Question 1207488: 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?
Found 3 solutions by ikleyn, Edwin McCravy, greenestamps:
Answer by ikleyn(52754)   (Show Source):
You can put this solution on YOUR website! .
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?
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The formulation in the post is absurdist.
Your question is irrelevant to the rest of the post.
Read, re-read, check and re-check.
This post looks like a (non-edited) message from a madhouse,
and does not correspond to high standards of this forum.
Please do not post nonsense to this forum.
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None other than both tutors, Edwin and @greenestamps, read in the head of visitors.
But I’m not so sure that they read correctly and not create something
irrelevant what occasionally comes to their mind.
From my side, I believe that the posted text is a mixture of two different
tasks mixed by mistake.
Dear colleagues, my position is simple: the post must be correct.
If it is incomplete or is ambiguous, or consists of logically
disconnected pieces, it can not be a subject of consideration.
It is a standard normal position in the world of common sense.
By the way, it is the RULE of this forum.
So, I require from visitors to be responsible for their writing. That's all.
You, on the contrary, encourage irresponsibility in order to appear kind.
Next, dear tutor @greenestamps, where do you see mocking from my side?
It is a kind of humor, to laugh together, nothing more. Is it not clear ?
Mocking was from the opposite side, when the visitor posted what he posted.
Answer by Edwin McCravy(20054)  (Show Source):
You can put this solution on YOUR website!
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
Answer by greenestamps(13195)  (Show Source):
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