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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.

Expert Answer

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:

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

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

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

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

\[ = \dfrac{5040}{240} (0.444)  (0.0289) \]

\[ = 21 (0.444)  (0.0289) \]

\[ = 0.02694 \]

Images/ Mathematical drawings are created with Geogebra. 

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