It's a question on sample space for dependent event and; as a precursor, I've first used independent events. So prepare for a lengthy question!

Suppose there is an urn with 3 white and 2 red balls. Two balls are picked randomly with replacement. To find out what’s the probability of getting two balls red, I decided to decompose picking of two balls into picking of 1st ball and picking of 2nd ball. I define two events A=the event that 1st ball is red and B=the event that 2nd ball is red. Now (assuming balls are implicitly numbered with color code) I’ve sample space of A and B as $S_A=S_B$={ W1, W2, W3, R1, R2}. The event spaces are $A$={R1, R2} and $B$={R1, R2}. To evaluate desired P(A and B) we can count number of sample points included in $(A\cap B)$ in the Cartesian product space $S_A\times S_B$. This works very smooth for me.

However when the picking of balls is without replacement, to evaluate probability of the same event, by above use of decomposed events A and B, I struggle to decide sample space for $B$. Should $S_B$ include 5 points as any one of the 5 balls in the urn can appear as 2nd ball or should I assume 5x4=20 points in $S_B$ viz. {W1W2, W1W3, ..R2R1} since all outcomes of 2nd ball depend on all outcomes of 1st ball? If I assume former approach where $S_B$ has 5 points, I see I can get a wrong answer for $P(A\cap B)$ as the product space $S_A\times S_B$ should not have $5\times 5=25$ points, rather just $5\times 4=20$ points. On the contrary if I take latter approach where $S_B$ has 20 points, then to solve P(A and B) we should consider the concept of product space is violated and the sample space for $(A\cap B)$ should be $S_B$ (and not $S_A\times S_B$).

Can anyone help me get over this confusion!!

  • $\begingroup$ If you really want to, you can use a $25$ member sample space for the second problem. However, the elements of this sample space will not be equally likely, since for example $(R_1,R_1)$ has probability $0$. I would use the natural $20$-element sample space. $\endgroup$ Dec 17, 2012 at 6:27
  • $\begingroup$ Thanks Andre...am also in favor of 20 elements in $S_B$. $\endgroup$
    – KGhatak
    Dec 20, 2012 at 2:38


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