Let p=0.1 representing the 10% probability that a person is left-handed. The probability of right-handedness is 1-p=0.9.
This is a binomial situation so we can use the expansion of the expression (p+(1-p))^8=1 to represent all the possibilities of handedness. The expansion is p^8+8(1-p)p^7+(8*7/2)(1-p)^2p^6+(8*7*6/(2*3))(1-p)^3p^5+...
Each term is a probability: 1st term is all 8 people are left-handed (none are right-handed); 2nd term is 1 person only is right-handed: 3rd term is 2L and 6R; 4th term is 3L and 5R; 5th term is 4L and 4R. The sum of these is all possibilities which is 100%.
The required probability is (8*7*6*5/(2*3*4))0.1^4*0.9^4=70*0.00006561=0.0045927 or 0.45927%.