Q.A shopkeeper offers a simple game as part of a promotional lottery: a customer pays ₹10 to roll a single fair die once. If the die shows a 6, the customer wins ₹60 (a net gain after the ₹10 fee); otherwise the customer wins nothing (a net loss of the ₹10 fee paid). Let $X$ denote the customer's net gain. Find the probability distribution of $X$, find $E(X)$, and state whether the game is favourable, unfavourable, or fair to the customer.

The customer's net gain takes only two values — a gain if a 6 appears (probability $\tfrac{1}{6}$) or a loss of the entry fee otherwise (probability $\tfrac{5}{6}$) — and $E(X)$ tells us the average result per game if it were played many times. …

Step 1 — identify the values of $X$ and their probabilities.

A fair die has 6 equally likely faces, so $P(\text{6 appears}) = \tfrac{1}{6}$ and $P(\text{6 does not appear}) = \tfrac{5}{6}$.

• If a 6 appears: the customer wins ₹60 but had paid ₹10 to play, so the net gain is $60-10 = 50$.
• If any other face appears: the customer wins nothing, but has still paid the ₹10 fee, so the net gain is $0-10=-10$ (a loss of ₹10).

This gives the probability distribution:

Check: $\tfrac{1}{6}+\tfrac{5}{6}=1$ and both probabilities are non-negative — a valid distribution.

Step 2 — find $E(X)$:

Cross-check using decimals: $50\times0.1\overline{6} = 8.3\overline{3}$ and $-10\times0.8\overline{3}=-8.3\overline{3}$; adding these gives $8.3\overline{3}-8.3\overline{3}=0$, confirming the fractional result exactly. …

Alternatively, compute the expected winnings (before subtracting the fee) as $60\times\tfrac{1}{6}+0\times\tfrac{5}{6}=10$, then subtract the fixed ₹10 fee using the property $E(X)=E(\text{winning …

A common mistake is to compute $E(X)$ using only the ₹60 prize and ₹0 outcomes, forgetting that the ₹10 entry fee must be subtracted from every outcome (since it is paid regardless of the result) to get the true net gain; this w …