# How can I prove this question using the Binomial Theorem of Newton?

I'm guessing I have to use the Binomial Theorem of Newton, but I'm getting stuck somewhere.

### Data

• $p$, $q$ are positive real numbers with $p+q=1$, and
• $k, n$ are natural numbers with $k\leq n$.

### Prove the following:

$$\sum_{j=k}^n{n\choose j}p^jq^{n-j}\leq {n\choose k}p^k.$$

I was trying to change the formula to match Newton's Binomium. So, I changed the counter to j=0 to n-k. But then I have to choose j+k from n. (and some other changes which aren't important) The counter goes to n-k and the choice is from n. But: In Newton's formula it is both n written here, so they should not be different values and n is different from n-k.

• "I'm getting stuck somewhere" Where exactly? – Did Oct 26 '14 at 12:39
• I was trying to change the formula to match Newton's Binomium. So, I changed the counter to j=0 to n-k. But then I have to choose j+k from n. (and some other changes which aren't important) The counter goes to n-k and the choice is from n. But: In Newton's formula it is both n written here, so they should not be different values and n is different from n-k.. – Shadowser Oct 26 '14 at 12:42

The left side is equal to the probability of at least $k$ successes in $n$ Bernoulli trials. The inequality is consequence of the fact: $P(A_1\cup A_2\cup\ldots)\le P(A_1)+P(A_2)+\ldots$