If n=100 and p=0.5, then find the probability that the number of successes X is greater than 60 ; i.e. p(x > 60)

Question

please help me

show all work thanks

Answer 1

Assuming each trial is independent and there are only successes and failures, then we can assume X is binomially distributed with p=0.5 and n=100. By complementarity, we can say that P(X>60) = 1-P(X<=60).

Then we can focus on finding P(X<=60), which includes the probability of X being equal to anything less than or equal to 60 successes. Since different numbers of successes represent different events without intersection, we can sum the probability of these events -- in other words, add the probability of X=0, probability of X=1, probability of X=2, all the way to the probability of X=60.

For each event, the probability that X takes a particular value is simply found using the binomial formula itself, in other words, the binomial coefficient (which I write as nCr(100,x) to be consistent with the way that Desmos's Graphing Calculator, desmos.com/calculator, writes it) multiplied by (1/2)^x multiplied by (1/2)^(100-x), so you have:

SUM for X from 0 to 60 of: nCr(100, x) * (0.5)^x * (0.5)^(100-x) which equals 0.982

You then subtract this from 1 to get your final answer: 0.018

Alternately you could just use the summation to directly compute P(X>60) by doing a summation of all events from X=61 through X=100, it would look like this:

SUM for X from 61 to 100 of: nCr(100, x) * (0.5)^x * (0.5)^(100-x) which equals 0.018