Question

5 cards are drawn at random from a well shuffled pack of 52 playing cards. If it is known that there will be at least 3 hearts, the probability that there are exactly 3 hearts is

• A. $\frac{{}^{13}{c}_4}{{}^{13}{c}_3{+}^{13}{c}_4{+}^{13}{c}_5}$
• B. $\frac{{}^{13}{c}_4}{{}^{13}{c}_3{\times }^{39}{c}_2{+}^{13}{c}_4{\times }^{39}{c}_1{+}^{13}{c}_5}$
• C. $\frac{{}^{13}{c}_3{\times }^{39}{c}_2}{{}^{13}{c}_3{\times }^{39}{c}_2{+}^{13}{c}_4{\times }^{39}{c}_1{+}^{13}{c}_5}$
• D. $\frac{{}^{13}{c}_4}{{}^{13}{c}_3{\times }^{13}{c}_4{\times }^{13}{c}_5}$

Answer 1

The correct option is C $\frac{{}^{13}{c}_3{\times }^{39}{c}_2}{{}^{13}{c}_3{\times }^{39}{c}_2{+}^{13}{c}_4{\times }^{39}{c}_1{+}^{13}{c}_5}$

The cases possible are:
(i) 3 H, 2 non-H: ${}_3^{13}C{\ast }_2^{39}C$
(ii) 4 H, 1 non-H: ${}_4^{13}C{\ast }_1^{39}C$
(iii) 5H: ${}_5^{13}C$
Hence, probability = $\frac{{}_3^{13}C{\ast }_2^{39}C}{{}_3^{13}C{\ast }_2^{39}C{+}_4^{13}C{\ast }_1^{39}C{+}_5^{13}C}$
Hence, (c) is correct.