- CFA Exams
- CFA Level I Exam
- Topic 1. Quantitative Methods
- Learning Module 3. Probability Concepts
- Subject 2. Unconditional, Conditional, and Joint Probabilities
CFA Practice Question
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?
B. 0.004
C. 0.072
A. 0.071
B. 0.004
C. 0.072
Correct Answer: C
p = (4/52) x (48/51) = 0.072398
User Contributed Comments 15
User | Comment |
---|---|
eejo | 4/52 x 48/51 = 192/2652 = .072 |
adenisov | for all situation relating P(B|A): we could say, than there is not "after taking 5", but situation C where just working with deck without on "5". Such we will have two independent outcomes. |
Ildy | can anyone care to explain why 48/51 and not 48/52? |
Ildy | the first card is not replaced :-) now i see it :-) |
Daddykay | P(picking a five the first time)=4/52. Since it is not replaced we now have 51 cards remaining. There are a total of 48 (52-4) cards in the pile that are not-five. Therefore if he makes another pick P(not five)=48/51. You now multiply 4/52*48/51 |
Welles | Why are the odds of picking a 5 the first go round 4/52 and not 4/48 (like from the question in the previous section regarding the odds of selecting a king being 4/48...chances for success/chances for failure)? |
sogah | look i dont play cards so quit using such examples |
GouldenOne | This says you have a 7 percent chance of drawing to 5's... im not buying it |
GouldenOne | Can someone explain the 48/51? The second time around there are 48 5's and not 3? |
banihas | A deck of cards has 52 total cards including four 5's. Now if we take one 5 out and do not replace it we are left with only three 5's left and a total of 51 cards remaining. Thus the probability of not picking a 5 from the deck of 51 remaining is 48/51 . Hence the probability of picking a card other than 5 GIVEN first picking a 5 is 4/52 * 48/51 = 0.072 |
fkigundu1 | Got it at last |
dbedford | General Formula is P(BA) = P(A)P(B|A) where P(A) = picking a 5 first time = 4/52 and P(B|A) = Not picking a 5 Given that you picked a 5 the first time = 48/51. The key is knowing that you are solving for P(BA) which is them saying AND, at least that's how I figured it out |
Zhenek | That feeling when I assume that the standard deck is 36 cards and not 52 ... |
zhefuli | I thought playing cards contain 2 jokers? So it'd be 54 cards right? |
MathLoser | I don't even know there are 52 cards in a deck. |