Let the event be:

• $L_S$: the lost card is a spade.
• $A$: the $n$ drawn cards are all spades.

We are given: $P(L_S\mid A)=\frac{11}{50}$

Use Bayes' theorem: $P(L_S\mid A)=\frac{P(A\mid L_S)P(L_S)}{P(A)}$

Compute the prior probability: Since one card is lost from a full deck of 52 cards, $P(L_S)=\frac{13}{52}=\frac14$ and $P(L_{\bar S})=\frac{39}{52}=\frac34$

Compute $P(A\mid L_S)$: If the lost card is a spade, then among the remaining 51 cards there are only 12 spades. So the probability that all $n$ drawn cards are spades is $P(A\mid L_S)=\frac{\binom{12}{n}}{\binom{51}{n}}$

Compute $P(A\mid L_{\bar S})$: If the lost card is not a spade, then all 13 spades are still present among the 51 cards. Hence $P(A\mid L_{\bar S})=\frac{\binom{13}{n}}{\binom{51}{n}}$

Now compute total probability $P(A)$: $P(A)=P(A\mid L_S)P(L_S)+P(A\mid L_{\bar S})P(L_{\bar S})$ $P(A)=\frac{\binom{12}{n}}{\binom{51}{n}}\cdot \frac14+\frac{\binom{13}{n}}{\binom{51}{n}}\cdot \frac34$

Substitute into Bayes' formula: $\frac{\frac{\binom{12}{n}}{\binom{51}{n}}\cdot \frac14}{\frac{\binom{12}{n}}{\binom{51}{n}}\cdot \frac14+\frac{\binom{13}{n}}{\binom{51}{n}}\cdot \frac34}=\frac{11}{50}$

Cancel $\binom{51}{n}$: $\frac{\binom{12}{n}\cdot \frac14}{\binom{12}{n}\cdot \frac14+\binom{13}{n}\cdot \frac34}=\frac{11}{50}$

Multiply numerator and denominator by 4: $\frac{\binom{12}{n}}{\binom{12}{n}+3\binom{13}{n}}=\frac{11}{50}$

Use the identity $\binom{13}{n}=\frac{13}{13-n}\binom{12}{n}$

So $\frac{\binom{12}{n}}{\binom{12}{n}+3\cdot \frac{13}{13-n}\binom{12}{n}}=\frac{11}{50}$

Cancel $\binom{12}{n}$: $\frac{1}{1+\frac{39}{13-n}}=\frac{11}{50}$

Simplify: $\frac{13-n}{52-n}=\frac{11}{50}$

**Solve for $n$: ** $50(13-n)=11(52-n)$ $650-50n=572-11n$ $78=39n$ $n=2$

Final answer: $\boxed{2}$

Comparison with stored answer: The derived answer is $2$, which matches the stored correct answer.