anonymous
  • anonymous
Coin can be tossed until a tail appears or until it has been tossed 3 times. Given that tail does not occur on the first toss, whats probability that the coin is tossed 3 times? I think I got answer by thinking but i don't know how to write it out So I need to use conditional probability and P(no tail on 1 try)=0.5 and how to get others?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
you already know that you didn't have a tail on the first toss, so change the words "3 times" to "two times"
anonymous
  • anonymous
sample space is now HHH HHT HT
anonymous
  • anonymous
Nice @ satellite 73

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
in two out of the 3 cases it is thrown three times, unless i misinterpreted the question
anonymous
  • anonymous
ty
anonymous
  • anonymous
given answer is actually 1/2
anonymous
  • anonymous
if you want a formula, you can write A= tail is tossed on first tryt B= coin is tossed 3 times and then you want \[P(B|A)=\frac{P(A\cap B)}{P(A)}\] but this begs the question because you still have to compute \[P(A\cap B)\] which just requires the same work, computing \[A\cap B=\{HHH, HHT\}\]
anonymous
  • anonymous
i had a typo above, it should have been A = heads tossed on first try ok first of all the answer, and it is \[\frac{2}{3}\] because out of the three events HHH HHT HT two are favorable
anonymous
  • anonymous
no \[P(A\cap B)\neq \frac{1}{2}\]
anonymous
  • anonymous
experiment is "Coin can be tossed until a tail appears or until it has been tossed 3 times." so lets write out a complete sample space, as it is small
anonymous
  • anonymous
shouldn't we write complete sample T, HT, HHT, HHH and given answer is 1/2 not 2/3 but it might be incorrect
anonymous
  • anonymous
{T, HT, HHT, HHH} only 4 possiblities
anonymous
  • anonymous
exactly what you wrote yes
anonymous
  • anonymous
so if we put the event A = first toss is heads, we know \[P(A)=\frac{1}{2}\]easily, but what we need is \[P(B|A)=\frac{P(A\cap B)}{P(A)}\]
anonymous
  • anonymous
\[\frac{3}{4}\div \frac{1}{2}>1\]
anonymous
  • anonymous
well i made a mistake of course, (not in the answer, in the computation)
anonymous
  • anonymous
P(A) is correct, how to get P(A∩B)?
anonymous
  • anonymous
if we look at full sample T, HT, HHT, HHH then HHT and HHH suits A and B, no? but it means 0.5 and 0.5/0.5=1 and again it's incorrect :/
anonymous
  • anonymous
ok i see the problem
anonymous
  • anonymous
i am sorry i am an idiot this morning, so lets go slow
anonymous
  • anonymous
first of all there are 4 possible outcomes, T HT HHT HHH but they are not EQUALLY LIKELY so we cannot use the uniform distribution
anonymous
  • anonymous
that is, each does not occur with probability 1/4
anonymous
  • anonymous
\[P(T)=\frac{1}{2}, P(HT)=\frac{1}{4},P(HHH)=\frac{1}{8},P(HHH)=\frac{1}{8}\]
anonymous
  • anonymous
ah, yeah..
anonymous
  • anonymous
ok now i understood lol
anonymous
  • anonymous
so i was completely wrong, and i apologize. should have a nice cup of coffee
anonymous
  • anonymous
ok so now let us finish the problem.
anonymous
  • anonymous
so \[ P(A \cap B)=\frac{1}{8}+\frac{1}{8}=\frac{1}{4}\]
phi
  • phi
|dw:1326555438102:dw|
anonymous
  • anonymous
we know \[A\cap B = \{HHT, HHH\}\] and so what you said, \[P(A\cap B)=\frac{1}{4}\]
anonymous
  • anonymous
now \[P(A)=\frac{1}{2}\] still (at least we knew that much!) and so your answer is \[\frac{1}{4}\div\frac{1}{2}=\frac{1}{2}\]
anonymous
  • anonymous
now let us make sense out of this our new sample space is HT HHT HHH that is we knew the first toss was an H we might as well have written T HT HH and now it is clear that on the second toss you get tails with probability 1/2 and if you don't get tails you will toss again and have three tosses!
anonymous
  • anonymous
sorry again for the error and taking up so much time being wrong

Looking for something else?

Not the answer you are looking for? Search for more explanations.