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How many number of times a coin must be tossed to get head with probability greater than $0.9$

[probability of getting Head/Tail is $\frac{1}{2}$]

i could think of $(\frac{1}{2})^N$ [$N$ no of times] but that'll never give the result!

how to approach this? [I'm beginner so need explanation from basic and an example would help a lot]

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2 Answers 2

I'll give you an example to illustrate. Suppose you throw the coin 3 times. The probability that ~at least one~ of the tosses yield head will be the number of cases at least one toss yields head divided by the total number of cases. You have 8 distinct cases (HHH, HHT, HTT, HTH, THH, TTH, THT, TTT) and amongst them you have 7 with at least one head.

See if that helps you.

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Can you see that the total number of cases will always be $2^N$ and that all but one case will have a head? –  José Siqueira Jul 11 '13 at 15:57

Suppose you toss a coin $n$ times; let $X_k$ be 1 if the result of the $k$-th toss is heads, and 0 otherwise. Now, let $Y_n = \sum_{k = 1}^n X_k$. What you're asking is basically, what's the probability that $Y_n \ne 0$. $Y_n = 0$ happens w.p. $\left(\frac{1}{2}\right)^n$. So you need to check when $\left(\frac12\right)^n< 0.1$. Solve this and you'll get an answer...

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