Step-by-step explanation:
I assume this means exactly one wrong number in the 10 attempts.
the probability to get a wrong number is 0.15 (15%).
therefore, the probability to get a correct number is
1 - 0.15 = 0.85
in 10 tries to get 1 wrong number (and 9 correct numbers) looks like this :
0.15×0.85⁹ = 0.034742542...
this would be the probability to e.g. get a wrong number at the first attempt, and all other attempts are correct.
how many different constellations can we have that way ?
10.
because the wrong number could be given at any of the 10 attempts.
so, the probability to get exactly 1 wrong answer is
10×0.15×0.85⁹ = 0.34742542... ≈ 35%
now, if the question is about the probability to get at least one number wrong, then we can say this is the opposite of the probability to get all 10 numbers correctly.
the probability to get all 10 numbers correctly is
0.85¹⁰ = 0.196874404... ≈ 20%
and therefore, the probability to get at least 1 number wrong is
1 - 0.196874404... = 0.803125596... ≈ 80%
FYI :
this would be, of course, also the result of we did it the complex way :
the sum of the probabilities of
exactly 1 number wrong
exactly 2 numbers wrong
...
exactly 9 numbers wrong
all 10 numbers wrong
and for each of the cases of n wrong numbers the probability is
(10! / (n! × (10-n)!)) × 0.15^n × 0.85^(10-n)
this represents the constellation of n numbers wrong (and 10-n numbers right) multiplied by the number of possible combinations of picking n numbers out of 10.
and again, all this summed up (1 <= n <= 10) is the same as
0.803125596...
