Question

Two cards are drawn successively with replacement from a well shuffled pack of 52 cards. The probability of drawing two aces is.
A. \[\dfrac{1}{13}\]
B. \[\dfrac{1}{13}\] \[\times \dfrac{1}{17}\]
C. \[\dfrac{1}{52}\] \[\times \dfrac{1}{51}\]
D. \[\dfrac{1}{13}\] \[\times \dfrac{1}{13}\]

Answer 1

Hint: In the problem we have to find the cards taken one after the other. This is a mutually exclusive event. We have to find the probability of the cards taken one at a time. And another one at a time.

Complete step-by-step answer:
Probability is a measure of the likelihood of an event to occur. Many events cannot be predicted with total certainty. We can predict only the chance of an event to occur i.e. how likely they are to happen, using it. Probability can range in between 0 to 1, where 0 means the event to be an impossible one and 1 indicates a certain event.
Deck contains 52 cards out of which 4 are aces.
Let A denotes the event that the first ace is drawn.
Let B denote the event that the second ace is drawn.
The probability that first ace is drawn \[P\left( A \right)\]= \[\dfrac{{}_{1}^{4}C}{{}_{1}^{52}C}\] . . . . . . . . . . . . . . . . . . . . . . . . (1)
\[\dfrac{{}_{1}^{4}C}{{}_{1}^{52}C}=\dfrac{4}{52}=\dfrac{1}{13}\]
The probability that second ace is drawn \[P\left( \dfrac{B}{A} \right)\] = \[\dfrac{{}_{1}^{3}C}{{}_{1}^{51}C}\] . . . . . . . . . . . . . . . . . (2)
\[\dfrac{{}_{1}^{3}C}{{}_{1}^{51}C}=\dfrac{3}{51}=\dfrac{1}{17}\]
The required probability is =\[\dfrac{1}{13}\] \[\times \dfrac{1}{17}\] = \[\dfrac{1}{221}\]
Therefore the answer is option B

Note: This is a direct problem with the straight definition of conditional probability. The term \[P\left( \dfrac{B}{A} \right)\] denotes the probability of occurrence of event B when event A has occurred. Take care while doing calculations.