Question

What is the probability that n tails in a row with a coin for which the probability of heads
is p? Given a fraction , with 0 < alpha < 1, show that the number of flips one must
perform to be  sure of obtaining at least one head is n =ln(1-alpha)/ln(1-p)



























 .

Answer 1

answer to first question is that if the probability of heads is \(p\) the then probability of tails is \(1-p\) and since the events are presumed to be independent, the probability of n tails in a row is \((1-p)^n\)

Answer 2

i don't understand what the second part of the question says. looks like something is missing

Answer 3

given a fraction alpha , with 0 <alpha<1, show that the number of flips one must perform to be alpha sure of obtaining atleast one head is
n=ln(1-alpha)/ln(1-p)

Answer 4

ok i am thinking out loud, so maybe this is wrong. probabiliy you get at least one head in \(n\) tosses is one minus probability you get all tails is \(1-(1-p)^n\)

Answer 5

if you put \(X_n\) as the number of heads in n tosses, then \(P(X_n\geq 1)=1-(1-p)^n\) and you want to solve
\[P(X\geq 1)\geq \alpha\] ( i am not sure why alpha has to be a fraction)

Answer 6

so i think this comes down to solving
\[1-(1-p)^n \geq \alpha\]

Answer 7

oh and it works exactly. not bad for first thing in the morning, huh?

Answer 8

first off did what i write make sense? and secondly can you solve this for \(n\)?

Answer 9

how abou introducing ln both sides

Answer 10

yes the solution starts with
\[(1-p)^n\leq 1-\alpha \] then the change of base formula gets it in one step

Answer 11

still not sure why \(\alpha\) has to be a "fraction"

Answer 12

maybe ase 's a probability and it is between 0 and 1

Answer 13

yes but no one says it has to be a "fraction"
it could be \[\sqrt{.5}\] for example. there are an infinite number of numbers between 0 and 1 that are not "fractions"
no matter, i think this is done right?

Answer 14

ya get you point

Answer 15

oh it occurs to me that there is one more thing to say, although it may be obvious
\[(1-p)^n\leq 1-\alpha\]
\[n\ln(1-p)\leq \ln(1-\alpha)\]change of base gives
\[(1-p)^n\leq 1-\alpha \iff n
\[n\geq \frac{\ln(1-\alpha)}{\ln(1-p)}\] switch the inequality because evidently \(\ln(1-p)\leq 0\)

Answer 16

\[(1-p)^n\leq 1-\alpha \iff n\geq \frac{\ln(1-\alpha)}{\ln(1-p)}\] is what i meant

Answer 17

will i be wromg if i put \[\le\]

Answer 18

hell yes

Answer 19

y

Answer 20

forget about the math for a second, think about what it is asking
how many tosses before you are more than \(\alpha\) certain of getting at least one head. clearly your inequality should look line \(n\geq \text{something}\)

Answer 21

in other words n has to be "at least" some number, not "less than" some number. that wouldn't make any sense

Answer 22

oh i get what u mean now

Answer 23

we start with
\[(1-p)^n\leq 1-\alpha\] and since the log is a one to one and increasing function the inequality is preserved when we take the log and get
\[n\ln(1-p)\leq \ln(1-\alpha)\]

Answer 24

but now to solve for \(n\) we must divide both sides by \(\ln(1-p)\) and since \(1-p<1\) we know
\(\ln(1-p)<0\) i.e. it is negative. when you divide by a negative number you have to change the sense of the inequality so you get
\[n\geq \frac{\ln(1-\alpha)}{\ln(1-p)}\]
good thing too, otherwise the answer wouldn't make any sense at all