Exams Joint Entrance Examination Mathematics Probability Question 79 Question 79 of 417 Medium NUMERICAL
A card from a pack of 52 cards is lost. From the remaining 51 cards, n cards are drawn and are found to be spades. If the probability of the lost card to be a spade is 11 50 \frac{11}{50} 50 11 , then n is equal to \_\_\_\_\_\_\_\_ .
Answer: 2
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P ( ) n ) P(\frac{{}_{)}}{n}) P ( n ) )
= P ( n s L s ) P ( L s ) P ( n s L s ) P ( L s ) + P ( n s L ⃗ s ) P ( L ⃗ s ) = 12 C n 51 C n × 1 4 12 C n 51 C n × 1 4 + 3 4 × 13 C n 51 C n = 1 1 + 3 ⋅ 13 C n 12 C n = 13 − n 52 − n ⇒ 13 − n 52 − n = 11 50 ⇒ n = 2 \begin{gathered} =\frac{P(\frac{{n}_{s}}{{L}_{s}})P({L}_{s})}{P(\frac{{n}_{s}}{{L}_{s}})P({L}_{s})+P(\frac{{n}_{s}}{{\vec{L}}_{s}})P({\vec{L}}_{s})} \\ =\frac{\frac{{}^{12}{C}_{n}}{{}^{51}{C}_{n}}\times \frac{1}{4}}{\frac{{}^{12}{C}_{n}}{{}^{51}{C}_{n}}\times \frac{1}{4}+\frac{3}{4}\times \frac{{}^{13}{C}_{n}}{{}^{51}{C}_{n}}}=\frac{1}{1+3\cdot \frac{{}^{13}{C}_{n}}{{}^{12}{C}_{n}}}=\frac{13-n}{52-n} \\ \Rightarrow \frac{13-n}{52-n}=\frac{11}{50} \\ \Rightarrow n=2 \end{gathered} = P ( L s n s ) P ( L s ) + P ( L s n s ) P ( L s ) P ( L s n s ) P ( L s ) = 51 C n 12 C n × 4 1 + 4 3 × 51 C n 13 C n 51 C n 12 C n × 4 1 = 1 + 3 ⋅ 12 C n 13 C n 1 = 52 − n 13 − n ⇒ 52 − n 13 − n = 50 11 ⇒ n = 2 Answer the question to unlock