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Tomas.A
 3 years ago
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?
Tomas.A
 3 years ago
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?

This Question is Closed

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.3you already know that you didn't have a tail on the first toss, so change the words "3 times" to "two times"

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.3sample space is now HHH HHT HT

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.3in two out of the 3 cases it is thrown three times, unless i misinterpreted the question

Tomas.A
 3 years ago
Best ResponseYou've already chosen the best response.1given answer is actually 1/2

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.3if you want a formula, you can write A= tail is tossed on first tryt B= coin is tossed 3 times and then you want \[P(BA)=\frac{P(A\cap B)}{P(A)}\] but this begs the question because you still have to compute \[P(A\cap B)\] which just requires the same work, computing \[A\cap B=\{HHH, HHT\}\]

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.3i 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

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.3no \[P(A\cap B)\neq \frac{1}{2}\]

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.3experiment 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

Tomas.A
 3 years ago
Best ResponseYou've already chosen the best response.1shouldn't we write complete sample T, HT, HHT, HHH and given answer is 1/2 not 2/3 but it might be incorrect

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.3{T, HT, HHT, HHH} only 4 possiblities

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.3exactly what you wrote yes

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.3so if we put the event A = first toss is heads, we know \[P(A)=\frac{1}{2}\]easily, but what we need is \[P(BA)=\frac{P(A\cap B)}{P(A)}\]

Tomas.A
 3 years ago
Best ResponseYou've already chosen the best response.1\[\frac{3}{4}\div \frac{1}{2}>1\]

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.3well i made a mistake of course, (not in the answer, in the computation)

Tomas.A
 3 years ago
Best ResponseYou've already chosen the best response.1P(A) is correct, how to get P(A∩B)?

Tomas.A
 3 years ago
Best ResponseYou've already chosen the best response.1if 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 :/

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.3ok i see the problem

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.3i am sorry i am an idiot this morning, so lets go slow

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.3first of all there are 4 possible outcomes, T HT HHT HHH but they are not EQUALLY LIKELY so we cannot use the uniform distribution

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.3that is, each does not occur with probability 1/4

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.3\[P(T)=\frac{1}{2}, P(HT)=\frac{1}{4},P(HHH)=\frac{1}{8},P(HHH)=\frac{1}{8}\]

Tomas.A
 3 years ago
Best ResponseYou've already chosen the best response.1ok now i understood lol

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.3so i was completely wrong, and i apologize. should have a nice cup of coffee

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.3ok so now let us finish the problem.

Tomas.A
 3 years ago
Best ResponseYou've already chosen the best response.1so \[ P(A \cap B)=\frac{1}{8}+\frac{1}{8}=\frac{1}{4}\]

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.3we know \[A\cap B = \{HHT, HHH\}\] and so what you said, \[P(A\cap B)=\frac{1}{4}\]

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.3now \[P(A)=\frac{1}{2}\] still (at least we knew that much!) and so your answer is \[\frac{1}{4}\div\frac{1}{2}=\frac{1}{2}\]

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.3now 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!

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.3sorry again for the error and taking up so much time being wrong
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