# "Compute the probability of x successes in the n independent trials of the experiment" ?

## I just need someone to show me how the by-hand work goes. I can do this via TI-84 since our professor teaches us to use that. Given parameters: part (A) $n = 7$ $p = 0.5$ $x \setminus > 3$ Answer: 0.5 part (B) $n = 12$ $p = 0.35$ $x \setminus \le 4$ Answer: $\setminus \approx 0.5833$

Jul 19, 2017

The formula is \sum_(x=4)^7 ("_x^n) * p^x * 1-p^(n-x)
and \sum_(x=0)^4 ("_x^n) * p^x * 1-p^(n-x)

#### Explanation:

this is a closed form solution but basically what your doing is checking all combinations meaning that for 0 success out of 12 you have only 1 way to accomplish this and that's if they all are failures. 1 success out of 12 you have 12 ways to accomplish this. Each one be success and the rest be failures.... and so on...

but remember that we have the probability of the path to consider and the rule of probabilities for each trial gives us $\left(p 1 , p 2 , p 3. . . , p 12\right) = p 1 \cdot p 2 \ldots . \cdot p 12$ if they are mutually exclusive.

so in the case of 1 we know $p \cdot \left(1 - p\right) \cdot \left(1 - p\right) \ldots = {p}^{1} \cdot {\left(1 - p\right)}^{11}$ and of course we get this 12 times so 12 * p^1 * (1-p)^11 = ("_1^12) * p^1 * 1-p^(12-1)

this generalization is only for the probability of exactly 1 success. The question is asking for additional scenarios that x can be so we just sum across those scenarios. These too follow probability rules such that p1 or p2 = $p 1 + p 2$in the mutually exclusive case.

Thus the binomial distribution is a pattern for following the rules of probability when we know that things are mutually exclusive or independent of each other.