GreenlightTestPrep wrote:
A bag contains 1 red chip and 9 white chips only. If the chips are randomly selected one at a time without replacement, what is the probability that the last chip selected is red?
A) \(\frac{1}{10^{10}}\)
B) \(\frac{9!}{10^{10}}\)
C) \(\frac{10!}{10^{10}}\)
D) \(\frac{9^9}{10^{10}}\)
E) \(\frac{1}{10}\)
I created this question to highlight the myth that the probability of randomly selecting a specific object decreases as more objects are selected.
In actuality, P(1st chip is red) = P(2nd chip is red) = P(3rd chip is red) = . . .. = P(9th chip is red) = P(10th chip is red) = \(\frac{1}{10}\)Answer: E
We can also perform the calculation as follows:
P(10th chip is red) = P(1st is white
AND 2nd is white
AND 3rd is white
AND 4th is white
AND.... 8th is white
AND 9th is white
AND 10th is RED)
= P(1st is white)
x P(2nd is white)
x P(3rd is white)
x P(4th is white)
x. . . .
x P(8th is white)
x P(9th is white)
x P(10th is RED)
= 9/10
x 8/9
x 7/8
x 6/7
x 5/6
x 4/5
x 3/4
x 2/3
x 1/2
x 1/1
A lot of numerators and denominators cancel out to get.....= 1/10