Two cards are drawn from an deck of 52 cards, without replacement. How do you find the probability that exactly one card is a spade?

Question

Savannah Anderson · Accepted Answer

The reduced fraction is #13/34#.

Let #S_n# be the event that card #n# is a spade. Then #notS_n# is the event that card #n# is not a spade. #"Pr(exactly 1 spade)"# #="Pr"(S_1)*"Pr"(notS_2|S_1)+"Pr"(notS_1)*"Pr"(S_2|notS_1)# #=13/52*39/51+39/52*13/51# #=2*1/4*39/51# #=39/102=13/34# Alternatively, #"Pr(exactly 1 spade)"# #=1-["Pr(both are spades)"+"Pr(neither are spades)"]# #=1-[(13/52*12/51)+(39/52*38/51)]# #=1-[1/4*12/51+3/4*38/51]# #=1-[(12+114)/(204)]# #=1-126/204# #=78/204=13/34# We could also look at it as #(("ways to draw 1 spade")*("ways to draw 1 non-spade"))/(("ways to draw any 2 cards"))# #=(""_13"C"_1*""_39"C"_1)/(""_52"C"_2)# #=((13!)/(12!1!)*(39!)/(38!1!))/((52!)/(50!2!))# #=(13*39)/[(52*51)//2]# #=(cancel(2)_1*cancel(13)^1*""^13cancel(39))/(cancel(52)_2^(cancel(4))*""^17cancel(51))# #=13/34# This last way is probably my favourite. It works for any group of items (like cards) that have subgroups (like suits), as long as the numbers left of the C's on top #(13 + 39)# add to the number left of the C on bottom #(52)#, and same for the numbers right of the C's #(1+1=2)#. What is the probability of randomly picking 3 boys and 2 girls for a committee, out of a classroom with 15 boys and 14 girls? Answer: #(""_15"C"_3*""_14"C"_2)/(""_29"C"_5)#

Genesis Allen · Answer

undefined The probability of drawing exactly one spade can be calculated by finding the probability of drawing one spade and one non-spade, and then adding the probabilities together.

1. Calculate the probability of drawing one spade and one non-spade:
   - Probability of drawing a spade: 13/52 (since there are 13 spades in a deck of 52 cards)
   - Probability of drawing a non-spade: 39/51 (since there are 39 non-spades left in the deck after removing one spade)

2. Multiply the two probabilities to get the probability of drawing one spade and one non-spade.

3. Since there are two ways this can happen (spade/non-spade or non-spade/spade), multiply the result by 2 to get the total probability of drawing exactly one spade.

4. Simplify the result to get the final probability.