A coin is biased so that the probability of obtaining a head is 0.6. ?

Method 1:

P(X = 1) + [P(X = 2) + P(X = 3) + …… + P(X = ∞)] = 1

Since P(X = 1) = 0.6
[P(X = 2) + P(X = 3) + …… + P(X = ∞)] = 1 - 0.6 = 0.4

P(X = 1) > [P(X = 2) + P(X = 3) + …… + P(X = ∞)]

The most likely number of tosses until a head is obtained = 1

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Method 2:

P(head) = 0.6
P(tail) = 1 - 0.6 = 0.4

X = n means the first (n - 1) tosses are all tails and the nth toss is head.
Hence, P(X = n) = 0.4ⁿ⁻¹ × 0.6

Obviously, the greater the value of n, the smaller the P(X = n) is.

Hence, P(X = n) is the greatest when n = 1
In other words,
the most likely number of tosses until a head is obtained = 1

Easy to do, hard to explain. No calculation is required once you understand what’s going on.  The answer must be 1.  See if this helps.

P(head) = 0.6 and P(tail) = 0.4

If you do k tosses, then the probability of k-1 tails followed by a head on the k-th throw is 0.4^(k-1)*0.6.

You want the value of k (the number of tosses) which makes 0.4^(k-1)*0.6 a maximum.  Just think about the possible values of 0.4^(k-1)*06:
k = 1 gives 0.4^(1-1)*0.6 = 1*0.6
k = 2 gives 0.4^(2-1)*0.6 = 0.4*0.6
k = 3 gives 0.4^(3-1)*0.6 = 0.16*0.6
The values continue to decrease as k increases.
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So the largest possible value of 0.4^(k-1)*0.6 is for k=1, one toss.

one toss, probability of that being a head is 0.6

two tosses, first must be a tail, second a head, so probability is 0.4•0.6 = 0.24

three tosses, first two must be a tail, third a head, so probability is 0.4•0.4•0.6 = 0.096

etc, each additional toss lowers the probability, so one toss is most likely.

This is easy.  The sum of the probabilities of all possible X values must be 1, and Pr(X=1) is 0.6.  All of the rest of the probabilities must sum to 1-0.6 = 0.4, so X=1 is clearly the most probable outcome.  That's going to be true whenever Pr(heads) >= 0.5.

It's still true when Pr(heads) < 0.5, too, but that's a bit less obvious.  Suppose Pr(heads) = p, with 0 < p < 1.  Then to get a random value X=n, you need (n-1) consecutive tosses of tails, followed by a toss of heads.  The probability of that is:

    Pr(X=n) = (1-p)^(n-1) * p

Since 0 < (1-p) < 1, then (1-p)^(n-1) < 1 for any n>1.  So:

   Pr(X=1) = p              .... since (1-p)^(1-1)  = (1-p)^0 = 1
   Pr(X=n) < p              .... for all n > 1

...and, again, X=1 is the most probable outcome.