- CFA Exams
- CFA Level I Exam
- Study Session 2. Quantitative Methods (1)
- Reading 8. Probability Concepts
- Subject 10. Principles of Counting

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**CFA Practice Question**

A coin is tossed five times. What is the probability of obtaining exactly three heads?

A. 3/16

B. 1/4

C. 5/16

**Explanation:**Probability of 3 events occurring in 5 trials:

Combination of 3 events in 5 trials * (outcome for each event)

^{x}* (outcome for each event)

^{(n-x)}

where:

Combination of 3 events in 5 trials = n! / (x! * (x-1)!)

= 5! / (3! * 2!) = 120 / 12 = 10

Probability = 10 * (1/2)

^{3}* (1/2)

^{(5-3)}= 5/16

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**User Contributed Comments**
9

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

chuong |
Bernoulli trials: (nCr)*(P^x)*[(1-p)^(n-x)] = (5!/2!3!)*0.5^3 x (1-0.5)^2 = 5/16 |

fahad |
SO the answer will be the same if the probability was to obtain exactly two heads instead of three? |

whoi |
Fahad: Yes, you don't a calculation to prove this, just remember that coin toss always is backed by probabilities of 50% for each outcome. therefore the one and only term that changes is the multiplicator (10). for exactly one head: 5 for excatly two heads: 10 for exactly three heads: 10 for exactly four heads: 5 for exactly five heads: 1 |

micheleus |
anybody can explain further? |

bashg |
Yes if each head is called H and each tail is T, then you want to achieve 3 coin tosses in 5. The combo would look like HHHTT. Now, there are a number of combinations of this eg HHTTH or THTHH or HTHTH etc. The first part of the formula calculates how many combinations of this you can get ie n!/x!(n-x)! Then you assign probability of independent events Probability of 3 heads is 0.5 x 0.5 x 0.5 in other words (1/2)^3 Probability of 2 tails is 0.5 x0.5 in other words (1/2)^2 Hope this helps |

aakidil |
Its actually 2 part question always and students attempt it wrong because answer of first part is also a choice in the list just do it carefully Part 1: identify number of combinations by nCr here n = 5 and r=3 plug in to your calc and get answer. Part 2: bionomial dist. asks about 3 thing a) no. of possible combinations b) prob. of success c) prob. of failure here a=10 (5C3 = 10) b= 50%^num. of events (12.5%) c=50%^(num. of trials - num. of events) (25%) so 10*12.5%*25% = 31.25% which is equal to 5/16 |

TiredHand |
As its a coin and its 50:50 either way the answer can be very quickly calculated as 0.5 to the power of 5 = 0.03125 |

fanDango |
0.03125 =/ 0.3125 |

birdperson |
1. how many possibilities are there (2^5) = 32 2. how many combinations are there with 3 heads (5C3) = 10 3. combinations with 3 heads / total possibilities = 10/32 = 5/16 |