{100}}$ $\cdot$ ${\frac{91}{99}}$ $\cdot$ ${\frac{90}{98}}$
P(w,w,w) = ${\frac{8}{100}}$ $\cdot$ ${\frac{7}{99}}$ $\cdot$ ${\frac{6}{98}}$
P(r,w,w) = ${\frac{92}{100}}$ $\cdot$ ${\frac{8}{99}}$ $\cdot$ ${\frac{7}{98}}$
P(r,r,w) = ${\frac{92}{100}}$ $\cdot$ ${\frac{91}{99}}$ $\cdot$ ${\frac{8}{98}}$
Hopefully you can see how this continues.
Part C1) No
Probability the first ball is red: ${\frac{92}{100}}$
Probability the second ball is red: ${\frac{91}{99}}$
Probability the third ball is white: ${\frac{8}{98}}$
Probability the fourth ball is white: ${\frac{7}{97}}$
Probability the fifth ball is white: ${\frac{6}{96}}$
To find the total probability of all these events happening, multiply each individual probability.
$\frac{(92\cdot91\cdot8\cdot7\cdot6)}{(100\cdot99\cdot98\cdot97\cdot96)}$ = ${\frac{2812992}{9034502400}}$ = $\frac{43953}{141164100}$ = 0.031136%
As you can see, this is extremely low. Please note that the order of the balls drawn does not matter. The numerator is always going to be $(r)$$(r-1)$$(w)$$(w-1)$$(w-2)$ where r represents the number of red balls and w represents the number of white balls.
The denominator will always be $(t)$$(t-1)$$(t-2)$$(t-3)$$(t-4)$ where t is the total number of balls.
Part C2) Yes
This problem is better to approach by finding the probability that the opposite will happen and subtracting that probability from 1.
What is the probability of exactly one red ball? Use the same method as part C1.
$\frac{8}{100}$$\cdot$$\frac{7}{99}$$\cdot$$\frac{6}{98}$$\cdot$$\frac{5}{97}$$\cdot$$\frac{92}{96}$ = 0.001710774907%
What is the probability of exactly zero red balls? Use the same method as part C1.
$\frac{8}{100}$$\cdot$$\frac{7}{99}$$\cdot$$\frac{6}{98}$$\cdot$$\frac{5}{97}$$\cdot$$\frac{4}{96}$ = 0.00007438151768%
Add these probabilities and subtract the total from 1.
$1-(1.785*10^{-5})$ or (100% - 0.001785%) = 99.9982%