prachi18oct wrote:I did it as follows:-
P(second ball as red) = Prob of first red ball * prob of second red ball + prob of first non-red ball * prob of second ball as red
=> 2/9 * 1/8 + 7/9 * 2/8 = 16/72 = 2/9
Your solution is perfect.
Will the probability of picking the second ball as red not change if the first ball picked is also red?
Yes.
As your solution illustrates, P(RR) ≠P(NR):
P(RR) = (2/9)(1/8) = 2/72.
P(NR) = (7/9)(2/8) = 14/72.
However, the SUM of these probabilities is equal to the probability of selecting a red marble on the first pick:
2/72 + 14/72 = 16/72 = 2/9.
As your solution proves, if we account for ALL of the ways to select a red marble on the NTH pick, the result will be equal to the probability of selecting a red marble on the FIRST pick:
P(red marble on the nth pick) = P(red marble on the 1st pick) = 2/9.
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