Probability 4 Question 22
22. Three players, $A, B$ and $C$, toss a coin cyclically in that order (i.e. $A, B, C, A, B, C, A, B, \ldots$ ) till a head shows. Let $p$ be the probability that the coin shows a head. Let $\alpha, \beta$ and $\gamma$ be, respectively, the probabilities that $A, B$ and $C$ gets the first head. Prove that $\beta=(1-p) \alpha$. Determine $\alpha, \beta$ and $\gamma$ (in terms of $p$ ).
$(1998,8$ M)
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Answer:
Correct Answer: 22. $\frac{23}{30}$
Solution:
- Let $q=1-p=$ probability of getting the tail. We have,
$\alpha=$ probability of $A$ getting the head on tossing firstly
$=P\left(H _1\right.$ or $T _1 T _2 T _3 H _4$ or $T _1 T _2 T _3 T _4 T _5 T _6 H _7$ or $\left.\ldots\right)$
$=P(H)+P(H) P(T)^{3}+P(H) P(T)^{6}+\ldots$
$=\frac{P(H)}{1-P(T)^{3}}=\frac{p}{1-q^{3}}$
Also,
$\beta=$ probability of $B$ getting the head on tossing secondly
$$ \begin{aligned} & =P\left(T _1 H _2 \text { or } T _1 T _2 T _3 T _4 H _5 \text { or } T _1 T _2 T _3 T _4 T _5 T _6 T _7 H _8 \text { or } \ldots\right) \\ & =P(H)\left[P(T)+P(H) P(T)^{4}+P(H) P(T)^{7}+\ldots\right] \\ & =P(T)\left[P(H)+P(H) P(T)^{3}+P(H) P(T)^{6}+\ldots\right] \\ & =q \alpha=(1-p) \alpha=\frac{p(1-p)}{1-q^{3}} \end{aligned} $$
Again, we have
$$ \begin{aligned} \alpha+\beta+\gamma & =1 \\ \gamma & =1-(\alpha+\beta)=1-\frac{p+p(1-p)}{1-q^{3}} \\ & =1-\frac{p+p(1-p)}{1-(1-p)^{3}} \\ & =\frac{1-(1-p)^{3}-p-p(1-p)}{1-(1-p)^{3}} \\ \gamma & =\frac{1-(1-p)^{3}-2 p+p^{2}}{1-(1-p)^{3}}=\frac{p-2 p^{2}+p^{3}}{1-(1-p)^{3}} \end{aligned} $$
Also,
$$ \alpha=\frac{p}{1-(1-p)^{3}}, \beta=\frac{p(1-p)}{1-(1-p)^{3}} $$
