- CFA Exams
- CFA Level I Exam
- Study Session 3. Quantitative Methods (2)
- Reading 9. Common Probability Distributions
- Subject 5. The Binomial Distribution

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**CFA Practice Question**

A set of four cards consists of two red cards and two black cards. The cards are shuffled thoroughly, and I am dealt two cards. I count the number of red cards X in these two cards. The random variable X has which of the following probability distributions?

B. The binomial distribution with parameters n = 2 and p = 0.5

C. None of the above

A. The binomial distribution with parameters n = 4 and p = 0.5

B. The binomial distribution with parameters n = 2 and p = 0.5

C. None of the above

Correct Answer: C

Because two cards are being selected, the number of observations is two. If the observations were independent, answer B would be correct. But the observations are not independent. If the first card is red, the probability that the second card dealt is red is 1/3. If the first card is black, the probability that the second card dealt is red is 2/3. These two probabilities would have to be the same if the observations were independent.

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**User Contributed Comments**
13

User |
Comment |
---|---|

danlan |
Good point. |

moma20 |
what a nice note |

n9845705 |
I don't see how it is 1/3 or 2/3 I get if 1st red then prob that 2nd red is 25/51=.4902 |

AyshaAli |
there id only 4 cards in total 2 red and 2 black. first attempt: probability of red is 2/4 (.5) second attenpt: probability of red is dependent on the first attempt. If red was dealt then the no. of red cards will be 1 and total of cards will be 3, hence, the prob is 1/3/ while if the card dealt was blak then there will be no change in total no. of red cards while the total cards are 3 in total now so prob is 2/3 |

aspazia |
do not see the difference? why are they not independent events? |

bansal |
imagine that there were only two cards to begin with - one red and one black...then try to see the difference.... |

gaur |
because there are only 4 cards...its a dependant event...if there were say 7000 cards then it would be independant |

o123 |
no...if the card was replaced it would be independant |

bobert |
For it to be independent it needs to be the same every time. If you roll a die, it is the same every time, 1/6, 1/6, 1/6...and so on. If you are changing the set though, as in this case taking two cards out of the deck, you will not have the same probability to receive the red cards. For all you know it is possible to get both red cards, then if you pick up the last two cards you know you have 0% chance of getting another red card. That is why this is a dependent event. If the cars were picked one at a time, and then put back, it would be independent because the picking of a card is the same each time. |

gazza77 |
It is dependant because the probability for each colour being drawn next will change depending on what was drawn previously |

bundy |
No Replacement so it can't be a Binomial distributin...ie the selction is no longer idependent after the first card is selected and not replaced |

johntan1979 |
Always assume dependent events (not replaced) if the probability is not stated in the question. |

deliawmx |
I think the problem is that whether these two cards are dealt together one-time or one by one. Misunderstanding exists under this situation. |