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I throw darts repeatedly. Assume that on each single throw, my chance of hitting the bullseye is 10%, independently of all other throws. I decide to throw until I have hit the bullseye 3 times. What is the chance that I throw exactly 30 times?

Mathematics
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you have this info: 30 throws, last throw is a BE (bullseye). P(BE)=0.1. But you don't know where the other two BE's are. So it's like that: (29 hits with 2BE's) and 1BE. Does this help?
ahh ok
you have to use a combintation "C(n,k)" factor in the computation of the probability of (29 throws with 2 BE's), then multiply by P(1BE).

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Other answers:

is the ans 40.6
No, the answer must be a number of [0,1]. but it should be \(P(\text{2 BE's in 29 throws})\times P(\text{1 BE}) = \binom{29}{2}0.1^{2}\,0.9^{27} \times 0.1\). I got 0.02360879322323428... as anwer.
is it the right ans ..... actually in frst attempt i got the the wrong ans
it's right.
thank you soooo much 4 helping me
yw
0.0236 is wrong
what's your answer ?
i got wrong when i entered this answer
@pakinam i gt it correct

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