Question 1016318
<pre>
P(T&6 or H&1&5 or H&2&4 or H&3&3 or H&4&2 or H&5&1) =

P(T&6) + P(H&1&5) + P(H&2&4) + P(H&3&3) + P(H&4&2) + P(H&5&1)

P(T&6) = (1/2)(1/6) = 1/12 =  6/72
P(H&1&5) = (1/2)(1/6)(1/6) =  1/72
P(H&2&4) = (1/2)(1/6)(1/6) =  1/72
P(H&3&3) = (1/2)(1/6)(1/6) =  1/72
P(H&4&2) = (1/2)(1/6)(1/6) =  1/72
P(H&5&1) = (1/2)(1/6)(1/6) =  1/72 
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                     Total = 11/72 

Edwin</pre>