A, B, C are three persons each having 3 coins. A has 3 normal coins, B has 2 normal coins and 1 DH coin and C has one normal coin and 2 DH coins. They toss their coins in order (A–B–C) indefinitely until first one who gets all 3 heads, wins the game. If P(A), P(B) and P(C) are their respective chances of winning the game then

Engineering

Mathematics

Total Probability Theorem and Bayes Theorem

Question

P(C) + P(A) < P(B)

P(A) + P(B) > P(C)

$P (A) = (\frac{8}{43})$

$P (C) = (\frac{22}{43})$

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P (AH) = probability of getting A all 3 heads

$= \frac{1}{8}$

$P (\bar{A} H) = \frac{7}{8}$

similarly

$\Rightarrow P (BH) = \frac{1}{4}, (\bar{B} H) = \frac{3}{4}$

$\Rightarrow p (CH) = \frac{1}{2}, p (C \bar{H}) = \frac{1}{2}$ ]

P(A) → A wins

$P (A) = \frac{1}{8} + \frac{7}{8} \times \frac{3}{4} \times \frac{1}{2} \times \frac{1}{8} + {(\frac{7}{8} \times \frac{3}{4} \times \frac{1}{2})}^{2} \times \frac{1}{8} . ............. = \frac{\frac{1}{8}}{1 - \frac{21}{64}} = \frac{8}{13}$

$P (B) = \frac{7}{8} \times \frac{1}{4} + (\frac{7}{8} \times \frac{3}{4} \times \frac{1}{2}) \times \frac{7}{8} \times \frac{1}{4} + . ....... = \frac{\frac{7}{8} \times \frac{1}{4}}{1 - \frac{21}{64}} = \frac{14}{43}$

$P(C)= \frac{7}{8} \times \frac{3}{4} + \frac{1}{2} + (\frac{7}{8} \times \frac{3}{4} \times \frac{1}{2}) \times \frac{7}{8} \times \frac{3}{4} \times \frac{1}{2} . .........$

$= \frac{\frac{21}{64}}{1 - \frac{21}{64}} = \frac{21}{43}$

P(A) + P(B) + P(C)