Question

With a standard deck of 52 playing cards, are your chances of drawing an ace and then a king better with or without "replacement"? Illustrate with computations.

Answer 1

solution:

we know that

Total No.of playing cards = 52

Total No.of aces = 4

Total No.of kings = 4

i) Consider a case without replacement :

Let A = event of drawing first card

n(A) = 4

Probability of drawing first card = P(A) = 4/52 = 1/13

Now, we have total of 51 cards remaining in which 4 kings are present

Let B|A = probability of drawing first card given that already first card is drawn  

P(B|A) = 4/51

Here , B A = event of drawing first card ace and second card is king

By Baye's rule,we have

P(B|A) = P(B A ) / P(A)

P(B A ) =   P(B|A) * P(A)

= 4/51 * 1/13

= 0.00603

ii) Consider a case without replacement :

Let A = event of drawing first card

n(A) = 4

Probability of drawing first card = P(A) = 4/52 = 1/13

Now, we have total of 52 cards as we replaced it again in which 4 kings are present

Let B|A = probability of drawing first card given that already first card is drawn  

P(B|A) = 4/52 = 1/13

Here , B A = event of drawing first card is ace and second card is king

By Baye's rule,we have

P(B|A) = P(B A ) / P(A)

P(B A ) =   P(B|A) * P(A)

= 1/13 * 1/13

= 0.00592

we can observe that the probability in first case(without replacement) is higher than second case (with replacement) as replacing one card again, increases more no.of cards resulting decreasing it's probability in second case.

Chances of drawing an ace and then a king is somewhat better without replacement