# Compute the following binomial probabilities directly from the formula for b(x, n, p):a) b(3, 8, .6)b) b(5, 8, .6)c) P(3 ≤ X ≤ 5) when n = 8 and p = .6d)P(1 ≤ X) when n = 12 and p = .1

Compute the following binomial probabilities directly from the formula for $$b(x, n, p)$$:

a) $$b(3,\ 8,\ 0.6)$$

b) $$b(5,\ 8,\ 0.6)$$

c) $$\displaystyle{P}{\left({3}≤{X}≤{5}\right)}$$

when $$n = 8$$ and $$p = 0.6$$

d)$$\displaystyle{P}{\left({1}≤{X}\right)}$$ when $$n = 12$$ and $$p = 0.1$$

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Step 1

Solution

a) Given:$$\displaystyle{n}={8},{p}={0.6},{x}={3}$$

$$X \sim$$ Binomeal $$\displaystyle{\left({x}={8},{p}={0.6}\right)}$$ I will use the excel formula "=BINOM.DIST(x, n, p, FALSE)" to find the probability. $$\begin{array}{|c|c|} \hline x & n & p & Excel formula & Probability, P(x) {Round to three decimals} \\ \hline 3 & 8 & 0.6 & =BINOM.DIST(3, 8, 0.6, FALSE) & 0.124 \\ \hline \end{array}$$

Therefore, this probability is $$\displaystyle{P}{\left({3}\right)}={0.124}$$

Step 2

b) Given: $$\displaystyle{n}={8},{p}={0.6},{x}={5}$$ $$X\sim$$Binomial $$\displaystyle{\left({x}={8},{p}={0.6}\right)}$$ I will use the excel formula "=BINOM.DIST(x, n, p, FALSE)" to find the probability.

$$\begin{array}{|c|c|} \hline x & n & p & Excel formula & Probability, P(x) {Round to three decimals} \\ \hline 5 & 8 & 0.6 & =BINOM.DIST(5, 8, 0.6, FALSE) & 0.279 \\ \hline \end{array}$$

Therefore, this probability is $$\displaystyle{P}{\left({5}\right)}={0.279}$$

Step 3

c) Given: $$\displaystyle{n}={8},{p}={0.6}$$

$$X\sim$$ Binomial $$\displaystyle{\left({n}={8},{p}={0.6}\right)}$$ find: $$\displaystyle{P}{\left({3}\leq{X}\leq{5}\right)}={P}{\left({3}\right)}+{P}{\left({4}\right)}+{P}{\left({5}\right)}$$ I will use the excel formula "=BINOM.DIST(x, n, p, FALSE)" to find the probability. $$\begin{array}{|c|c|} \hline x & n & p & Excel formula & Probability, P(x) Round\ to\ three\ decimals \\ \hline 3 & 8 & 0.6 & =BINOM.DIST(3, 8, 0.6, FALSE) & 0.124 \\ \hline 4 & 8 & 0.6 & =BINOM.DIST(4, 8, 0.6, FALSE) & 0.232 \\ \hline 5 & 8 & 0.6 & =BINOM.DIST(5, 8, 0.6, FALSE) & 0.279 \\ \hline \end{array}$$

Therefore, this probability is $$\displaystyle{P}{\left({3}\leq{X}\leq{5}\right)}={0.124}+{0.232}+{0.279}$$
$$\displaystyle{P}{\left({3}\leq{X}\leq{5}\right)}={0.635}$$

Step 4

d) Given: $$\displaystyle{n}={12},{p}={0.1}$$

$$X\sim$$ Binomial $$\displaystyle{\left({n}={12},{p}={0.1}\right)}$$ find: $$\displaystyle{P}{\left({1}\leq{X}\right)}=?$$ Use complement rule, $$\displaystyle{P}{\left({1}\leq{X}\right)}={1}-{P}{\left({0}\right)}$$ I will use the excel formula "=BINOM.DIST(x, n, p, FALSE)" to find the probability. $$\begin{array}{|c|c|} \hline x & n & p & Excel formula & Probability, P(x) Round\ to\ three\ decimals \\ \hline 0 & 12 & 0.1 & =BINOM.DIST(0, 12, 0.1, FALSE) & 0.282 \\ \hline \end{array}$$

Therefore, this probability is $$\displaystyle{P}{\left({1}\leq{X}\right)}={1}-{0.282}$$
$$\displaystyle{P}{\left({1}\leq{X}\right)}={0.718}$$ Answer: 0.718