Here's the question:
A spinner is constructed so that the numbers 1 to 10 are possible and the probability of seeing a number k+1 is twice as likely as seeing the number k (for k=1,2,....9).
Find the probability of getting A) 1 B) 10 (answers correct to 3 decimal places).
I had to read this a few time before I got some inkling of it but I'm not sure I'm correct. I thought if the k+1th term is twice as likely as the kth term then we are dealing with a geometric series which increases by a ration of 2:1 so that would mean that the probability of 10 would be 2^9 x the first probability. All the probabilities must add up to 1. So I then thought I could divide 1 by 2^9 to get the first value which came out as 0.001953...which I left as 0.001 (should I have rounded it up to 0.002?) . I then multiplied this number by 2^5 to get 0.512. The answers in the book were 0.001 and 0,500 which looks they've rounded to significant figures rather than three decimal places but I might be wrong.
Not all all sure I'm correct here, so help appreciated!
A spinner is constructed so that the numbers 1 to 10 are possible and the probability of seeing a number k+1 is twice as likely as seeing the number k (for k=1,2,....9).
Find the probability of getting A) 1 B) 10 (answers correct to 3 decimal places).
I had to read this a few time before I got some inkling of it but I'm not sure I'm correct. I thought if the k+1th term is twice as likely as the kth term then we are dealing with a geometric series which increases by a ration of 2:1 so that would mean that the probability of 10 would be 2^9 x the first probability. All the probabilities must add up to 1. So I then thought I could divide 1 by 2^9 to get the first value which came out as 0.001953...which I left as 0.001 (should I have rounded it up to 0.002?) . I then multiplied this number by 2^5 to get 0.512. The answers in the book were 0.001 and 0,500 which looks they've rounded to significant figures rather than three decimal places but I might be wrong.
Not all all sure I'm correct here, so help appreciated!