People Born on Same Day of The Week

What is the probability that, of three people selected at random, at least two were born on the same day of the week?
Source: NCTM Mathematics Teacher, April 2008

Solution
If the first two people were born on Monday, the 3rd person could be born on Tuesday, Wednesday, Thursday, Friday, Saturday, or Sunday. We illustrate this outcome by the triplet $\mathrm{(M,M,X)}$ , where $\mathrm{M}$ represents Monday and $\mathrm{X}$ ={Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday}. The number of possible outcomes for $\mathrm{(M,M,X)}=6$ . There are $3$ ways to select two people out of three: $\mathrm{(M,M,X),(M,X,M)}$ , and $\mathrm{(X,M,M)}$ . The number of possible outcomes for two people born on Monday equals $6+6+6=18$ .
The same result applies to Tuesday, Wednesday, etc. Since there are $7$ days in a week, the number of possible outcomes for two people born on the same day equals $7\times 18=126$ . In addition the number of possible outcomes for three people born on the same day equals $7$ .
Number of desired outcomes
$126+7=133$
Number of possible outcomes
$7^3=343$
Probability that at least two people were born on the same day
$133/343=19/49$

Answer: $19/49$

Alternative solution 1
We introduce the following notations
$\mathrm{P(M,M,X)}$ = probability of event $\mathrm{M,M,X}$
$\mathrm{P(M)}$ = probability of event $\mathrm{M}$
$\mathrm{P(X)}$ = probability of event $\mathrm{X}$ .
Since birthdays are independent events,
$\mathrm{P(M,M,X)}=\mathrm{P(M)}\times\mathrm{P(M)}\times\mathrm{P(X)}$
$=1/7\times 1/7\times 6/7$
$=6/343$
$P\mathrm{(M,X,M)}=P\mathrm{(M)}\times \mathrm{P(X)}\times\mathrm{P(M)}$
$=1/7\times 6/7\times 1/7$
$=6/343$
$\mathrm{P(X,M,M)}=\mathrm{P(X)}\times \mathrm{P(M)}\times\mathrm{P(M)}$
$=6/7\times 1/7 \times 1/7$
$=6/343$
Probability of two people born on Monday
$6/343+6/343+6/343=18/343$
Probability two people born on Tuesday
$6/343+6/343+6/343=18/343$
$\cdots$
Probability of two people born on the same day
$7(18/343)=18/49$
Probability of three people born on Monday
$\mathrm{P(M,M,M)}=\mathrm{P(M)}\times\mathrm{P(M)}\times\mathrm{P(M)}$
$=1/7\times 1/7 \times 1/7$
$=1/343$
Probability of three people born on Tuesday
$\mathrm{P(T,T,T)}=1/343$
$\cdots$
Probability of three people born on same day
$7(1/343)=1/49$
Probability of at least two people born on the same day
$18/49+1/49=19/49$

Alternative solution 2
$\mathrm{P}$ (at least two people born on same day) = $1-\mathrm{P}$ (all three born on different days)
$=1-(7/7 \times 6/7 \times 5/7)$
$=1-30/49$
$=19/49$