You are at a European beach with 60 other visitors. 36 of them speak English. If you randomly meet two people on the beach, what is the probability that they both speak English

Assuming I'm one of the 36 English speakers and the other 24 speak Spanish for illustration purposes.  The problem can be modeled as 59 marbles with 35 E and 24 S marbles as N = 59 possible outcomes = n(E) + n(S) = 35 + 24.

So I reach into the pile of marbles (on the beach) and the probability that it's p(E) = n(E)/N = 35/59 = 0.593220339 when I meet the one person. ANS

I assume I remember that first person; so I remove him from the marbles (by avoiding him on the beach) and now my probability is p(E and E) = n(E)/N * n(E)-1/(N - 1) = 35/59*34/58 = 0.347749854 ANS

Following the same logic p(E and E and E) = 35/59*34/58*33/57 = 0.201328863 ANS

This last one is different from the first three.  This one is p(E >= 1|4 attempts).  We can trace out a probability tree to identify those branches that contain at least one E event.  So:

p() = 35/59 * 34/58 * 33/57 * 32/56 =  

EEES p() = 35/59 * 34/58 * 33/57 * 24/56 =

EESE p() = 35/59 * 34/58 * 24/57 * 33/56 =

ESEE p() = 35/59 * 24/58 * 34/57 * 33/56 =

SEEE p() = 24/59 * 35/58 * 34/57 * 33/56 =

EESS p() = 35/59 * 34/58 * 24/57 * 23/56 =

ESES p() = 35/59 * 24/58 * 34/57 * 23/56 =

SEES

SESE

SSEE

ESSS  And so on, but...a big BUT...why do all this when

SESS

SSES

SSSE

p(E>=1|4) = 1 - p(S and S and S and S) = 1 - 24/59 * 23/58 * 22/57 * 21/56 = 0.976652619   ANS.  In other words we find the probability of not meeting an Englishman and take 1 minus that value to find the probability of meeting at least one.

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Step-by-step explanation:

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