Question

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?

Answer 1

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?

Answer 2

ahh ok

Answer 3

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).

Answer 4

is the ans 40.6

Answer 5

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.

Answer 6

is it the right ans ..... actually in frst attempt i got the the wrong ans

Answer 7

it's right.

Answer 8

thank you soooo much 4 helping me

Answer 9

yw

Answer 10

0.0236 is wrong

Answer 11

what's your answer ?

Answer 12

i got wrong when i entered this answer

Answer 13

@pakinam i gt it correct