The first two cases are without replacement and the next two cases are with replacement.
Case 1:
Let’s say there is an urn with 5 red balls and 50 green balls. What is the probability of picking one red ball from this urn?
This one is easy – there are 5 red balls and the total number of balls are 55 – so, 5/55 = 1/11 = 9%
Case 2:
Now, let’s say instead of taking one ball you take two balls at once. What is the probability that both the balls are red?
i). Taking two balls at once is the same as taking one ball first and without replacing it taking another ball again. So, the probability for the first ball to be red is 5/55 and the probability for the second ball to be red is 4/54. Therefore, it is (5/55)*(4/54) = 0.0067 = 0.6%
ii). The probability of two being balls being red is 5c2/55c2
iii). The other way of looking at it is: the probability of the first ball being red p(a) is 5/55, and that is straightforward. The probability of the second ball being red given that the first ball is red will follow the simple conditional probability below.
p(b | a) = p(a and b)/p(a)
p(red ball | first ball red) = p(a)p(b)/p(a) = p(b)
This means a and b are independent events.
Total probability is p(a)*p(b) = (5/55)*(4/54)
Case 3:
Now, let’s say you take one ball first and then replace it and then take a second ball. What is the probability of taking two red balls?
The probability of picking the first red ball is 5/55 and the probability of picking the second ball is again 5/55. So the answer is (5/55)*(5/55)
Case 4:
Now, let’s say you take one ball first and then replace it and then take a second ball. What is the probability of taking one red ball and one green ball?
Since the order is not mentioned, it could be red first or red second.
The probability of picking red ball first and green ball second is: (5/55)*(50/55)
The probability of picking green ball first and red ball second is: (50/55)*(5/55)
Probability of picking one red ball and one green ball is: (5/55)*(50/55) + (50/55)*(5/55)
Hope this is useful, thank you.
