Answer:
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:
EEEE 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
SSSS
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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Explanation: