Probability question?

Question

miguel tosses 10 coins per each turn in infinite turns. What is the probability that 10 heads will not appear in any 1 turn?

Answer 1

total number of results in each turn 2^10= 1024
favourable results(10 heads) = 1
Probability of not getting 10 heads in a turn = 1-(1/1024) = 1023/1024
=> Probability of not getting 10H in 1st turn= 1023/1024
=> Probability of not getting 10H in 2nd turn= (1023/1024)^2
Similarily probability of not getting 10H in nth turn= (1023/1024)^n,
which approaches 0 when n approaches ∞

Answer 2

If they take infinite terms, the probability of not having 10 heads out of 10 is equal to 0.

Answer 3

well, it is probably easier to first calculate the probability that 10 heads will appear in a single turn

that would, of course, be

.5*.5* .5*.5*.5* .5*.5*.5*.5* .5*.5=.5^10=.000977

the probability that that wouldn't happen includes all the other outcomes so it can be calculated as 1-.000977= .9990

so the odds are 99.90 % for any one turn that you will not get all heads

Answer 4

if he repeats the trial infinately many times, then the probability that it will never happen goes to 0

Answer 5

use nCr formula or do the branching thing you will get ur answer