Given that p is the prob. that coin shows a head then 1 – p will be the prob. that coin shows a tail.
Now α = P (A gets the 1st head in 1st try)
⇒ α = P (H) + P (T) P (T) P (T) P (H) + P (T) P (T) P (T) P (T) P (T) P (T) P (H)
= p + (1 – p) 3 p + (1 – p) 6 p+ . . . . . . . . . . . . . . . . . . . . .
= p [1 + (1 – p) 3 (1 – p) 6 + . . . . . . . . . . . . . . . . . . . . . . . ]
= p/1 – (1 – p) 3
NOTE THIS STEP: . . . . . . . . . . . . . . . . . . . . (i)
Similarly β = P (B gets the 1st head in 1st try) + P (B gets the 1st head in 2nd try) + . . . . . . . . . . . . . . . . . .
= P (T) P (H) + p (T) p (T) p (T) p (T) p (T) P (H) + . . . . . . . . . . . . . . . . . . . .
= (1 – p) p + (1 – p) 4 + . . . . . . . . . . . . . . . . . . . .
= (1 – p) p/1 – (1 – p) 3 . . . . . . . . . . . . . . . . . . . . . (ii)
From (i) and (ii) we get β = (1 – p) α
Also (i) and (ii) give expression for α and β in terms of p.
Also α + β + γ = 1(exhaustive events and mutually exclusive events)
⇒ γ = 1 – α – β = 1 - α – (1 – p) α
= 1 – (2 – p) α = 1 – (2 – p) p/1 – (1 - p) 3
= 1 – (1 – p) 3 – (2p – p2) /1 – (1 – p) 3
= 1 – 1 + p3 + 3p (1 – p) – 2p + p2/1 – (1 – p) 3
= p3 – 2p2 + p/1 – (1 – p) 3 = p (p2 – 2p + 1)/1 – (1 – p) 3 = p (1 – p) 2/1 – (1 – p) 3
