ProbabilityHardBloom L3
Question
Cards are drawn one by one at random without replacement from a well shuffled pack of 52 cards until 2 aces are obtained for the first time. The probability that 18 draws are required for this is
Options
A.$\frac{\ ^{48}C_{16} \times \ ^{4}C_{2}}{\ ^{52}C_{17}} \times \frac{1}{35}$
B.$\frac{\ ^{48}C_{16} \times \ ^{4}C_{1}}{\ ^{52}C_{17}} \times \frac{3}{35}$
C.$\frac{1}{78}$
D.$\frac{17 \times \ ^{34}C_{2}}{\ ^{52}C_{4}}$
Solution
Sol. [Key Concept: Cards drawn until the 2nd ace: 18 draws need exactly one ace in the first 17, then an ace.]
18 draws are needed iff exactly one ace lies in the first 17 cards and the 18th card is an ace:
$P = \frac{C(4,1)\ C(48,16)}{C(52,17)} \times \frac{3}{35}$ — option (B); rearranged this equals $\frac{17\ C(34,2)}{C(52,4)}$ — option (D). → (B), (D).
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