mSKR wrote:
Please help me to identify the value which I get is actually what?
Out of 5 times coin flipped : Kate can have $11 or $13 , more than 10$ and less than 15$.
The values that she can have are: 5$, 7$, 9$, 11$, 13$,15$ ( if Kate wins 0, 1,2,3,4,5 times respectively)
Favored cases = 2/6 ( total cases) = 1/3
You might imagine using the same approach to answer the question "what is the probability I will win the lottery?" You might then think there are two cases -- you win or you lose -- and that the probability is 1/2. But that's clearly not right. The issue is, the two cases in the lottery question are not equally likely. If we're going to solve a probability question by counting outcomes, those outcomes need to be equally likely to happen. In the question in this thread, Kate only gets $15 when the coin lands on Tails five times, which can only happen in one way (TTTTT). But she gets $11 in the end if the coin lands on Tails three times and on Heads twice, which is a much more likely outcome -- it can happen in ten ways (TTTHH, TTHTH, THTTH, HTTTH, etc).
mSKR wrote:
In other words, what would be the question exactly, if the answer is 1/3.
please suggest.
I can quickly modify the question above so 1/3 becomes the right answer, but I think this question setup is very awkward (I'd never write a question like this personally!) :
David and Kate each have $10. David and Kate play a game where they flip a coin once, and if the coin lands on Tails, David will give Kate a randomly selected even number of dollars between $2 and $6 inclusive, and if it lands on Heads, Kate will give David a randomly selected even number of dollars between $2 and $6 inclusive. What is the probability Kate will have more than $10 but less than $15 if she plays this game once?Then there are six equally likely outcomes (-$6, -$4, -$2, +$2, +$4, +$6), two of which produce the result the question asks for, so the answer is 2/6 = 1/3.