conditional probability quesiton?

NOTE:

It's not true that knowing number of heads obtained will not affect probability that 1st toss is heads (as mentioned previously).

To understand this, let's look at extreme cases:
When r = 0 (0 heads were obtained), then P(1st toss = heads) = 0
When r = n (all tosses were heads), the P(1st toss = heads) = 1
The more head are obtained, then the greater the probability that any one toss (including 1st toss) will be heads.

1.

P(head on 1st toss | r heads)
= P(head on 1st toss and r heads) / P(r heads)
= P(head on 1st toss and (r−1) heads on remaining (n−1) tosses) / P(r heads in n tosses)
= [1/2 * C(n−1,r−1)/2^(n−1)] / [C(n,r)/2^n]
= C(n−1,r−1) / C(n,r)
= (n−1)!/((r−1)!(n−r)!) * (r!(n−r)!)/n!
= r/n

Here's an example:

n = 3 tosses
r = 2 heads
Sample space:
H H T
H T H
T H H

3 outcomes, 2 with Heads in 1st toss.
P(Head on 1st toss | 2 heads) = r/n = 2/3

2.

Assuming we are as likely to choose patch 1 as we are to choose patch 2, then:

P(patch is occupied)
= P(patch 1 is chosen and occupied) + P(patch 2 is chosen and occupied)
= (1/2 * 1/5) + (1/2 * 1/8)
= 1/10 + 1/16
= 13/80

P(patch 2 chosen | patch is occupied)
= P(patch 2 chosen and occupied) / P(patch is occupied)
= (1/16) / (13/80)
= 5/13