Step 1

Let A be the event that THH comes before TTH:

According to law of total probability:

\(\displaystyle{P}{\left\lbrace{A}\right\rbrace}={P}{\left\lbrace{A}{\mid}{H}\right\rbrace}{P}{\left\lbrace{H}\right\rbrace}+{P}{\left\lbrace{A}{\mid}{T}\right\rbrace}{P}{\left\lbrace{T}\right\rbrace}\)

As: \(\displaystyle{P}{\left\lbrace{H}\right\rbrace}={P}{\left\lbrace{T}\right\rbrace}={\frac{{{1}}}{{{2}}}}\)

\(\displaystyle{P}{\left\lbrace{A}\right\rbrace}={\frac{{{1}}}{{{2}}}}{P}{\left\lbrace{A}{\mid}{H}\right\rbrace}+{\frac{{{1}}}{{{2}}}}{P}{\left\lbrace{A}{\mid}{T}\right\rbrace}\)

Also \(\displaystyle{P}{\left\lbrace{A}{\mid}{H}\right\rbrace}={P}{\left\lbrace{A}\right\rbrace}\), Therefore \(\displaystyle{P}{\left\lbrace{A}{\mid}{H}\right\rbrace}={P}{\left\lbrace{A}{\mid}{T}\right\rbrace}\)

\(\displaystyle{P}{\left\lbrace{A}{\mid}{T}\right\rbrace}={P}{\left\lbrace{A}{\mid}\top\right\rbrace}{P}{\left\lbrace{T}\right\rbrace}+{P}{\left\lbrace{A}{\mid}{T}{H}\right\rbrace}{P}{\left\lbrace{T}\right\rbrace}\ldots\ldots..{\left({1}\right)}\)

\(\displaystyle{P}{\left\lbrace{A}{\mid}{T}\right\rbrace}={0}\times{\frac{{{1}}}{{{2}}}}+{P}{\left\lbrace{A}{\mid}{T}{H}\right\rbrace}\times{\frac{{{1}}}{{{2}}}}\)

\(\displaystyle{2}{P}{\left\lbrace{A}{\mid}{T}\right\rbrace}={P}{\left\lbrace{A}{\mid}{T}{H}\right\rbrace}\ldots\ldots\ldots..{\left({2}\right)}\)

Step 2

Now for second condition:

\(\displaystyle{P}{\left\lbrace{A}{\mid}{T}{H}\right\rbrace}={P}{\left\lbrace{A}{\mid}{T}{H}{H}\right\rbrace}{P}{\left\lbrace{H}\right\rbrace}+{P}{\left\lbrace{A}{\mid}{T}{H}{T}\right\rbrace}{P}{\left\lbrace{T}\right\rbrace}\)

\(\displaystyle{P}{\left\lbrace{A}{\mid}{T}{H}\right\rbrace}={1}\times{\frac{{{1}}}{{{2}}}}+{\frac{{{1}}}{{{2}}}}{P}{\left\lbrace{A}{\mid}{T}\right\rbrace}\)

\(\displaystyle{2}{P}{\left\lbrace{A}{\mid}{T}\right\rbrace}={\frac{{{1}}}{{{2}}}}+{\frac{{{1}}}{{{2}}}}{P}{\left\lbrace{A}{\mid}{T}\right\rbrace}\)....From equation (2)

\(\displaystyle{P}{\left\lbrace{A}{\mid}{T}\right\rbrace}={\frac{{{1}}}{{{3}}}}\) and \(\displaystyle{P}{\left\lbrace{A}{\mid}{T}\right\rbrace}={P}{\left({A}\right)}\ldots\ldots{\left({3}\right)}\)

Therefore:

\(\displaystyle{P}{\left({A}\right)}={\frac{{{1}}}{{{3}}}}\)

So, TTH is more likely to appear first and it appears 2/3 of time.

Let A be the event that THH comes before TTH:

According to law of total probability:

\(\displaystyle{P}{\left\lbrace{A}\right\rbrace}={P}{\left\lbrace{A}{\mid}{H}\right\rbrace}{P}{\left\lbrace{H}\right\rbrace}+{P}{\left\lbrace{A}{\mid}{T}\right\rbrace}{P}{\left\lbrace{T}\right\rbrace}\)

As: \(\displaystyle{P}{\left\lbrace{H}\right\rbrace}={P}{\left\lbrace{T}\right\rbrace}={\frac{{{1}}}{{{2}}}}\)

\(\displaystyle{P}{\left\lbrace{A}\right\rbrace}={\frac{{{1}}}{{{2}}}}{P}{\left\lbrace{A}{\mid}{H}\right\rbrace}+{\frac{{{1}}}{{{2}}}}{P}{\left\lbrace{A}{\mid}{T}\right\rbrace}\)

Also \(\displaystyle{P}{\left\lbrace{A}{\mid}{H}\right\rbrace}={P}{\left\lbrace{A}\right\rbrace}\), Therefore \(\displaystyle{P}{\left\lbrace{A}{\mid}{H}\right\rbrace}={P}{\left\lbrace{A}{\mid}{T}\right\rbrace}\)

\(\displaystyle{P}{\left\lbrace{A}{\mid}{T}\right\rbrace}={P}{\left\lbrace{A}{\mid}\top\right\rbrace}{P}{\left\lbrace{T}\right\rbrace}+{P}{\left\lbrace{A}{\mid}{T}{H}\right\rbrace}{P}{\left\lbrace{T}\right\rbrace}\ldots\ldots..{\left({1}\right)}\)

\(\displaystyle{P}{\left\lbrace{A}{\mid}{T}\right\rbrace}={0}\times{\frac{{{1}}}{{{2}}}}+{P}{\left\lbrace{A}{\mid}{T}{H}\right\rbrace}\times{\frac{{{1}}}{{{2}}}}\)

\(\displaystyle{2}{P}{\left\lbrace{A}{\mid}{T}\right\rbrace}={P}{\left\lbrace{A}{\mid}{T}{H}\right\rbrace}\ldots\ldots\ldots..{\left({2}\right)}\)

Step 2

Now for second condition:

\(\displaystyle{P}{\left\lbrace{A}{\mid}{T}{H}\right\rbrace}={P}{\left\lbrace{A}{\mid}{T}{H}{H}\right\rbrace}{P}{\left\lbrace{H}\right\rbrace}+{P}{\left\lbrace{A}{\mid}{T}{H}{T}\right\rbrace}{P}{\left\lbrace{T}\right\rbrace}\)

\(\displaystyle{P}{\left\lbrace{A}{\mid}{T}{H}\right\rbrace}={1}\times{\frac{{{1}}}{{{2}}}}+{\frac{{{1}}}{{{2}}}}{P}{\left\lbrace{A}{\mid}{T}\right\rbrace}\)

\(\displaystyle{2}{P}{\left\lbrace{A}{\mid}{T}\right\rbrace}={\frac{{{1}}}{{{2}}}}+{\frac{{{1}}}{{{2}}}}{P}{\left\lbrace{A}{\mid}{T}\right\rbrace}\)....From equation (2)

\(\displaystyle{P}{\left\lbrace{A}{\mid}{T}\right\rbrace}={\frac{{{1}}}{{{3}}}}\) and \(\displaystyle{P}{\left\lbrace{A}{\mid}{T}\right\rbrace}={P}{\left({A}\right)}\ldots\ldots{\left({3}\right)}\)

Therefore:

\(\displaystyle{P}{\left({A}\right)}={\frac{{{1}}}{{{3}}}}\)

So, TTH is more likely to appear first and it appears 2/3 of time.