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Sagot :
To construct a binomial probability distribution for [tex]\( n = 6 \)[/tex] and [tex]\( p = 0.6 \)[/tex], we need to find the probabilities for each value of [tex]\( x \)[/tex] from [tex]\( 0 \)[/tex] to [tex]\( 6 \)[/tex].
The probabilities are given by:
[tex]\[ P(X = x) = \binom{n}{x} p^x (1 - p)^{n - x} \][/tex]
For [tex]\( n = 6 \)[/tex] and [tex]\( p = 0.6 \)[/tex], we have the following probabilities for each [tex]\( x \)[/tex]:
[tex]\[ \begin{array}{c|l|l} x & P(X = x) \\ \hline 0 & 0.0041 \\ \hline 1 & 0.0369 \\ \hline 2 & 0.1382 \\ \hline 3 & 0.2765 \\ \hline 4 & 0.3110 \\ \hline 5 & 0.1866 \\ \hline 6 & 0.0467 \\ \hline \end{array} \][/tex]
These probabilities have been rounded to four decimal places.
So, the binomial probability distribution is:
[tex]\[ \begin{array}{c|c} x & P(X = x) \\ \hline 0 & 0.0041 \\ \hline 1 & 0.0369 \\ \hline 2 & 0.1382 \\ \hline 3 & 0.2765 \\ \hline 4 & 0.3110 \\ \hline 5 & 0.1866 \\ \hline 6 & 0.0467 \\ \hline \end{array} \][/tex]
The probabilities are given by:
[tex]\[ P(X = x) = \binom{n}{x} p^x (1 - p)^{n - x} \][/tex]
For [tex]\( n = 6 \)[/tex] and [tex]\( p = 0.6 \)[/tex], we have the following probabilities for each [tex]\( x \)[/tex]:
[tex]\[ \begin{array}{c|l|l} x & P(X = x) \\ \hline 0 & 0.0041 \\ \hline 1 & 0.0369 \\ \hline 2 & 0.1382 \\ \hline 3 & 0.2765 \\ \hline 4 & 0.3110 \\ \hline 5 & 0.1866 \\ \hline 6 & 0.0467 \\ \hline \end{array} \][/tex]
These probabilities have been rounded to four decimal places.
So, the binomial probability distribution is:
[tex]\[ \begin{array}{c|c} x & P(X = x) \\ \hline 0 & 0.0041 \\ \hline 1 & 0.0369 \\ \hline 2 & 0.1382 \\ \hline 3 & 0.2765 \\ \hline 4 & 0.3110 \\ \hline 5 & 0.1866 \\ \hline 6 & 0.0467 \\ \hline \end{array} \][/tex]
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