Probability Formula

[turophile]

Here's the problem:

According to the theory of probability, if p is the probability of success in a single trial of an experiment then P, the probability of exactly r successes in a series of n independent trials (0 ? r ? n), is given by:

P = C[sub:15ismcwq]n , r[/sub:15ismcwq]p[sup:15ismcwq]r[/sup:15ismcwq](1 – p)[sup:15ismcwq]n - r[/sup:15ismcwq],

where C[sub:15ismcwq]n , r[/sub:15ismcwq] depends on n and r but not on p (explicitly, C[sub:15ismcwq]n , r[/sub:15ismcwq] = n!/[r!(n – r)!]). P itself is a function of the three variables; but for fixed n and r it is a function of p alone. Prove that for fixed n and r, P is a maximum when p = r/n.

My work so far:

Since P is a function of p alone:

P = C[sub:15ismcwq]n , r[/sub:15ismcwq]p[sup:15ismcwq]r[/sup:15ismcwq](1 – p)[sup:15ismcwq]n - r[/sup:15ismcwq]

? P' = C[sub:15ismcwq]n , r[/sub:15ismcwq] ? [p[sup:15ismcwq]r[/sup:15ismcwq] ? (n – r) ? (1 – p)[sup:15ismcwq]n - r - 1[/sup:15ismcwq] ? (– 1) + (1 – p)[sup:15ismcwq]n - r[/sup:15ismcwq] ? r ? p[sup:15ismcwq]r - 1[/sup:15ismcwq]]

= C[sub:15ismcwq]n , r[/sub:15ismcwq] ? [p ? p[sup:15ismcwq]r - 1[/sup:15ismcwq] ? (n – r) ? (1 – p)[sup:15ismcwq]n - r - 1[/sup:15ismcwq] ? (– 1) + (1 – p) ? (1 – p)[sup:15ismcwq]n - r - 1[/sup:15ismcwq] ? r ? p[sup:15ismcwq]r - 1[/sup:15ismcwq]]

= C[sub:15ismcwq]n , r[/sub:15ismcwq] ? p[sup:15ismcwq]r - 1[/sup:15ismcwq] ? (1 – p)[sup:15ismcwq]n - r - 1[/sup:15ismcwq] ? [r ? (1 – p) – (n – r) ? p] ? 0

? r ? (1 – p) = p ? (n – r)
? r – p ? r = p ? n – p ? r
? p ? n – p ? r + p ? r = r
? p ? n = r
? p = r/n

So I've got p = r/n, but how do I show p is a maximum rather than a minimum? Also, aren't p = 0 and p = 1 also critical points? How do I test them?

 

[Deleted member 4993]

Here's the problem:

According to the theory of probability, if p is the probability of success in a single trial of an experiment then P, the probability of exactly r successes in a series of n independent trials (0 ? r ? n), is given by:

P = C[sub:2e6irbji]n , r[/sub:2e6irbji]p[sup:2e6irbji]r[/sup:2e6irbji](1 – p)[sup:2e6irbji]n - r[/sup:2e6irbji],

where C[sub:2e6irbji]n , r[/sub:2e6irbji] depends on n and r but not on p (explicitly, C[sub:2e6irbji]n , r[/sub:2e6irbji] = n!/[r!(n – r)!]). P itself is a function of the three variables; but for fixed n and r it is a function of p alone. Prove that for fixed n and r, P is a maximum when p = r/n.

My work so far:

Since P is a function of p alone:

P = C[sub:2e6irbji]n , r[/sub:2e6irbji]p[sup:2e6irbji]r[/sup:2e6irbji](1 – p)[sup:2e6irbji]n - r[/sup:2e6irbji]

? P' = C[sub:2e6irbji]n , r[/sub:2e6irbji] ? [p[sup:2e6irbji]r[/sup:2e6irbji] ? (n – r) ? (1 – p)[sup:2e6irbji]n - r - 1[/sup:2e6irbji] ? (– 1) + (1 – p)[sup:2e6irbji]n - r[/sup:2e6irbji] ? r ? p[sup:2e6irbji]r - 1[/sup:2e6irbji]]

= C[sub:2e6irbji]n , r[/sub:2e6irbji] ? [p ? p[sup:2e6irbji]r - 1[/sup:2e6irbji] ? (n – r) ? (1 – p)[sup:2e6irbji]n - r - 1[/sup:2e6irbji] ? (– 1) + (1 – p) ? (1 – p)[sup:2e6irbji]n - r - 1[/sup:2e6irbji] ? r ? p[sup:2e6irbji]r - 1[/sup:2e6irbji]]

= C[sub:2e6irbji]n , r[/sub:2e6irbji] ? p[sup:2e6irbji]r - 1[/sup:2e6irbji] ? (1 – p)[sup:2e6irbji]n - r - 1[/sup:2e6irbji] ? [r ? (1 – p) – (n – r) ? p] ? 0

? r ? (1 – p) = p ? (n – r)
? r – p ? r = p ? n – p ? r
? p ? n – p ? r + p ? r = r
? p ? n = r
? p = r/n

So I've got p = r/n, but how do I show p is a maximum rather than a minimum? Also, aren't p = 0 and p = 1 also critical points? How do I test them?

Also, aren't p = 0 and p = 1 also critical points? How do I test them - easiest thing is to evaluate P at those points.

As a back-up, calculate (p')' or p" - which will be <0 for maximum.