Question

how would you find the probability of an independent event (probability 0.21) happening on or after the seventh attempt in a total number of 10 trials?

Answer 1

Binomial experiment? This would mean that it did not occur the first six times...

Answer 2

yes, it did not occur the first six times. The other event, probability 0.79 occurred, but the event with probability 0.21 occurred on or after the seventh try at least once.

Answer 3

P(not occurring in the first attempt) = 0.79
P(not occurring in the first two attempts) = 0.79 + 0.79^2

Answer 4

If it occurred the other attempts, then the answer is 0.21, since it is independent.

Answer 5

p(7and x=8,9,10)=p(7)*p(8)*p(9)*p(10)
if the above condition is satisfied then that particular event is independent. i mean it may b p(7and8) r p(8 and 9) etc..,

Answer 6

P =(4C1)* (.79)^9*(.21) +(4C2)* (.79)^8*(.21)^2 + (4C3)(.79)^7*(.21)^3 +(4C4)(.79)^6*(.21)^4

Answer 7

its the Binomial Distribution
xCy is x choose y (combinations)

Answer 8

you can also do it this way

\((.79)^6\) for the first 6. then you want at least one 'success' in the last 4. which is the same as 1 minus the probability of no successes in the last 4. this gives

\[(.79)^6(1-(.79)^4)\approx.1484\]