Determine P(X=k) for a certain k and for X following a given binomial distribution, using the formula Statistics & Probabilities

The random variable X is the count of the number of successes in the n trials.

The probability that exactly k successes are obtained in n trials of a binomial random variable is:

P\left(X=k\right)=\begin{pmatrix} n \cr\cr k \end{pmatrix}\cdot p^{k}\cdot\left(1-p\right)^{n-k}=\dfrac{n!}{k!\cdot\left(n-k\right)!}\cdot p^{k}\cdot\left(1-p\right)^{n-k}

In our problem, X follows \mathcal{B}\left(4{,}0.2\right) . Therefore:

P\left(X=3\right)=\begin{pmatrix} 4 \cr\cr 3 \end{pmatrix}\cdot \left(0.2\right)^{3}\cdot\left(1-0.2\right)^{1}

P\left(X=3\right)=\dfrac{4!}{3!\cdot1!}\cdot\left(0.2\right)^{3}\cdot\left(0.8\right)^{1}

P\left(X=3\right)=4\cdot\dfrac{1}{5}\cdot\dfrac{1}{5}\cdot\dfrac{1}{5}\cdot\dfrac{4}{5}

The random variable X is the count of the number of successes in the n trials.

The probability that exactly k successes are obtained in n trials of a binomial random variable is:

P\left(X=k\right)=\begin{pmatrix} n \cr\cr k \end{pmatrix}\cdot p^{k}\cdot\left(1-p\right)^{n-k}=\dfrac{n!}{k!\cdot\left(n-k\right)!}\cdot p^{k}\cdot\left(1-p\right)^{n-k}

In our problem, X follows \mathcal{B}\left(3{,}0.5\right) . Therefore:

P\left(X=2\right)=\begin{pmatrix} 3 \cr\cr 2 \end{pmatrix}\cdot \left(0.5\right)^{2}\cdot\left(1-0.5\right)^{1}

P\left(X=2\right)=\dfrac{3!}{2!\cdot1!}\cdot\left(0.5\right)^{2}\cdot\left(0.5\right)^{1}

P\left(X=2\right)=3\cdot\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot\dfrac{1}{2}=\dfrac{3}{8}

The random variable X is the count of the number of successes in the n trials.

The probability that exactly k successes are obtained in n trials of a binomial random variable is:

P\left(X=k\right)=\begin{pmatrix} n \cr\cr k \end{pmatrix}\cdot p^{k}\cdot\left(1-p\right)^{n-k}=\dfrac{n!}{k!\cdot\left(n-k\right)!}\cdot p^{k}\cdot\left(1-p\right)^{n-k}

In our problem, X follows \mathcal{B}\left(4{,}0.3\right) . Therefore:

P\left(X=1\right)=\begin{pmatrix} 4 \cr\cr 1 \end{pmatrix}\cdot \left(0.3\right)^{1}\cdot\left(1-0.3\right)^{3}

P\left(X=1\right)=\dfrac{4!}{1!\cdot3!}\cdot\left(0.3\right)^{1}\cdot\left(0.7\right)^{3}

P\left(X=1\right)=4\cdot\dfrac{3}{10}\cdot\dfrac{7}{10}\cdot\dfrac{7}{10}\cdot\dfrac{7}{10}=\dfrac{1\ 029}{2\ 500}

The random variable X is the count of the number of successes in the n trials.

The probability that exactly k successes are obtained in n trials of a binomial random variable is:

P\left(X=k\right)=\begin{pmatrix} n \cr\cr k \end{pmatrix}\cdot p^{k}\cdot\left(1-p\right)^{n-k}=\dfrac{n!}{k!\cdot\left(n-k\right)!}\cdot p^{k}\cdot\left(1-p\right)^{n-k}

In our problem, X follows \mathcal{B}\left(5{,}0.4\right) . Therefore:

P\left(X=3\right)=\begin{pmatrix} 5 \cr\cr 3 \end{pmatrix}\cdot \left(0.4\right)^{3}\cdot\left(1-0.4\right)^{2}

P\left(X=3\right)=\dfrac{5!}{3!\cdot2!}\cdot\left(0.4\right)^{3}\cdot\left(0.6\right)^{2}

P\left(X=3\right)=10\cdot\dfrac{2}{5}\cdot\dfrac{2}{5}\cdot\dfrac{2}{5}\cdot\dfrac{3}{5}\cdot\dfrac{3}{5}=\dfrac{144}{625}

The random variable X is the count of the number of successes in the n trials.

The probability that exactly k successes are obtained in n trials of a binomial random variable is:

P\left(X=k\right)=\begin{pmatrix} n \cr\cr k \end{pmatrix}\cdot p^{k}\cdot\left(1-p\right)^{n-k}=\dfrac{n!}{k!\cdot\left(n-k\right)!}\cdot p^{k}\cdot\left(1-p\right)^{n-k}

In our problem, X follows \mathcal{B}\left(4{,}0.1\right) . Therefore:

P\left(X=2\right)=\begin{pmatrix} 4 \cr\cr2 \end{pmatrix}\cdot \left(0.1\right)^{2}\cdot\left(1-0.1\right)^{2}

P\left(X=2\right)=\dfrac{4!}{2!\cdot2!}\cdot\left(0.1\right)^{2}\cdot\left(0.9\right)^{2}

P\left(X=2\right)=6\cdot\dfrac{1}{10}\cdot\dfrac{9}{10}\cdot\dfrac{9}{10}=\dfrac{243}{5\ 000}

The random variable X is the count of the number of successes in the n trials.

The probability that exactly k successes are obtained in n trials of a binomial random variable is:

P\left(X=k\right)=\begin{pmatrix} n \cr\cr k \end{pmatrix}\cdot p^{k}\cdot\left(1-p\right)^{n-k}=\dfrac{n!}{k!\cdot\left(n-k\right)!}\cdot p^{k}\cdot\left(1-p\right)^{n-k}

In our problem, X follows \mathcal{B}\left(6{,}0.5\right) . Therefore:

P\left(X=4\right)=\begin{pmatrix} 6 \cr\cr4 \end{pmatrix}\cdot \left(0.5\right)^{4}\cdot\left(1-0.5\right)^{2}

P\left(X=4\right)=\dfrac{6!}{4!\cdot2!}\cdot\left(0.5\right)^{4}\cdot\left(0.5\right)^{2}

P\left(X=4\right)=15\cdot\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot\dfrac{1}{2}=\dfrac{15}{64}