krusta80 wrote:
The probability of selecting X on the Nth pick is equal to the probability of selecting X on the FIRST pick. <--- ONLY APPLIES TO SELECTION WITH REPLACEMENT
The rule does not require that selections be made with replacement.
Please see my post above, in which my calculations assume no replacement.
Given 3 red marbles and 5 blue marbles, P(R) on any given pick (with or without replacement) = 3/8.
In a box with 10 blocks, 3 of which are red, what is probability of picking out a red block on each of your first two tries? Consider no replacement.
Solution :
P(R) on first pick - 3/10
P(R) on second pick - 2/9 (DOMINO EFFECT - If red block is chosen on first pick, then the number of blocks in box reduced to 9 from 10. Additionally, the number of red blocks now decreased to 2 from 3. So P(R) on second pick is different from P(R) on first pick. )
Therefore, P(R) on both picks is 3/10 x 2/9 = 6/90= 1/15.
The solution above determines the probability that BOTH blocks are red.
Thus, there are TWO events that must happen: red on the first pick AND red on the second pick.
As the solution shows, the probability that BOTH of these events happen = 1/15.
The probability of selecting X on the Nth pick is equal to the probability of selecting X on the FIRST pick.
This rule applies to ONE EVENT ONLY: the probability of selecting X on any given pick.
It does not apply to the block problem, which asks for the probability of TWO EVENTS: red on the first pick AND red on the second pick.
While the probability of selecting red on BOTH picks is 1/15, the probability of selecting red on the SECOND pick -- with no restrictions on the first pick -- is greater.
The reason is that there is MORE THAN ONE WAY to get red on the second pick:
Case 1: First block is not red, second block is red
P(first block is not red) = 7/10. (Of the 10 blocks, 7 are not red.)
P(second block is red) = 3/9. (Of the 9 remaining blocks, 3 are red).
Since we want both of these events to happen, we multiply the fractions:
7/10 * 3/9 = 7/30.
Case 2: Both blocks are red
As we saw above, P(both blocks are red) = 1/15.
Since each case is a way to get red on the second pick, we add the fractions:
P(red on the second pick) = 7/30 + 1/15 = 9/30 = 3/10.
Notice that the calculations for each case take into account the "domino" effect.
Notice also that -- when all of the different ways to get red on the second pick are considered -- the probability of selecting red on the second pick, WITHOUT replacement, is equal to the probability of selecting red on the FIRST pick: 3/10.
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