Colleen times her morning commute such that there is an equal likeliho

Question

Colleen times her morning commute such that there is an equal likelihood that she will arrive early or late to work on any given day. If she always arrives either early or late, what is the probability that Colleen will arrive late to work no more than twice during a five-day workweek?

(A) 1/4
(B) 15/32
(C) 1/2
(D) 1732
(E) 3/4

My Answer (B) 
I made it look like Head and Tails questions and looked for answer, not more than two Heads
HTTTT + HHTTT
 = 5c1/32 + 5c2/32 = 15/32 (But not correct answer)

quantumliner · Accepted Answer

Case 1: Colleen was late on 0 out of 5 days.

==> 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 1/32

Case 2: Colleen was late on 1 out of 5 days.

==> 1/2 * 1/2 * 1/2 * 1/2 * 1/2 * 5C1

==> 1/32 * 5 = 5/32

Case 3: Colleen was late on 2 out of 5 days.

==> 1/2 * 1/2 * 1/2 * 1/2 * 1/2 * 5C2

==> 1/32 * 10 = 5/16

Probability that Colleen will arrive late to work no more than twice during a five-day workweek

= 1/32 + 5/32 + 5/16 = 16/32 = 1/2

Answer is C

AaronPond · Answer

This problem asks you to calculate the probability that Colleen will arrive late to work no more than twice during a five-day workweek (so that is none, once, or twice). Since Colleen has an equal chance of being early or late (she is never on time), you essentially have a coin flip: equal chance between E and L on each day.

There are two ways to solve this problem. One of them is calculational, one of them is conceptual. Let's start with the calculational:

Method #1: Do the Math  
Any probability question basically breaks down a single concept: 
  P=\frac{	ext{# of situations that fit the criteria}}{	ext{Total # of situations}}

Calculating the total number of possibilities is relatively straightforward using your knowledge that there are two possibilities each day. With  5  days having  2  options each, the total number of possible situations is   2^5 = 32  .

Now you need the favorable options. First calculate the easy ones:

There is only one arrangement when she is early every day: (  EEEEE  ).

There are five arrangements where she can be late once per week: (  L  EEEE  ,   E  L  EEE  ,   EE  L  EE  ,   EEE  L  E  ,   EEEE  L  ).

The number of arrangements where Colleen is late twice requires a bit of combinatorics. In this case, you are dealing with a permutation with repeated elements (three   E  s and two   L  s). Therefore, the number of arrangements is equal to:

A=\frac{N!}{R_E!∗R_L!}=\frac{5!}{3!∗2!}=\frac{5∗4∗3!}{3!∗2}=\frac{5∗4}{2}=10

Combining all of these options together, the total number of ways Colleen can arrange her week and still be late "no more than two times" is   1 + 5 + 10 = 16  . To calculate the probability, we divide this number of the total number of possible arrangements ( 32 ):
  P=\frac{	ext{# of situations that fit the criteria}}{	ext{Total # of situations}}=\frac{1+5+10}{32}=\frac{16}{32}=\frac{1}{2}

The answer is “C”.

Method #2: Conceptualize  
If you can see it, there is a much quicker way to conceptually get to the answer. As we went through Method #1, did you notice the natural symmetry between being (1)   always early   and (2)   always late  , being (1)   early once   and (2)   late once  , and being (1)   early twice   and (1)   late twice  ? Think about it: for every situation where Colleen is early a certain number of times, there is an  equal  number of options where she is late the same number of times. Since she can be late 0, 1, 2, 3, 4, or 5 times in a given week and "no more than two times" is half of those symmetrically-sized options, then she has a  50\%  chance (or  3  in  6 ) chance of meeting the criteria given in the problem. The answer is still "C", but the question can be solved in less than 30 seconds.

IanStewart · Answer

Your thinking is right here, but you missed one case. If you think of it as a coinflip problem, it is also possible that all 5 flips are tails: TTTTT. Since that has a (1/2)^5 = 1/32 chance of happening, that's the reason your answer was too low by exactly 1/32.

There's a slightly faster way to do the problem. The probability she will be late at most 2 times is exactly the same as the probability she will be early at most 2 times. But those are all the possible scenarios, so each must happen 1/2 the time.

sandeepmanocha · Answer

Thanks for pointing, and yes AT MOST means, 0 or more upto given limit. It is going into my Tips

Although I am not sure about this comment. Please can you elaborate on this? Thanks

IanStewart · Answer

Say you flip a coin 5 times, and are asked two questions:

1. what's the probability you get more Heads than Tails?
2. what's the probability you get more Tails than Heads?

the answer to those two questions must be the same, because, since it's equally likely you get Heads or Tails on each toss, there's no logical reason why the answer to question 1 should be lower or higher than the answer to question 2.

But the situations in 1 and 2 are the only things that can possibly happen - we can't get an equal number of H and T if we flip a coin an odd number of times. And the probabilities of all the things that can happen must add up to 1. So we have two equal probabilities that add to 1, and both probabilities must be 1/2.

Notice that question 2, "what's the probability you get more Tails than Heads" is exactly the question you were answering: it's the same question as: "what's the probability you get at most 2 Heads if you flip a coin 5 times?"

So that's one way to see that the answer is 1/2 here without looking at any cases.

mbaprep2016 · Answer

probability that Colleen will arrive late to work no more than twice 
what does that mean, shall i consider at-least one till max 2 than answer will be 15/32

but if we can consider even a zero case of late than answer will be be C 16/32= 1/2

please give the correct explanation

ankitmani2004 · Answer

Yes, you are right. At most 2 includes "not late even once" so the answer would be 15/32 + 1/32 = 1/2.

DhanyaAbhirami · Answer

Being late and being early have equal chance. But how can we be sure that she is never on time as it is not mentioned?

Krunaal · Answer

Please read the question carefully, it is mentioned:  "... she always arrives either early or late,..."

devdattakhoche · Answer

Okay, let's break down the problem using basic counting principles, focusing on the possible scenarios where Colleen is late no more than twice in a 5-day week. Since each day has two possibilities (late or early), we can represent the outcomes as sequences of L's (late) and E's (early).
 Total Possible Outcomes: 
Each day has 2 possibilities (early or late). Over 5 days, the total number of possible outcomes is  2^5 = 32 .
 Favorable Outcomes (Late no more than twice): 
We need to count the cases where Colleen is late 0, 1, or 2 times.
 
  0 times late (all early):  There's only one way this can happen: EEEEE. 
  1 time late:  Colleen can be late on any of the 5 days. So, there are 5 possibilities:
 LEEEE, ELEEE, EELEE, EEELE, EEEEL 
  2 times late:  We need to choose 2 days out of 5 for Colleen to be late. The number of ways to do this is  \binom{5}{2} = \frac{5!}{2!3!} = \frac{5 	imes 4}{2 	imes 1} = 10 .  
 Total Favorable Outcomes: 
1 (0 times late) + 5 (1 time late) + 10 (2 times late) = 16
 Probability: 
The probability of Colleen being late no more than twice is the number of favorable outcomes divided by the total number of possible outcomes:
 \frac{16}{32} = \frac{1}{2} 
So, the probability is  \frac{1}{2} .

Colleen times her morning commute such that there is an equal likeliho

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Colleen times her morning commute such that there is an equal likeliho

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