n men and m women are standing in a line (randomly).Find the expectancy of the number of men that stand beside a women (at least one side - left or right)Harder question: Now solve it, but they stand in a circle and not a line.I wanted to use an indicator: X = number of men standing beside at least one women. X i = 1 if standing besides a women 0 elseAnd so: E [ X ] = E [ ∑ i = 1 n X i ] = ∑ i = 1 n E [ X i ]the problem is that I am having a time computing what is the probability a random men will stand next to a women (left or right or both)

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n men and m women are standing in a line (randomly).Find the expectancy of the number of men that stand beside a women (at least one side - left or right)Harder question: Now solve it, but they stand in a circle and not a line.I wanted to use an indicator:  X  = number of men standing beside at least one women.      X    i    =  1           if standing besides a women                                  0           elseAnd so:   E  [  X  ]  =  E  [      ∑          i      =      1        n        X    i    ]  =      ∑          i      =      1        n    E  [      X    i    ]the problem is that I am having a time computing what is the probability a random men will stand next to a women (left or right or both)

robegarj · Accepted Answer

E      X    1    =  P  (      X    1    =  1  )  =  1  −  P  (  man 1 has male neighbors  )  =  1  −                                      (                                      n            −            1                    2                                  )                                              (                          m          0                                  )                                          (                              n          +          m          −          1                2                          )              =  1  −            n      −      1              n      +      m      −      1                  n      −      2              n      +      m      −      2      Moreover by symmetry we have   E      X    i    =  E      X    1    for every   i  ∈  {  1  ,  …  ,  n  } so that:  E  X  =  n  E      X    1    =  n      (    1    −                  n        −        1                    n        +        m        −        1                            n        −        2                    n        +        m        −        2              )  To find   E      X    1    =  P  (      X    1    =  1  ) in the first problem is a bit harder.If E denotes the event then man 1 is at utmost left or utmost right position then we must split up:We must split up in:   P  (      X    1    =  1  )  =  P  (      X    1    =  1  ∣  E  )  P  (  E  )  +  P  (      X    1    =  1  ∣      E      ∁    )  P  (      E      ∁    )Here   P  (  E  )  =      2          n      +      m       and   P  (      X    1    =  1  ∣  E  )  =      m          n      +      m      −      1      For   P  (      X    1    =  1  ∣      E      ∁    ) we find what we found for   P  (      X    1    =  1  ) in the first problem.

n men and m women are standing in a line (randomly). Find the expectancy of the number of men that

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