The key to this is to understand that in half of all cases where two aces were drawn, the second random step will pick the heart 1⁄2 the time.
Thus—the first step selects 5 out of 6 possible outcomes. In the second step, 4 of the 5 cases have fixed outcomes; the other (Aa) has two posssible outcomes.
Thus there are six possible final states:
1 a D showing a (1/5)
2 a d showing a (1/5)
3 A D showing A (1/5)
4 A d showing A (1/5)
6 A a showing a (1/10)
6 A a showing A (1/10)
The answer to the second question is yes in cases 2, 3, and 6. The collective probability of these cases is 1⁄5 + 1⁄5 + 1⁄10 = 1⁄2.
The key to this is to understand that in half of all cases where two aces were drawn, the second random step will pick the heart 1⁄2 the time.
Thus—the first step selects 5 out of 6 possible outcomes. In the second step, 4 of the 5 cases have fixed outcomes; the other (Aa) has two posssible outcomes.
Thus there are six possible final states:
1 a D showing a (1/5) 2 a d showing a (1/5) 3 A D showing A (1/5) 4 A d showing A (1/5) 6 A a showing a (1/10) 6 A a showing A (1/10)
The answer to the second question is yes in cases 2, 3, and 6. The collective probability of these cases is 1⁄5 + 1⁄5 + 1⁄10 = 1⁄2.
1⁄10 over 1⁄2 is 1⁄5.