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Mathematics · Probability

JEE Main 2025 — 4 April, Evening Shift — Question 39

A card from a pack of 52 cards is lost. From the remaining 51 cards, nn cards are drawn and are found to be spades. If the probability of the lost card to be a spade is 1150\frac{11}{50}, then nn is equal to \qquad -

Answer: 2

Numerical answer — enter this value.

Step-by-step solution

P( Lost (spade) n cards are spade )P\left(\frac{\text { Lost }_{\text {(spade) }}}{\mathrm{n} \text { cards are spade }}\right)

=P(nsLs)P(Ls)P(nsLs)P(Ls)+P(nsLs‾)P(Ls‾)= \frac{P\left(\frac{n_s}{L_s}\right) P(L_s)}{P\left(\frac{n_s}{L_s}\right) P(L_s) + P\left(\frac{n_s}{\overline{L_s}}\right) P(\overline{L_s})} =12Cn51Cn×1412Cn51Cn×14+13Cn51Cn×34=12Cn412Cn4+3⋅13Cn4=12Cn12Cn+3⋅13Cn= \frac{\frac{{}^{12}C_n}{{}^{51}C_n} \times \frac{1}{4}}{\frac{{}^{12}C_n}{{}^{51}C_n} \times \frac{1}{4} + \frac{{}^{13}C_n}{{}^{51}C_n} \times \frac{3}{4}} = \frac{\frac{{}^{12}C_n}{4}}{\frac{{}^{12}C_n}{4} + \frac{3 \cdot {}^{13}C_n}{4}} = \frac{{}^{12}C_n}{{}^{12}C_n + 3 \cdot {}^{13}C_n} =11+3⋅13Cn12Cn=11+3⋅13−n12−n=12−n12−n+3(13−n)=12−n12−n+39−3n=12−n51−4n= \frac{1}{1 + 3 \cdot \frac{{}^{13}C_n}{{}^{12}C_n}} = \frac{1}{1 + 3 \cdot \frac{13-n}{12-n}} = \frac{12-n}{12-n + 3(13-n)} = \frac{12-n}{12-n + 39 - 3n} = \frac{12-n}{51 - 4n} 12Cn12Cn+3⋅13Cn=12!n!(12−n)!12!n!(12−n)!+3⋅13!n!(13−n)!=112−n112−n+3⋅1313−n=13−n13−n+39(12−n)\frac{{}^{12}C_n}{{}^{12}C_n + 3 \cdot {}^{13}C_n} = \frac{\frac{12!}{n!(12-n)!}}{\frac{12!}{n!(12-n)!} + 3 \cdot \frac{13!}{n!(13-n)!}} = \frac{\frac{1}{12-n}}{\frac{1}{12-n} + 3 \cdot \frac{13}{13-n}} = \frac{13-n}{13-n + 39(12-n)} 12Cn12Cn+3⋅13Cn=11+3⋅13Cn12Cn=11+3⋅13−n12−n=12−n12−n+39−3n=12−n51−4n\frac{{}^{12}C_n}{{}^{12}C_n + 3 \cdot {}^{13}C_n} = \frac{1}{1 + 3 \cdot \frac{{}^{13}C_n}{{}^{12}C_n}} = \frac{1}{1 + 3 \cdot \frac{13-n}{12-n}} = \frac{12-n}{12-n + 39 - 3n} = \frac{12-n}{51 - 4n} 12Cn51Cn×14÷(12Cn51Cn×14+13Cn51Cn×34)=12Cn12Cn+3⋅13Cn=11+3⋅13−n12−n=12−n12−n+39−3n=12−n51−4n\frac{{}^{12}C_n}{{}^{51}C_n} \times \frac{1}{4} \div \left( \frac{{}^{12}C_n}{{}^{51}C_n} \times \frac{1}{4} + \frac{{}^{13}C_n}{{}^{51}C_n} \times \frac{3}{4} \right) = \frac{{}^{12}C_n}{{}^{12}C_n + 3 \cdot {}^{13}C_n} = \frac{1}{1 + 3 \cdot \frac{13-n}{12-n}} = \frac{12-n}{12-n + 39 - 3n} = \frac{12-n}{51 - 4n} 12Cn12Cn+3⋅13Cn=11+3⋅1313−n⋅12−n12=12(13−n)12(13−n)+39(12−n)\frac{{}^{12}C_n}{{}^{12}C_n + 3 \cdot {}^{13}C_n} = \frac{1}{1 + 3 \cdot \frac{13}{13-n} \cdot \frac{12-n}{12}} = \frac{12(13-n)}{12(13-n) + 39(12-n)} 11+3⋅13−n12−n=12−n12−n+39−3n=12−n51−4n\frac{1}{1 + 3 \cdot \frac{13-n}{12-n}} = \frac{12-n}{12-n + 39 - 3n} = \frac{12-n}{51 - 4n} 12−n51−4n=13−n52−n\frac{12-n}{51 - 4n} = \frac{13-n}{52 - n} (12−n)(52−n)=(13−n)(51−4n)(12-n)(52-n) = (13-n)(51-4n) 624−12n−52n+n2=663−52n−13n+4n2624 - 12n - 52n + n^2 = 663 - 52n - 13n + 4n^2 624−64n+n2=663−65n+4n2624 - 64n + n^2 = 663 - 65n + 4n^2 3n2−n+39=03n^2 - n + 39 = 0 11+3⋅13−n12−n=12−n12−n+3(13−n)=12−n12−n+39−3n=12−n51−4n\frac{1}{1 + 3 \cdot \frac{13-n}{12-n}} = \frac{12-n}{12-n + 3(13-n)} = \frac{12-n}{12-n + 39 - 3n} = \frac{12-n}{51 - 4n} 12−n51−4n=13−n52−n\frac{12-n}{51 - 4n} = \frac{13-n}{52 - n} (12−n)(52−n)=(13−n)(51−4n)(12-n)(52-n) = (13-n)(51-4n) 624−64n+n2=663−65n+4n2624 - 64n + n^2 = 663 - 65n + 4n^2 3n2−n+39=03n^2 - n + 39 = 0 11+3⋅13Cn12Cn=11+3⋅13−n12−n=12−n12−n+39−3n=12−n51−4n\frac{1}{1 + 3 \cdot \frac{{}^{13}C_n}{{}^{12}C_n}} = \frac{1}{1 + 3 \cdot \frac{13-n}{12-n}} = \frac{12-n}{12-n + 39 - 3n} = \frac{12-n}{51 - 4n} 12−n51−4n=13−n52−n\frac{12-n}{51 - 4n} = \frac{13-n}{52 - n} (12−n)(52−n)=(13−n)(51−4n)(12-n)(52-n) = (13-n)(51-4n) 624−64n+n2=663−65n+4n2624 - 64n + n^2 = 663 - 65n + 4n^2 3n2−n+39=03n^2 - n + 39 = 0   ⟹  13−n52−n=1150\implies \frac{13-n}{52-n} = \frac{11}{50} 50(13−n)=11(52−n)50(13-n) = 11(52-n) 650−50n=572−11n650 - 50n = 572 - 11n 78=39n78 = 39n   ⟹  n=2\implies n = 2

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
4 April, Evening Shift
Subject
Mathematics
Chapter
Probability
Topic
Conditional Probability and Multiplication Theorem