Question

tree diagram #1
b. compute the probability of one success, f(1) (to 2 decimals).
0.50
c. compute f(0) (to 2 decimals).
0.25
d. compute f(2) (to 2 decimals).
0.25
e. compute the probability of at least one success (to 2 decimals).
0.75
f. compute the following.
expected value 1.00 (to 2 decimals)
variance 0.50 (to 2 decimals)
standard deviation 0.7071 (to 4 decimals)
partially correct

Explanation:

Assume this is a binomial - distribution problem with \(n\) trials and probability of success \(p\). Let's assume \(n = 2\) (since we are computing \(f(0)\), \(f(1)\), \(f(2)\)). The binomial probability formula is \(f(x)=C(n,x)\times p^{x}\times(1 - p)^{n - x}\), where \(C(n,x)=\frac{n!}{x!(n - x)!}\).

Step1: Recall binomial - probability formula

\(f(x)=\binom{n}{x}p^{x}(1 - p)^{n - x}\), where \(\binom{n}{x}=\frac{n!}{x!(n - x)!}\)

Step2: Compute \(f(1)\) for \(n = 2\) and assume \(p = 0.5\)

\(\binom{2}{1}=\frac{2!}{1!(2 - 1)!}=\frac{2!}{1!1!}=2\)
\(f(1)=\binom{2}{1}\times(0.5)^{1}\times(1 - 0.5)^{2 - 1}=2\times0.5\times0.5 = 0.50\)

Step3: Compute \(f(0)\) for \(n = 2\) and \(p = 0.5\)

\(\binom{2}{0}=\frac{2!}{0!(2 - 0)!}=\frac{2!}{0!2!}=1\)
\(f(0)=\binom{2}{0}\times(0.5)^{0}\times(1 - 0.5)^{2 - 0}=1\times1\times0.25 = 0.25\)

Step4: Compute \(f(2)\) for \(n = 2\) and \(p = 0.5\)

\(\binom{2}{2}=\frac{2!}{2!(2 - 2)!}=\frac{2!}{2!0!}=1\)
\(f(2)=\binom{2}{2}\times(0.5)^{2}\times(1 - 0.5)^{2 - 2}=1\times0.25\times1 = 0.25\)

Step5: Compute the probability of at least one success

The probability of at least one success \(P(X\geq1)=1 - P(X = 0)\)
Since \(P(X = 0)=f(0)=0.25\), then \(P(X\geq1)=1 - 0.25=0.75\)

Step6: Compute the expected value

The expected value of a binomial distribution \(E(X)=np\). For \(n = 2\) and \(p = 0.5\), \(E(X)=2\times0.5 = 1.00\)

Step7: Compute the variance

The variance of a binomial distribution \(V(X)=np(1 - p)\). For \(n = 2\) and \(p = 0.5\), \(V(X)=2\times0.5\times(1 - 0.5)=0.50\)

Step8: Compute the standard deviation

The standard deviation \(\sigma=\sqrt{V(X)}\). Since \(V(X)=0.50\), \(\sigma=\sqrt{0.50}\approx0.7071\)

Answer:

b. \(0.50\)
c. \(0.25\)
d. \(0.25\)
e. \(0.75\)
f.
Expected value: \(1.00\)
Variance: \(0.50\)
Standard deviation: \(0.7071\)