Two cards are drawn one by one from a pack of cards. The probability of getting first card an ace and second a coloured one is (before drawing second card first card is not placed again in the pack):

Question

Two cards are drawn one by one from a pack of cards. The probability of getting first card an ace and second a coloured one is (before drawing second card first card is not placed again in the pack): (A)  4/13  (B)  1/26  (C)  5/52  (D)  5/221

Tardigrade Admin · Accepted Answer

Let E1= Event of getting an ace card ∴ P(E1)=(4/52) And E2= Event of getting an coloured card P(E2)=(15/51) ∴ Required probability =(4/52) × (15/51) =(5/221)

Q. Two cards are drawn one by one from a pack of cards. The probability of getting first card an ace and second a coloured one is (before drawing second card first card is not placed again in the pack):

$4/13$

$1/26$

$5/52$

$5/221$

Let $E_{1} =$ Event of getting an ace card
$∴ P (E_{1}) = \frac{4}{52}$
And $E_{2} =$
Event of getting an coloured card
$P (E_{2}) = \frac{15}{51}$
$∴$ Required probability $= \frac{4}{52} \times \frac{15}{51}$
$= \frac{5}{221}$

Q. Two cards are drawn one by one from a pack of cards. The probability of getting first card an ace and second a coloured one is (before drawing second card first card is not placed again in the pack):

4/13

1/26

5/52

5/221

Let E1​= Event of getting an ace card ∴P(E1​)=524​ And E2​= Event of getting an coloured card P(E2​)=5115​ ∴ Required probability =524​×5115​ =2215​

$4/13$

$1/26$

$5/52$

$5/221$

Let $E_{1} =$ Event of getting an ace card
$∴ P (E_{1}) = \frac{4}{52}$
And $E_{2} =$
Event of getting an coloured card
$P (E_{2}) = \frac{15}{51}$
$∴$ Required probability $= \frac{4}{52} \times \frac{15}{51}$
$= \frac{5}{221}$