Binomial Probability Solutions

• A    binomial probability     is a type of independent probability with n trials in which the trials are either a    success   , p, or a    failure   , q.

• Because the outcome is either success or failure, the sum of p and q is one.
  $p+q=1$, or $p=1–q$

• The formula for binomial probability is: $\mathrm{P}\left(r\right)=nCr\left(n, r\right)·p^r·q^{n–r}$

  • n:    the number of trials    

  • r:    the number of successes   
  • 
p:    probability of success   
  • 
q:    probability of failure    

Example 4

Steve forgot to study for a ten-question, multiple-choice test. He needs a grade of 80%. What are the chances of correctly guessing exactly eight questions if each question has four choices and only one correct answer?

$$\begin{gathered} n=10, r=8, p=0.25, q=0.75 \\ \mathrm{P}\left(80\%\right)=\mathit{nCr}\left(10, 8\right)·{\left(0.25\right)}^8·{\left(0.75\right)}^{\left(10–8\right)} \\ \mathrm{P}\left(80\%\right)=3.86238098\times {10}^{–4} \end{gathered}$$

Example 5

A coin is tossed three times. What is the likelihood that the coin lands tails up at least twice?

   P(getting tails up two times     or      getting tails up three times)

$$\begin{gathered} n=3,\mathit{ }r=2,\mathit{ }p=0.5,\mathit{ }q=0.5\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }n=3,\mathit{ }r=3,\mathit{ }p=0.5,\mathit{ }q=0.5 \\ \mathit{nCr}\left(3, 2\right)·{\left(0.5\right)}^2·{\left(0.5\right)}^{\left(3–2\right)}+\mathit{nCr}\left(3, 3\right)·{\left(0.5\right)}^3·{\left(0.5\right)}^{\left(3–3\right)} \\ 0.375+0.125 \\ =0.5 \\ 50\% \end{gathered}$$  

There is a 50% chance the coin flip will land tails up at least twice.        

Example 6

At County High School, 3% of the students have earned a merit scholarship. If 40 students are selected at random, find the probability that four or five of the selected students have earned a merit scholarship.

$$\begin{gathered} \mathrm{P}(\mathrm{four} \mathrm{or} \mathrm{five} \mathrm{merit} \mathrm{scholarships}) \\ \mathrm{Four}: n=40, r=4, p=0.03, q=0.97 \\ \mathrm{Five}: n=40, r=5, p=0.03, q=0.97 \end{gathered}$$

$$\begin{gathered} \mathit{nCr}\left(40, 4\right)·{\left(0.03\right)}^4·{\left(0.97\right)}^{\left(40–4\right)}+\mathit{nCr}\left(40, 5\right)·{\left(0.03\right)}^5·{\left(0.97\right)}^{\left(40–5\right)} \\ 0.0247+0.0055 \\ 0.0302 \\ \approx 3\% \end{gathered}$$

There is a 3% chance that 4 or 5 students out of 40 will have earned a merit scholarship.