Question

A man wins a rupee for head and loses a rupee for tail when a coin is tossed. Suppose that he tosses once and quits if he wins but tries once more if he loses on the first toss. Find the probability distribution of the number of rupees the man wins.

Answer 1

Let X be the number of rupees the man wins.

\[\text{ First, let us assume that he gets head in the first toss }  . \]
\[\text{ Probability would be } \frac{1}{2} . \text{ Also, he wins Re } . 1 . \]
\[\text{ The second possibility is that he gets a tail in the first toss } . \]
\[\text{ Then he tosses again } . \]
\[\text{ Suppose he gets ahead in thesecond toss }  . \]
\[\text{ Then, he wins Re 1 in the second toss but loses Re 1 in the first toss } . \]
\[\text{ So, the money won = Rs . 0 } \]
\[\text{ Probability for winning Rs . 0 = Getting tail in } I^{st} \text{ toss x Getting head in }  {II}^{nd} \text{ toss } \]
\[ = \frac{1}{2} \times \frac{1}{2}\]
\[ = \frac{1}{4}\]
\[\text{ The third possibility is getting tail in the}  I^{st} \text{ toss and also tail in the }  {II}^{nd} \text{ toss }  . \]
\[\text{ Then, the money that he would win = - 2 (As he loses Rs }  . 2)\]
\[\text{ Probability for the third possibility } = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\]
\[\text{ Probability distribution of X is as follows } . \]
     X   1   0 - 2
\[P(X) \frac{1}{2} \frac{1}{4} \frac{1}{4}\]