Answer :
The conditions are suitable to be modelled using a binomial distribution.
p=0.43
n=15
x=5
P(x=5)
=C(15,5)*0.43^5*(1-0.43)^(15-5)
=15!/(5!(15-5)!)*0.43^5*(1-0.43)^(15-5)
=3003*0.0147008*0.003620333
=0.1598
p=0.43
n=15
x=5
P(x=5)
=C(15,5)*0.43^5*(1-0.43)^(15-5)
=15!/(5!(15-5)!)*0.43^5*(1-0.43)^(15-5)
=3003*0.0147008*0.003620333
=0.1598