Show that p(y) p(y − 1) = (n − y + 1) p yq > 1 if y < (n + 1)p. This establishes that p(y) > p(y − 1) if y is small (y < (n + 1)p) and p(y) < p(y − 1) if y is large (y > (n + 1) p). Thus, successive binomial probabilities increase for a while and decrease from then on. (n + 1)p > y

Question is not well presented

Consider the binomial distribution with n trials and P(S) = p.

Show that

p(y) /p(y − 1) = (n − y + 1)p /yq > 1 if y < (n + 1)p.

This establishes that p(y) > p(y − 1) if y is small (y < (n + 1)p) and p(y) < p(y − 1) if y is large (y > (n + 1) p). Thus, successive binomial probabilities increase for a while and decrease from then on.

See Explanation Below

Step-by-step explanation:

Given that

p(y) /p(y − 1) = (n − y + 1)p /yq > 1 if y < (n + 1)p.

First, we make the following assumption

p(y) /p(y − 1) = (n − y + 1)p /yq = 1;

So, we have

(n − y + 1)p /yq = 1

(n − y + 1)p = yq

Note that p + q = 1;.

So, q = 1 - p

Substitute 1-p for q in the above expression

(n − y + 1)p = y(1-p)

np - py + p = y - py

Solve for y

Collect like terms

np + p = y - py + py

np + p = y;

So, y = np + p

y = (n + 1)p

From the above,

We know that

p(y) /p(y − 1) = 1

if y = (n + 1)p.

Similarly, we've also obtain that

p(y) /p(y − 1) > 1 if y < (n + 1)p.