Answer to Question #255323 in Statistics and Probability for A. A

Question

Three men A, B and C agree to meet at the theatre. The man A cannot remember whether they agree to meet at the palace or the queens and tosses a coin to decide which theatre to go to. The man B also tosses a coin to decide between the queens and royalty. The man C tosses a coin to decide whether to go to the palace or not and in this latter case he tosses again to decide between the queens and royal. Find A and B meet, B and C meet, A, B and C all meet, A, B and C all go to different place, and at least two meet.

Answer 1

A and B meet

There is one situation when A and B can meet - if both of them go to queens

P(A and B) = P(A-queens)*P(B-queens) = 0.5 * 0.5 = 0.25

B and C meet

There are two situations when B and C can meet - if both of them go to queens or both go to royalty

P(B and C) = P(B-royal)*P(C-royal) + P(B-queens) + P(C-queens) = 0.5 * 0.5 * 0.5 + 0.5 * 0.5 * 0.5 = 0.25

A, B and C meet

There is one situation when A, B and C can meet - if all of them go to queens

P(A, B and C) = P(A-queens) * P(B-queens) * P(C-queens) = 0.5 * 0.5 * 0.5 * 0.5 = 0.0625

A, B and C all go to the different places

There are two situations when all of them go to the different places - A certainly go to palace, B go to queens and C go to royals or B to royals and C to queens

P(all diff) = P(A-palace)*(P(B-queens) * P(C-royal) + P(B-royal) + P(C-queens)) = 0.5*(0.5 * 0.5 * 0.5 +

+ 0.5 * 0.5 * 0.5) = 0.125

 At least two meet

"At least two meet" and "no one meet" is a complete event groups, so

P(At least two) = 1 - P(no one)

There are some situations when no one meets:

A certainly go to palace, B go to queens and C go to royals or B to royals and C to queens (0.125)

A go to palace, B go anywere, C stays home

A go to queens, B go to royal, C stays home

P(A-palace, B-any, C-home) = 0.5 * 1 * 0.5 = 0.25

P(A-queens, B-royal, C-home) = 0.5 * 0.5 * 0.5 = 0.125

P(at least two) = 1 - 0.125 - 0.25 - 0.125 = 0.5