Question

1. compute the probability of x successes, using table b in appendix a.

a. n = 15, p = 0.80, x = 12
b. n = 17, p = 0.05, x = 0
c. n = 20, p = 0.50, x = 10
d n = 16, p = 0.20, x = 3

1. compute the probability of x successes, using the binomial formula.

a. n = 6, x = 3, p = 0.03
b. n = 4, x = 2, p = 0.18
c. n = 5, x = 3, p = 0.63

1. compute the probability of x successes, using the binomial formula.

a. n = 9, x = 0, p = 0.42
b. n = 10, x = 5, p = 0.37
for exercises 7 through 16, assume all variables are binomial. (note: if values are not found in table b of appendix a, use the binomial formula.)

1. today’s marriages a television commercial claims that 1 out of 5 of “today’s marriages” began as an online relationship. assuming that this is true, calculate the following for eight randomly selected “today’s marriages”.

Explanation:

Step1: Recall binomial probability formula

The binomial probability formula is $P(X)={n\choose X}p^{X}(1 - p)^{n - X}$, where ${n\choose X}=\frac{n!}{X!(n - X)!}$.

Step2: Solve 5a

For $n = 6$, $X = 3$, $p=0.03$:
First, calculate ${6\choose 3}=\frac{6!}{3!(6 - 3)!}=\frac{6!}{3!3!}=\frac{6\times5\times4}{3\times2\times1}=20$.
Then, $(1 - p)=1 - 0.03 = 0.97$, $p^{X}=(0.03)^{3}=0.03\times0.03\times0.03 = 0.000027$, $(1 - p)^{n - X}=(0.97)^{3}=0.97\times0.97\times0.97\approx0.912673$.
$P(3)={6\choose 3}(0.03)^{3}(0.97)^{3}=20\times0.000027\times0.912673\approx0.000493$.

Step3: Solve 5b

For $n = 4$, $X = 2$, $p = 0.18$:
${4\choose 2}=\frac{4!}{2!(4 - 2)!}=\frac{4!}{2!2!}=\frac{4\times3}{2\times1}=6$.
$(1 - p)=1 - 0.18 = 0.82$, $p^{X}=(0.18)^{2}=0.0324$, $(1 - p)^{n - X}=(0.82)^{2}=0.82\times0.82 = 0.6724$.
$P(2)={4\choose 2}(0.18)^{2}(0.82)^{2}=6\times0.0324\times0.6724\approx0.131$.

Step4: Solve 5c

For $n = 5$, $X = 3$, $p = 0.63$:
${5\choose 3}=\frac{5!}{3!(5 - 3)!}=\frac{5!}{3!2!}=\frac{5\times4}{2\times1}=10$.
$(1 - p)=1 - 0.63 = 0.37$, $p^{X}=(0.63)^{3}=0.63\times0.63\times0.63\approx0.250047$, $(1 - p)^{n - X}=(0.37)^{2}=0.37\times0.37 = 0.1369$.
$P(3)={5\choose 3}(0.63)^{3}(0.37)^{2}=10\times0.250047\times0.1369\approx0.342$.

Step5: Solve 6a

For $n = 9$, $X = 0$, $p = 0.42$:
${9\choose 0}=\frac{9!}{0!(9 - 0)!}=1$.
$(1 - p)=1 - 0.42 = 0.58$, $p^{X}=(0.42)^{0}=1$, $(1 - p)^{n - X}=(0.58)^{9}=0.58\times0.58\times\cdots\times0.58\approx0.00647$.
$P(0)={9\choose 0}(0.42)^{0}(0.58)^{9}=1\times1\times0.00647 = 0.00647$.

Step6: Solve 6b

For $n = 10$, $X = 5$, $p = 0.37$:
${10\choose 5}=\frac{10!}{5!(10 - 5)!}=\frac{10\times9\times8\times7\times6}{5\times4\times3\times2\times1}=252$.
$(1 - p)=1 - 0.37 = 0.63$, $p^{X}=(0.37)^{5}=0.37\times0.37\times0.37\times0.37\times0.37\approx0.00754$, $(1 - p)^{n - X}=(0.63)^{5}=0.63\times0.63\times0.63\times0.63\times0.63\approx0.09925$.
$P(5)={10\choose 5}(0.37)^{5}(0.63)^{5}=252\times0.00754\times0.09925\approx0.187$.

Answer:

5a: $P(3)\approx0.000493$
5b: $P(2)\approx0.131$
5c: $P(3)\approx0.342$
6a: $P(0)=0.00647$
6b: $P(5)\approx0.187$