In the 2nd problem there are 32 (2^5) elements in the outcome set, not 6.

Let's look at a simpler variant of this problem. If you were tossing a coin 3 times, there would be 8 (2^3) elements in the outcome set:

O = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.

Then

P(at most half the coin tosses are heads)

= P(exactly 0 heads) + P(exactly 1 head)

= 1/8 + 3/8

= 1/2

i) For tossing a coin 5 times, the outcome set starts like this:

O = {HHHHH, HHHHT, HHHTH, HHHTT...}

Then

P(at most half the coin tosses are heads)

= P(exactly 0 heads) + P(exactly 1 head) + P(exactly 2 heads)

= ?/32 + ?/32 + ?/32

...

Can you take it from here?

ii) This is the easier one. The answer would be 1/32.

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