Moises has been randomly assigned two days in November to se

Question

Moises has been randomly assigned two days in November to see an otolaryngologist and a podiatrist. If he has been assigned an odd-numbered day to see the podiatrist and an even-numbered day other than November 2 to see the otolaryngologist, what is the probability that his appointment with the podiatrist will be the first appointment?

(A) \(\frac{7}{15}\)
(B) \(\frac{1}{2}\)
(C) \(\frac{8}{15}\)
(D) \(\frac{17}{30}\)
(E) \(\frac{3}{5}\)

(A)  \frac{7}{15} 
(B)  \frac{1}{2} 
(C)  \frac{8}{15} 
(D)  \frac{17}{30} 
(E)  \frac{3}{5}

MistyNova · Answer

kevincan  can you provide solution for this?

chetan2u · Answer

Odd days are 1,3,.....29 and even days are 14 of them 4,6,....30

Let us look at each odd number day..
1) If the odd numbered day is 1, then all even numbers are greater..  P = \frac{1}{15}*\frac{14}{14} 
2) If the odd numbered day is 3, then all even numbers are greater again.  P = \frac{1}{15}*\frac{14}{14} 
3) If the odd numbered day is 5, then all even numbers except number 4 are greater..  P = \frac{1}{15}*\frac{13}{14} 
4) If the odd numbered day is 7, then all even numbers except 4 and 6 are greater..  P = \frac{1}{15}*\frac{12}{14} 
And so on

Over all probability =  \frac{1}{15} *(\frac{14}{14} +\frac{14}{14}+\frac{13}{14}+...\frac{1}{14}) =\frac{1}{15}*(\frac{14}{14}+\frac{14+13+..1}{14})=\frac{1}{15}*(1+\frac{14*15}{2*14}) =\frac{1}{15}*(1+\frac{15}{2})=\frac{1}{15}*\frac{17}{2}=\frac{17}{30}

D

findingmyself · Answer

Not 100% sure of this solution, but here is the path taken.

Odd days are: 1, 3....30
Even days (excluding 2 are ): 4,6...30

Number of odd days is (29-1)/2 +1 =15
Even days are (30-4)/2 +1 =14

Probability of Podiatrist (P) first and otolaryngologist (O) second is: Probability of selecting a date of either 1 or 3 (Since even days starts from 4, if the dates allocated are 1 or 3, then Podiatrist will definitely be first) + Probability of selecting an odd date from 4th of the month to 30th of the month

P (1 or 3)= 2/30=1/15
P (Odd date between 4th to 30th of month)= 1/2

Combined is 1/15+1/2= 17/30

kevincan · Answer

You got the right result, but there’s a couple of mistakes in your calculations

kevincan · Answer

There are 15 × 14 = 210 equally likely assignments of the two appointments. For each possible podiatrist date, count how many otolaryngologist dates come afterward:

Podiatrist 1 → 14
Podiatrist 3 → 14
Podiatrist 5 → 13
Podiatrist 7 → 12
...
Podiatrist 29 → 1

Thus the favorable assignments are:

14 + 14 + 13 + 12 + ... + 1
= 14 + (1 + 2 + ... + 14)
= 14 + 14(15)/2 = 119.

Therefore, P = 119/210 = 17/30.

egmat · Answer

Hi MistyNova,

Happy to walk through this one. It's really a  counting  problem dressed up as probability: list the possible days for each doctor, then count how often the podiatrist day comes first.

Step 1 - Count the possible days

-  Podiatrist  = odd-numbered day:  1 ,  3 ,  5 , ...,  29  -  15  choices.
-  Otolaryngologist  = even-numbered day except Nov  2 :  4 ,  6 ,  8 , ...,  30  -  14  choices.

An odd day and an even day can never collide, so every pairing is valid and equally likely. Total equally likely assignments =  15 × 14 = 210 .

Step 2 - Count the favorable pairs (podiatrist first)

We want podiatrist's odd day  <  otolaryngologist's even day. The clean trick is to go through each otolaryngologist day  o  and count how many odd days fall below it - that count is just  o/2 :

-  o = 4  - odds { 1 ,  3 } =  2 
-  o = 6  - odds { 1 ,  3 ,  5 } =  3 
- ... continuing ...
-  o = 30  - odds { 1 ,  3 , ...,  29 } =  15

So the favorable counts are  2, 3, 4, ..., 15 . Adding them:

2 + 3 + ... + 15 = (2 + 15) × 14 / 2 = 119

Step 3 - Divide

P = 119/210 = 17/30

That's answer  (D) .

Why it's not (B) 1/2:  it's tempting to say "one goes first or the other, so  50/50 ." That symmetry only holds if both lists cover the same range. But  Nov 2  is removed from the even side, which tilts the otolaryngologist days slightly  later  - so the podiatrist ends up first a bit more than half the time. That small asymmetry is exactly what pushes  15/30  up to  17/30 .

Answer: D

Moises has been randomly assigned two days in November to se

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Moises has been randomly assigned two days in November to se

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