I'm working on two bionmial distribution questions about unfair coin, where I got confused for one of em by its wording.

Unfair coin: You have a coin with which you are 2 times more likely to get heads than tails. You flip the coin 100 times. What is the probability of getting 20 tails? What is the probability of getting at least one heads?

for 20 tails:

P(x=20) = 100!/(20!)(80!) * (1/3)^20 * (2/3)^80

anything wrong here?

for 'at least one head':

This is where I feel little lost. do I get the sum of probability from getting 1 head to getting 100 heads? Is there any other trick to find the answer?


Your expression for $P(x = 20)$ looks good to me.

Hint For the other question: The complement of the event "at least one head" is the event "no heads".

  • $\begingroup$ P(x=100) = 100!/100! * (1/3)^100 = 1/(3^100) = no heads. Therefore, 1 - 1/(3^100) ?? $\endgroup$ – Daniel Kim Dec 19 '18 at 5:27
  • $\begingroup$ Looks good to me! $\endgroup$ – Travis Dec 19 '18 at 5:45
  • $\begingroup$ thank you Travis! $\endgroup$ – Daniel Kim Dec 19 '18 at 6:22
  • $\begingroup$ You're welcome, I'm glad you found it useful! $\endgroup$ – Travis Dec 19 '18 at 6:43
  • $\begingroup$ Travis, do you mind to check my question about exponential distribution as well? math.stackexchange.com/questions/3046085/… $\endgroup$ – Daniel Kim Dec 19 '18 at 6:47

What you can do is calculate the probability there is absolutely no head, then do one minus the probability of no head. $$P(\text{no head}) = \left(\frac13\right)^{100},$$ so $$P(\text{at least 1 head}) = 1 - P(\text{no head}).$$

  • $\begingroup$ thank you! I was also working on same thing via Travis's hint. $\endgroup$ – Daniel Kim Dec 19 '18 at 5:28

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