#### Answer

See below

#### Work Step by Step

The standard formula: $P(number-of-successes)=\frac{n!}{(n-k)!k!}.p^k(1-p)^{n-k}$
The probability that someone is Rh+ is $p=0.37+0.34+0.1+0.04=0.85$
It means that 5,6,7,8,9,10 can be type Rh+.
Substituting $p=0.85,n=10,k=5$ we have:
$P(5)=\frac{10!}{(10-5)!5!}.(0.85)^5(1-0.85)^{10-5}\approx0.0085$
Substituting $p=0.5,n=10,k=6$ we have: $P(6)=0.0401$
Substituting $p=0.5,n=10,k=7$ we have: $P(7)=0.1298$
Substituting $p=0.5,n=10,k=8$ we have: $P(8)=0.2759$
Substituting $p=0.5,n=10,k=9$ we have: $P(9)=0.3474$
Substituting $p=0.5,n=10,k=10$ we have: $P(10)=0.1969$
Hence,
$P(5,6,7,8,9,10)=0.0085+0.0401+0.1298+0.2759+0.3474+0.1969=0.9986$
The probability that at least 5 people are Rh+ is $0.9986$