 # Assume that a procedure yields a binomial distribution. With n=6 trials and a probability of success of p=0.5 . Use a binomial probability table to find the probability that the number of successes x is exactly 3.

The target of this question is to find the probability using a binomial distribution table. With the given number of trials and probability of success, the exact probability of a number is calculated.

Moreover, this question is based on the concepts of statistics. Trails are a single performance of well-defined experiments such as the flipping of a coin. Probability is simply how likely something is to happen, for example a head or a tail after the coin is flipped.

Finally, a binomial distribution can be thought of as the probability of a SUCCESS or FAIL result in an experiment or survey that is conducted several times. For a discrete variable “X”, the formula of a binomial distribution is as follows:

$P(X = x) = \binom{n}{x}p^x (1-p)^{n-x}; x = 0, 1, … , n$

where,

$n$ = number of trials,

$p$ = probability of success, and

$q$ = probability of failure obtained as $q = (1 – p)$.

We have all the above information given in the question as:

$n = 6$,

$p = 0.5$, and

$q = 0.5$.

Therefore, using the binomial distribution probability for the number of success x exactly 3, this can be calculated as follows:

$P(X = 3) = \binom{6}{3}(0.5)^3 (1 – 0.5)^{6 – 3}; as x = 3$

$= \dfrac{6!}{3! (6 – 3)!}(0.5)^3(0.5)^3$

$= \dfrac{6!}{3! (3)!}(0.5)^3 (0.5)^3$

$= \dfrac{720}{36}(0.5)^6$

$= 20 (0.5)^6$

$= 20 (0.0156)$

$= 0.313$

Therefore, $P(X = x) = 0.313$.

## Numerical Results

The probability that the amount of successes equals $x$ is exactly 3, using the binomial distribution table is:

$P(X = x) = 0.313$

## Example

Suppose that a procedure yields a binomial distribution with a trial repeated $n = 7$ times. Use the binomial probability formula to find the probability of $k = 5$ successes given the probability $p = 0.83$ of success on a single trial.

Solution

As we have all the given information, we can use the binomial distribution formula:

$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}; x = 0, 1, … , n$

$P(X = 5) = \binom{7}{5} (0.83)^5 (1 – 0.83)^{7 – 5}$

$= \dfrac{7!}{5!(7 – 5)!} (0.83)^5 (0.17)^2$

$= \dfrac{7!}{5! (2)!} (0.83)^5 (0.17)^2$

$= \dfrac{5040}{240} (0.444) (0.0289)$

$= 21 (0.444) (0.0289)$

$= 0.02694$

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