Question

From shuffled deck of cards, 4 cards are randomly selected without replacement from shuffled deck of 52 cards. What is the probability to get at least one ace?

Answer 1

Note that:

P(Getting at least one ace) = 1 - P(getting no aces on 4 draws) .............(1)

Now, we find the probability of the event 'getting no aces' when 4 cards are randomly selected without replacement from shuffled deck of 52 cards. Let the 4 cards be drawn one by one (this does not alter the answer).

When the first card is drawn, there are a total of 52 cards out of which 4 are aces and 48 are not aces. Thus:

P(not getting an ace on the first draw) = 48/52

Now, given that the first card was not an ace, we are left with 51 cards out of which 4 are aces and 47 are not aces. Thus:

P(not getting an ace on the second draw | the first card was not an ace) = 47/51

Now, given that the first two cards were not aces, we are left with 50 cards out of which 4 are aces and 46 are not aces. Thus:

P(not getting an ace on the third draw | the first two cards were not aces) = 46/50

Now, given that the first three cards were not aces, we are left with 49 cards out of which 4 are aces and 45 are not aces. Thus:

P(not getting an ace on the fourth draw | the first three cards were not aces) = 45/49

Thus:

P(getting no aces on 4 draws) = P(not getting an ace on the first draw)*P(not getting an ace on the second draw | the first card was not an ace)*P(not getting an ace on the third draw | the first two cards were not aces)*P(not getting an ace on the fourth draw | the first three cards were not aces)

= (48/52)*(47/51)*(46/50)*(45/49)

= 4669920/6497400

= 0.718737 .....................(2)

From equations (1) and (2), we get:

P(Getting at least one ace) = 1 - 0.718737 = 0.281263 [ANSWER]

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