Flipping a biased coin

Question

A bias coin is flipped four times. If the probability of ‘tails’ occurring is twice the probability of ‘heads’ occurring, find the probability that 3 ‘heads’ occur.

Answer 1

P(T) =2P(H)

P(TvH) = 1

P(H)= 1/3 = about .33

P(T) =2/3 = about .67

3 heads can occur in 4 flips 4 different ways

HTTT, THTT, TTHT or TTTH

each time P(3T and 1H in any order) = (.67)(.33)^3

multiply by 4C3 = 4!/3!1! = 4

P(3H and 1T) = 4C3(P(T)(P(H))^3

P(3H^1T) = 4(2/3)(1/3)^3

= 8/81 = about 0.098765431

= about 9.9% chance of 3 Tails and 1 Head

P(0H) = 4C0(2/3)^4(1/3)^0 = 16/81

P(1H) = 4C1(2/3)^3(1/3)^1= 32/81

P(2H) = 4C2(2/3)^2(1/3)^2 = 24/81

P(3H) = 4C3(2/3)^1(1/3)^3= 8/81

P(4H) = 4C4(2/3)^0(1/3)^4= 1/3^4 = 1/81

16/81+32/81+24/81+8/81+1/81 = 81/81 =1 = 100%

19.8%+39.5% + 29.6% + 9.9% + 1.2% = about 100%