Question 1207488
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I think you just left off parts (a) an (b).  I'll assume the entire problem was:

A coin is biased, so that the head is twice as likely to occur as tail. If the
coin is tossed 3 times, what is the probability of getting an even number
(a) of heads?
(b) of tails?

0 is an even number since it is divisible by 2.

P(H)+P(T)=1
2P(T)+P(T)=1
3P(T)=1
P(T)=1/3
P(H)=2P(T)=2/3

The sample space

1. P(HHH) = (2/3)(2/3)(2/3) =  8/27  <--odd number of heads, 3 
2. P(HHT) = (2/3)(2/3)(1/3) =  4/27  <--even number of heads, 2
3. P(HTH) = (2/3)(1/3)(2/3) =  4/27  <--even number of heads, 2
4. P(HTT) = (2/3)(1/3)(1/3) =  2/27  <--odd number of heads, 1
5. P(THH) = (1/3)(2/3)(2/3) =  4/27  <--even number of heads, 2
6. P(THT) = (1/3)(2/3)(1/3) =  2/27  <--odd number of heads, 1
7. P(TTH) = (1/3)(1/3)(2/3) =  2/27  <--odd number of heads, 1
8. P(TTT) = (1/3)(1/3)(1/3) =  1/27  <--even number of heads, 0
                      Total = 27/27 = 1

(a) P(even number of heads) = P(case 2 or 3 or 5 or 8) = 4/27+4/27+4/27+1/27 = 13/27.

(b) P(even number of tails) = P(case 1 or 4 or 6 or 7) = 8/27+2/27+2/27+2/27 = 14/27.

I could have gotten (b) just be subtracting 13/27 from 1, but that might not be
100% obvious to you. So I figured it the long way.

Since heads is twice as likely as tails, you might expect (a) to be the larger
probability, but as we see, it's the smaller!!!

Edwin</pre>