Find the probability p

Question

A man is dealt 4 spade cards from an ordinary pack of 52 cards. If he is given three

more cards, find the probability p that at least one of the additional cards is also a spade.

Answer 1

4 out of 52 leaves (13-4) = 9 spades left out of (52-4) = 48 cards and 39 cards not spades

39/48 = chance of no spade on the next draw

38/47 = chance of no spade on 2nd draw

37/46 = chance of no spade on 3rd draw

use the multiplication rule to find probability of no spades on 3 more draws

3 more cards not spades have probability = (39/48)(38/47)(37/46)

= (39)(38)(37)/48(47)(46)

=13(19)(37)/16(47)(23)

probability of at least one spade = 1 minus the probability of no spades

= 1 - 13(19)(37)/16(47)(23)

=1- .5283880666

about 52.8% chance of not getting a spade

about 100%-52.8% = 47.2% probability of at least one spade in next 3 draws

= probability of 1, 2 or all spades in next 3 draws

You could separately calculate the probability of 1 spade, then of 2 spades, then of 3 spades

and sum those probabilities to get the probability of at least 1 spade in 3 more draws

but that's a little more tedious

P(3 spades) = 9/48)(8/48)(7/48)= (1/16)(1/6)(7/16)= 7/96(16)= about .005

P(2 spades) = 3C2(9/48)(39/47)(8/46)= (9/16)(39/47)(4/23)= about .081

P(1 spade= 3C1(9/48)(39/47)(38/46) = (3/16)(39/47)(38/15)= about .394

then sum them = .480= 48% which is very close to the 47.2% in the 1st method above

rounding approximations likely explain that fraction of 1% difference