Two cards are randomly selected, without replacement, from a standard deck of playing cards. What is the probability of first picking a five and then picking a card other than a five?
A. 0.071 B. 0.004 C. 0.072
I thought this was P(Other | 5 ), so I rearranged to say
[(4/52) x (48/51)] / (4/52)
The question seems to suggest that you seek the probability of picking a 5, and then picking anything other than a five afterwards.
The answer gives: p = (4/52) x (48/51) = 0.072398.
Is this simply because both events are independent and not happening at the same time? Any tips on grasping these concepts, do I just need to practice them every day until it all clicks? Will it?
P( 5 on first card AND not 5 on second card) = P(5 on first card) * P(not 5 on second card / 5 on first card)
= (4/52) * (48/51) as given
Both events are not independent: you are conditioning the second card picked based on getting a 5 as the first draw, so you only have 51 cards from which to draw, only 48 of which meet the condition for the second card. Whenever you see “without replacement”, that’s usually a dead giveaway that you have to use conditional probability.
Are you comfortable with the formula P(AB) = P(A) × P(B|A)?
If you’re not, we have a lot of work to do.
If you are, then we’ll define the events A and B as:
A is “picking a 5 from a 52-card deck”
B is “picking something other than a 5 from a 51-card deck”
P(A) is easy: 4/52 (= 1/13).
P(B) is more complicated: it depends on whether the 51-card deck contains four fives (and, therefore, 47 non-fives), or three fives (and, therefore, 48 non-fives). Here, however, our original formula says that we want not P(B), but P(B|A): the probability of picking a non-five from the (51-card) deck, given that we picked a 5 first; thus, we have a 51-card deck with three fives and 48 non-fives, so P(B|A) = 48/51.
Putting this all together, P(AB) = P(A) × P(B|A) = 4/52 × 48/51 = 7.2398%.
