Question

Cards are drawn one by one without replacement a pack of 52 playing cards until all Ace's are drawn out. Then the probability that 2 cards are left unturned when the last ace is drawn

Let A = event of drawing 3 aces in the first 49 draws

• A.
• B. $\frac { 4 { } ^ { 48 } \mathrm { C } _ { 2 } } { { } ^ { 52 } \mathrm { C } _ { 3 } } \cdot \frac { 1 } { 3 }$
• C. ⁴⁸C₂/⁵²C₃
• D. ⁴⁸C₂/3.⁵²C₃

Answer 1

Answer: (B)

$\frac { 4 { } ^ { 48 } \mathrm { C } _ { 2 } } { { } ^ { 52 } \mathrm { C } _ { 3 } } \cdot \frac { 1 } { 3 }$

Answer 2

B = event of drawing last ace in 50th draw

P(1) = . P(B/A) = $\frac { 1 } { 3 }$

∴ P(A∩B) = P(1). P(2)