A box contains only the letters A,B,C,D and E. Here are the probabilities for A-D: P(a)=1/7 P(b)=1/8 P(c)=1/6 and P(d)=1/10. P(e) is unknown.

Question

A box contains only the letters A,B,C,D and E. Here are the probabilities for A-D: P(a)=1/7 P(b)=1/8 P(c)=1/6 and P(d)=1/10. P(e) is unknown. What is P(e)?

Answer 1

"contains only the letters A,B,C,D and E"  means that when you draw one letter, you will get one of these -- that's for certain [meaning P(a-e) = 1].

 

So,

  P(a) + P(b) + P(c) + P(d) + P(e) =1         [that is, you must get one of A-E]

  1/7  + 1/8  + 1/6  + 1/10  + P(e) = 1

     [note: the denominators are 7, 2*2*2, 2*3, 2*5;  the LCM=2*2*2*3*5*7 = 840]

  120  + 105 + 140 +  84   + 840*P(e)  =  840        [multiply by LCM=840]

         449 + 840*P(e) = 840                              [collect terms]

         P(e) = 391/840     [P(e)=0.45]

 

Check:

  Does    1/7  +  1/8  +  1/6  +  1/10  +  391/840  = 1   ?

             120  +  105 +  140  +   84   +  391  = 840   ?      [multiply by 840]

                        840 = 840   ?yes