Number of outcomes is all that matters

The 1st Case is true but in the 2nd case, there are 4 outcomes​ i.e. HH, HT, TH, TT. so there is 1/4 chances to get HH or TT and 1/2 chance to get one head and one tail. So 2nd case is false

You could get two heads, two tails, first one head then a tail, or first a tail then a head. 2 ways out of 4 ways is a $\boxed{\dfrac{1}{2}}$ chance of getting a tail and a head.

Let denote head by H and tail by T then possible outcomes of tossing a fair coin twice are

HH, TT, HT or TH

Hence the probability of obtaining HT is $\boxed{1/2}$

There are not one, but two possibilities for a head and a tail because we have two coins, meaning we can choose either one to be the head (which equivalently means the other coin would be the tail). So in all, we have four possibilities, not three. The rest of the result follows easily.

The first toss becomes a given (heads or tails). Then we toss again to get the opposite (tails or heads) for which the probability is 50%.

There exists the possibility, however small, that any tossed coin may land on its side.

Regardless, the second ignores one possibility (a H/T combination)

• coin 1:H coin2:H
• coin 1:H coin2:T
• coin 1:T coin2:H
• coin 1:T coin2:T

Each of the four are "equally likely" and two of the four have 1H 1T so probability is still 1/2.

You may get TH or HT so this makes the probability 2/4...or 1/2

$P(H \land T) = P(H) \times P(T) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ , where H and T are mutually exclusive.

It just didn't​ seem right to me.

Keep it simple: It is irrelevant whether the first toss is a head or a tail, the chance of getting the opposite from the second toss is simple - 50% ie: 1 in 2 !

James,

Vancouver BC

If you consider the possibility of the coin landing on its rim, then all the cases are false