Answer
See below
Work Step by Step
The standard formula: $P(number-of-successes)=\frac{n!}{(n-k)!k!}.p^k(1-p)^{n-k}$
Substituting $p=0.36,n=10,k=3$ we have:
$P(3)=\frac{10!}{(10-3)!3!}.(0.36)^3(1-0.36)^{10-3}\approx0.2426$
Substituting $p=0.36,n=10,k=4$ we have:
$P(4)=\frac{10!}{(10-4)!4!}.(0.36)^4(1-0.36)^{10-4}\approx0.24239$
Substituting $p=0.36,n=10,k=6$ we have:
$P(6)=\frac{10!}{(10-6)!6!}.(0.36)^6(1-0.36)^{10-6}\approx0.07669$
Substituting $p=0.36,n=10,k=7$ we have:
$P(7)=\frac{10!}{(10-7)!7!}.(0.36)^7(1-0.36)^{10-7}\approx0.02465$
We can see that 3 is the most likely to be the number of successes. Hence, the correct option is A.
